UNIT 4
Quadratic Equations and Models CONTENTS COMMON CORE
A-SSE.B.3a A-REI.B.4b A-SSE.B.3a
COMMON CORE
A-REI.B.4b A-SSE.B.3b A-REI.B.4a A-REI.B.4b A-REI.C.7
COMMON CORE
S-ID.B.6b F-IF.C.7e F-IF.C.7e A-CED.A.2 F-LE.A.1b
303A
Unit 4
MODULE 8
Using Factors to Solve Quadratic Equations
Lesson 8.1 Lesson 8.2 Lesson 8.3
Solving Equations by Factoring x 2 + bx + c . . . . . . . . . . . . . . 307 Solving Equations by Factoring ax 2 + bx + c . . . . . . . . . . . . . . 319 Using Special Factors to Solve Equations . . . . . . . . . . . . . . . . 331
MODULE 9
Using Square Roots to Solve Quadratic Equations
Lesson 9.1 Lesson 9.2 Lesson 9.3 Lesson 9.4 Lesson 9.5
Solving Equations by Taking Square Roots . . . . . . . Solving Equations by Completing the Square . . . . . Using the Quadratic Formula to Solve Equations . . . Choosing a Method for Solving Quadratic Equations. Solving Nonlinear Systems . . . . . . . . . . . . . . . . .
MODULE 10 Lesson 10.1 Lesson 10.2 Lesson 10.3 Lesson 10.4 Lesson 10.5
Linear, Exponential, and Quadratic Models
. . . . .
Fitting a Linear Model to Data . . . . . . . . . . . . . . . . Graphing Exponential Functions. . . . . . . . . . . . . . . Modeling Exponential Growth and Decay . . . . . . . . . Modeling with Quadratic Functions. . . . . . . . . . . . . Comparing Linear, Exponential, and Quadratic Models
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355 367 381 395 411
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429 445 459 473 489
UNIT 4
Unit Pacing Guide 45-Minute Classes Module 8 DAY 1
DAY 2
DAY 3
DAY 4
DAY 5
Lesson 8.1
Lesson 8.1
Lesson 8.2
Lesson 8.2
Lesson 8.3
DAY 1
DAY 2
DAY 3
DAY 4
DAY 5
Lesson 9.1
Lesson 9.1
Lesson 9.2
Lesson 9.2
Lesson 9.3
DAY 6
DAY 7
DAY 8
DAY 9
DAY 10
Lesson 9.3
Lesson 9.4
Lesson 9.5
Lesson 9.5
Module Review and Assessment Readiness
DAY 1
DAY 2
DAY 3
DAY 4
DAY 5
Lesson 10.1
Lesson 10.2
Lesson 10.2
Lesson 10.3
Lesson 10.3
DAY 6
DAY 7
DAY 8
DAY 9
DAY 10
Lesson 10.4
Lesson 10.4
Lesson 10.5
Module Review and Assessment Readiness
Unit Review and Assessment Readiness
DAY 1
DAY 2
DAY 3
Lesson 8.1
Lesson 8.2
Lesson 8.3 Module Review and Assessment Readiness
DAY 1
DAY 2
DAY 3
DAY 4
DAY 4
Lesson 9.1
Lesson 9.2
Lesson 9.3
Lesson 9.4 Lesson 9.5
Lesson 9.5 Module Review and Assessment Readiness
DAY 1
DAY 2
DAY 3
DAY 4
DAY 5
Lesson 10.1
Lesson 10.2
Lesson 10.3
Lesson 10.4
Module Review and Assessment Readiness
Lesson 10.2
Lesson 10.3
Lesson 10.4
Lesson 10.5
Unit Review and Assessment Readiness
DAY 6
Module Review and Assessment Readiness Module 9
Module 10
90-Minute Classes Module 8
Module 9
Module 10
Unit 4
303B
Program Resources PLAN
ENGAGE AND EXPLORE
HMH Teacher App Access a full suite of teacher resources online and offline on a variety of devices. Plan present, and manage classes, assignments, and activities.
Real-World Videos Engage students with interesting and relevant applications of the mathematical content of each module.
Explore Activities Students interactively explore new concepts using a variety of tools and approaches.
ePlanner Easily plan your classes, create and view assignments, and access all program resources with your online, customizable planning tool.
Professional Development Videos Authors Juli Dixon and Matt Larson model successful teaching practices and strategies in actual classroom settings. QR Codes Scan with your smart phone to jump directly from your print book to online videos and other resources. DO NOT EDIT--Changes must be made through "File info" CorrectionKey=NL-A;CA-A
Teacher’s Edition Support students with point-of-use Questioning Strategies, teaching tips, resources for differentiated instruction, additional activities, and more. NOT EDIT--Changes must made through "File info" DODO NOT EDIT--Changes must bebe made through "File info" CorrectionKey=NL-A;CA-A CorrectionKey=NL-A;CA-A
Name Name
Isosceles and Equilateral Triangles
NOT EDIT--Changes must made through "File info" DODO NOT EDIT--Changes must bebe made through "File info" CorrectionKey=NL-A;CA-A CorrectionKey=NL-A;CA-A
Class Class
straightedge to draw segment to draw lineline segment BCBC . . CCUseUsethethestraightedge
Investigating Isosceles Triangles
G-CO.C.10
Explore Explore
Prove theorems about triangles.
INTEGRATE TECHNOLOGY
Mathematical Practices COMMON CORE
angles have base a side base angles. TheThe angles thatthat have thethe base as aasside areare thethe base angles.
ENGAGE
work in the space provided. a straightedge to draw angle. in the space provided. UseUse a straightedge to draw an an angle. AADoDoyouryourwork
QUESTIONING STRATEGIES
Possible answer Triangle m∠A 70°; m∠B ∠55°; m∠C 55°. Possible answer forfor Triangle 1: 1: m∠A == 70°; m∠B == ∠55°; m∠C == 55°.
a different each time. is aisdifferent sizesize each time.
A A
Reflect Reflect
Explain Explain 1 1 Proving Provingthe theIsosceles Isosceles Triangle Theorem Triangle Theorem
sides of the angle. Label points B and sides of the angle. Label thethe points B and C. C.
andItsItsConverse Converse and
A A
In the Explore, made a conjecture base angles of an isosceles triangle congruent. In the Explore, youyou made a conjecture thatthat thethe base angles of an isosceles triangle areare congruent. This conjecture proven it can stated a theorem. This conjecture cancan be be proven so so it can be be stated as aastheorem. C C
Isosceles Triangle Theorem Isosceles Triangle Theorem
© Houghton Mifflin Harcourt Publishing Company
Make a Conjecture Looking at your results, what conjecture made about base angles, 2. 2. Make a Conjecture Looking at your results, what conjecture cancan be be made about thethe base angles, ∠C? ∠B∠B andand ∠C? The base angles congruent. The base angles areare congruent. Using a compass, place point vertex draw intersects a compass, place thethe point onon thethe vertex andand draw an an arcarc thatthat intersects thethe BBUsing
B B
If two sides a triangle congruent, then angles opposite sides If two sides of aoftriangle areare congruent, then thethe twotwo angles opposite thethe sides areare congruent. congruent. This theorem is sometimes called Base Angles Theorem stated as “Base angles This theorem is sometimes called thethe Base Angles Theorem andand cancan alsoalso be be stated as “Base angles of an isosceles triangle congruent. of an isosceles triangle areare congruent. ” ” Module Module 22 22
must be EDIT--Changes A;CA-A DO NOT CorrectionKey=NL-
made through
Lesson Lesson 2 2
1097 1097
"File info"
Module Module 22 22
1098 1098
Lesson Lesson 2 2
Date Class
al and Equilater 22.2 Isosceles Triangles
Name
Essential
COMMON CORE
IN1_MNLESE389762_U8M22L2 IN1_MNLESE389762_U8M22L2 10971097
Question:
G-CO.C.10
relationships the special What are triangles? and equilateral
Prove theorems
triangle is
The congruent The angle The side
a triangle
sides are
formed by
opposite
with at least
called the
the legs is
the vertex
among
in isosceles
Resource Locker
HARDCOVER PAGES 10971110 HARDCOVER PAGES 10971110
PROFESSIONALDEVELOPMENT DEVELOPMENT PROFESSIONAL
about triangles.
Investigating
Explore An isosceles
sides angles and
legs of the
the vertex
angle is the
Isosceles
two congruent
Triangles
sides.
Legs
Vertex angle
triangle.
Base
angle.
Base angles
base.
the as a side are
base angles.
other potential
the base and investigate that have triangles isosceles you will construct special triangles. angle. es of these In this activity, to draw an characteristics/properti Use a straightedge space provided. figure. work in the in the Do your as shown angle ∠A, A Label your
The angles
Check students’
construtions.
The side opposite the vertex angle is the base. The angles that have the base as a side are the base angles. In this activity, you will construct isosceles triangles and investigate other potential characteristics/properties of these special triangles.
How know triangles constructed isosceles triangles? 1. 1. How do do youyou know thethe triangles youyou constructed areare isosceles triangles? ――. ―― The compass marks equal lengths both sides ∠A; therefore, ABAB ≅≅ ACAC . The compass marks equal lengths onon both sides of of ∠A; therefore,
Check students’ construtions. Check students’ construtions.
4/19/14 12:10 4/19/14 12:10 PM PM
Watch the hardcover Watch forfor the hardcover student edition page student edition page numbers this lesson. numbers forfor this lesson.
IN1_MNLESE389762_U8M22L2 IN1_MNLESE389762_U8M22L2 10981098
LearningProgressions Progressions Learning
this lesson, students add their prior knowledge isosceles and equilateral InIn this lesson, students add toto their prior knowledge ofof isosceles and equilateral
4/19/14 12:10 4/19/14 12:10 PM PM
Legs
The angle formed by the legs is the vertex angle.
How could you draw isosceles triangles without using a compass? Possible answer: Draw ∠A and plot point B on one side of ∠A. Then _ use a ruler to measure AB and plot point C on the other side of ∠A so that AC = AB.
Repeat steps A–D at least more times record results in the table. Make sure steps A–D at least twotwo more times andand record thethe results in the table. Make sure ∠A∠A EERepeat
Vertex angle
The congruent sides are called the legs of the triangle.
What must be true about the triangles you construct in order for them to be isosceles triangles? They must have two congruent sides.
m∠B m∠B
Label your angle ∠A, as shown in the figure. Label your angle ∠A, as shown in the figure.
© Houghton Mifflin Harcourt Publishing Company
Triangle Triangle 4 4
m∠C m∠C
In this activity, construct isosceles triangles investigate other potential In this activity, youyou willwill construct isosceles triangles andand investigate other potential characteristics/properties of these special triangles. characteristics/properties of these special triangles.
© Houghton Mifflin Harcourt Publishing Company
Triangle Triangle 3 3
© Houghton Mifflin Harcourt Publishing Company
View the Engage section online. Discuss the photo, explaining that the instrument is a sextant and that long ago it was used to measure the elevation of the sun and stars, allowing one’s position on Earth’s surface to be calculated. Then preview the Lesson Performance Task.
Triangle Triangle 2 2
m∠mA∠A
Base Base Base angles Base angles
opposite vertex angle is the base. TheThe sideside opposite thethe vertex angle is the base.
PREVIEW: LESSON PERFORMANCE TASK
Triangle Triangle 1 1
angle formed is the vertex angle. TheThe angle formed by by thethe legslegs is the vertex angle.
Explain to a partner what you can deduce about a triangle if it has two sides with the same length.
In an isosceles triangle, the angles opposite the congruent sides are congruent. In an equilateral triangle, all the sides and angles are congruent, and the measure of each angle is 60°.
Legs Legs
congruent sides called of the triangle. TheThe congruent sides areare called thethe legslegs of the triangle.
MP.3 Logic
Language Objective
Essential Question: What are the special relationships among angles and sides in isosceles and equilateral triangles?
DD
Vertex angle Vertex angle
Investigating Isosceles Triangles
An isosceles triangle is a triangle with at least two congruent sides.
Students have the option of completing the isosceles triangle activity either in the book or online.
a protractor to measure each angle. Record measures in the table under column UseUse a protractor to measure each angle. Record thethe measures in the table under thethe column Triangle forfor Triangle 1. 1.
InvestigatingIsosceles Isosceles Triangles Investigating Triangles
isosceles triangle a triangle with at least congruent sides. AnAn isosceles triangle is aistriangle with at least twotwo congruent sides.
G-CO.C.10 Prove theorems about triangles.
Explore
C C
B B Resource Resource Locker Locker
The student is expected to: COMMON CORE
Resource Locker
EXPLORE
A A
Essential Question: What special relationships among angles and sides in isosceles Essential Question: What areare thethe special relationships among angles and sides in isosceles and equilateral triangles? and equilateral triangles?
Common Core Math Standards
COMMON CORE
__
Date Date
22.2 Isosceles Isoscelesand andEquilateral Equilateral 22.2 Triangles Triangles
Date
Essential Question: What are the special relationships among angles and sides in isosceles and equilateral triangles?
NOT EDIT--Changes must made through "File info" DODO NOT EDIT--Changes must bebe made through "File info" CorrectionKey=NL-A;CA-A CorrectionKey=NL-A;CA-A
EXPLAIN 1
Proving the Isosceles Triangle Theorem and Its Converse
Do your work in the space provided. Use a straightedge to draw an angle. Label your angle ∠A, as shown in the figure. A
CONNECT VOCABULARY Ask a volunteer to define isosceles triangle and have students give real-world examples of them. If possible, show the class a baseball pennant or other flag in the shape of an isosceles triangle. Tell students they will be proving theorems about isosceles triangles and investigating their properties in this lesson.
hing Company
22.2
Class
22.2 Isosceles and Equilateral Triangles
DONOT NOTEDIT--Changes EDIT--Changesmust mustbebemade madethrough through"File "Fileinfo" info" DO CorrectionKey=NL-A;CA-A CorrectionKey=NL-A;CA-A
DO NOT EDIT--Changes must be made through "File info" CorrectionKey=NL-A;CA-A
LESSON
Name
Base Base angles
PROFESSIONAL DEVELOPMENT
TEACH
ASSESSMENT AND INTERVENTION
Math On the Spot video tutorials, featuring program authors Dr. Edward Burger and Martha Sandoval-Martinez, accompany every example in the textbook and give students step-by-step instructions and explanations of key math concepts.
Interactive Teacher Edition Customize and present course materials with collaborative activities and integrated formative assessment.
C1
Lesson 19.2 Precision and Accuracy
Evaluate
1
Lesson XX.X ComparingLesson Linear, Exponential, and Quadratic Models 19.2 Precision and Accuracy
teacher Support
1
EXPLAIN Concept 1
Explain
The Personal Math Trainer provides online practice, homework, assessments, and intervention. Monitor student progress through reports and alerts. Create and customize assignments aligned to specific lessons or Common Core standards. • Practice – With dynamic items and assignments, students get unlimited practice on key concepts supported by guided examples, step-by-step solutions, and video tutorials. • Assessments – Choose from course assignments or customize your own based on course content, Common Core standards, difficulty levels, and more. • Homework – Students can complete online homework with a wide variety of problem types, including the ability to enter expressions, equations, and graphs. Let the system automatically grade homework, so you can focus where your students need help the most! • Intervention – Let the Personal Math Trainer automatically prescribe a targeted, personalized intervention path for your students. 2
3
4
Question 3 of 17
Concept 2
Determining Precision
ComPLEtINg thE SquArE wIth EXPrESSIoNS Avoid Common Errors Some students may not pay attention to whether b is positive or negative, since c is positive regardless of the sign of b. Have student change the sign of b in some problems and compare the factored forms of both expressions. questioning Strategies In a perfect square trinomial, is the last term always positive? Explain. es, a perfect square trinomial can be either (a + b)2 or (a – b)2 which can be factored as (a + b)2 = a 2 + 2ab = b 2 and (a – b)2 = a 2 + 2ab = b 2. In both cases the last term is positive. reflect 3. The sign of b has no effect on the sign of c because c = ( b __ 2 ) 2 and a nonzero number squared is always positive. Thus, c is always positive. c = ( b __ 2 ) 2 and a nonzero number c = ( b __ 2 ) 2 and a nonzero number
5
6
7
View Step by Step
8
9
10
11 - 17
Video Tutor
Personal Math Trainer
Textbook
X2 Animated Math
Solve the quadratic equation by factoring. 7x + 44x = 7x − 10
As you have seen, measurements are given to a certain precision. Therefore,
x=
the value reported does not necessarily represent the actual value of the measurement. For example, a measurement of 5 centimeters, which is
,
Check
given to the nearest whole unit, can actually range from 0.5 units below the reported value, 4.5 centimeters, up to, but not including, 0.5 units above it, 5.5 centimeters. The actual length, l, is within a range of possible values:
Save & Close
centimeters. Similarly, a length given to the nearest tenth can actually range from 0.05 units below the reported value up to, but not including, 0.05 units above it. So a length reported as 4.5 cm could actually be as low as 4.45 cm or as high as nearly 4.55 cm.
?
!
Turn It In
Elaborate
Look Back
Focus on Higher Order Thinking Raise the bar with homework and practice that incorporates higher-order thinking and mathematical practices in every lesson.
Differentiated Instruction Resources Support all learners with Differentiated Instruction Resources, including • Leveled Practice and Problem Solving • Reading Strategies • Success for English Learners • Challenge Calculate the minimum and maximum possible areas. Round your answer to
Assessment Readiness
the nearest square centimeters.
The width and length of a rectangle are 8 cm and 19.5 cm, respectively.
Prepare students for success on high stakes tests for Integrated Mathematics 2 with practice at every module and unit
Find the range of values for the actual length and width of the rectangle.
Minimum width =
7.5
cm and maximum width
0, b>0
Sign of Constant Terms in Binomial Factors of x 2 + bx + c both constants positive
-1 and 30
-1 + 30 = 29
-2 and 15
-2 + 15 = 13
c > 0, b 0, the factors will have the same sign. Factors of 64
Sum of Factors 1 + 64 = 65
© Houghton Mifflin Harcourt Publishing Company
1 and 64
2 + 32 = 34
2 and 32
4 + 16 = 20
4 and 16
8 + 8 = 16
8 and 8 -1 and -64
-1 - 64 = -64
-4 and -16
-4 - 16 = -20
-2 and -32 -8 and -8
-2 - 32 = -34 -8 - 8 = -16
The possible values of b are the sums of the factors of 64: ±65, ±34, ±20, and ±16.
Module 8
IN2_MNLESE389830_U4M08L1.indd 317
317
Lesson 8.1
317
Lesson 1
4/2/14 12:57 AM
Lesson Performance Task
QUESTIONING STRATEGIES
The area of the roof of a large building is modeled by the function f R(x) = 2x 2 - 251x + 80,000 and the area of the roof that is devoted to mechanical support is given by the function f M(x) = x 2 + 224x + 31,250. Given that the area of the green space is 123,750 square feet, find its dimensions. The area of the green space is f G(x).
How can you find a polynomial that represents the area of the green space? Subtract the polynomial representing the area devoted to mechanical support from the polynomial representing the total area of the roof.
f R(x) = f M(x) + f G(x) f G(x) = f R(x) - f M(x)
Using that polynomial, how can you find the values of x that give the correct area for the green space? Set the polynomial equal to 123,750, then subtract 123,750 from both sides of the equation so that one side is 0. Next, factor the other side, set each factor equal to 0, and solve.
f G(x) = 2x 2 - 251x + 80,000 - (x 2 + 224x + 31,250) = 2x - 251x + 80,000 - x + 224x + 31,250 2
2
= 2x 2 - x - 251x - 224x + 80,000 - 31,250 2
= x 2 - 475x + 48,750 Solve f G(x) = 123,750 for x.
123,750 = x 2 - 475x + 48,750 0 = x 2 - 475x - 75,000
After finding the values of x, how can you find the dimensions of the green space? Since the area of a rectangle is equal to the product of the length and the width, factor the polynomial representing the area of the green space, then substitute the values for x in each factor and evaluate. Discard any negative solutions.
0 = (x - 600)(x + 125) x = 600 and x = -125
Since negative numbers do not apply in the context of this problem, use x = 600. Now, factor f G(x) = x 2 - 475x + 48,750.
f G(x) = x 2 - 475x + 48,750 = (x - 150)(x - 325)
Substitute x = 600 from the previous step. = (600 - 150)(600 - 325)
© Houghton Mifflin Harcourt Publishing Company
= (450)(275)
Therefore, the green section of the roof is 450 feet by 275 feet.
Module 8
318
Lesson 1
EXTENSION ACTIVITY IN2_MNLESE389830_U4M08L1.indd 318
Have students research historical examples of green roofs, such as the Hanging Gardens of Babylon or the sod roofs used by Native Americans and American pioneers. Have them describe how the construction of modern green roofs is different from the construction of their ancient counterparts.
4/2/14 12:57 AM
Scoring Rubric 2 points: Student correctly solves the problem and explains his/her reasoning. 1 point: Student shows good understanding of the problem but does not fully solve or explain his/her reasoning. 0 points: Student does not demonstrate understanding of the problem.
Solving Equations by Factoring x 2 + bx + c
318
LESSON
8.2
Name
Solving Equations by Factoring ax 2 + bx + c
8.2
Resource Locker
Factoring ax 2 + bx + c When c > 0
When you factor a quadratic expression in standard form (ax2 + bx + c), you are looking for two binomials, and possibly a constant numerical factor whose product is the original quadratic expression.
A-REI.B.4b
Recall that the product of two binomials is found by applying the Distributive Property, abbreviated sometimes as FOIL:
Solve quadratic equations by ... factoring, as appropriate to the initial form of the equation … Also A-SSE.A.2, A-SSE.B.3a
(2x + 5)(3x + 2) = 6x 2 + 4x + 15x + 10 = 6x 2 + 19x + 10
Mathematical Practices
F
F O⎫ ⎬ I⎭
MP.5 Using Tools
Language Objective
L
Explain to a partner how to factor a trinomial in the form ax 2 + bx + c.
O
I
L
The sum of the coefficients of the first terms is a. The sum of the coefficients of the outer and inner products is b. The product of the last terms is c.
Because the a and c coefficients result from a single product of terms from the binomials, the coefficients in the binomial factors will be a combination of the factors of a and c. The trick is to find the combination of factors that results in the correct value of b. Follow the steps to factor the quadratic 4x 2 - 26x + 42.
ENGAGE
A
Essential Question: How can you use factoring to solve quadratic equations in standard form for which a ≠ 1?
B
First, factor out the largest common factor of 4, 26, and 42 if it is anything other than 1. 4x 2 + 26x + 42 = 2
© Houghton Mifflin Harcourt Publishing Company
Factor the quadratic equation. Set each linear factor equal to 0. Solve each linear equation. The solutions are the solutions of the original quadratic equation.
Solving Equations by Factoring ax 2 + bx + c
Explore
The student is expected to:
COMMON CORE
Date
Essential Question: How can you use factoring to solve quadratic equations in standard form for which a ≠ 1?
Common Core Math Standards COMMON CORE
Class
(2x 2 + 13x + 21)
Next, list the factor pairs of 2:
1 and 2
C
List the factor pairs of 21:
1 and 21, 3 and 7
D
Make a table listing the combinations of the factors of a and c, and find the value of b that results from summing the outer and inner products of the factors.
Outer + inner
Factors of 2
Factors of 21
PREVIEW: LESSON PERFORMANCE TASK
1 and 2
1 and 21
1 and 2
3 and 7
(1)(7) + (2)(3) = 13
View the Engage section online. Have students demonstrate how a car accelerating from a stopped position could first be passed by a bus traveling at a constant speed, then overtake and pass the bus. Then preview the Lesson Performance Task.
1 and 2
7 and 3
(1)(3) + (2)(7) = 17
1 and 2
21 and 1
(1)(1) + (2)(21) = 43
Module 8
(1)(21) + (2)(1) = 23
Lesson 2
319
gh “File info”
made throu
be ges must EDIT--Chan DO NOT Key=NL-A;CA-A Correction
Date Class
ns by ng Equatio Solvi c 2 ax + bx +
Name
8.2
Factoring
form Resource in standard Locker equations quadratic of the ing to solve initial form use factor can you riate to the ion: How ng, as approp a ≠ 1? by ... factori for which tic equations … Solve quadra A-SSE.B.3a A-REI.B.4b 0 > A-SSE.A.2, c n Also and … ials, Whe binom equation 2 g for two + bx + c ), you are lookin oring ax 2 + bx + c Fact sion. x a ( tic expres rd form Explore sion in standa ct is the original quadra iated tic expres ty, abbrev produ utive Proper factor a quadra ical factor whose the Distrib When you numer applying constant is found by possibly a binomials two of ct the produ + 10 Recall that x2 + 19x + 10 = 6 as FOIL: 2 sometimes + 4x + 15x L + 2) = 6x I O (2x + 5)(3x F is a. first terms ients of the is b. products of the coeffic and inner The sum F of the outer coefficients sum of the O⎫ ients in the ⎬ The is c. ials, the coeffic of factors that terms I⎭ last the binom nation ct of the terms from the combi The produ product of is to find L from a single of a and c. The trick ients result factors a and c coefficcombination of the Because the be a 1. factors will b. ial of 42. binom other than 2 t value 26x + anything the correc tic 4x 42 if it is results in the quadra of 4, 26, and steps to factor factor the on Follow largest comm
Quest Essential
HARDCOVER PAGES 319 319330
COMMON CORE
IN2_MNLESE389830_U4M08L2 319
First, factor
y g Compan Publishin
out the
+ 2 4x + 26x
Next, list
Harcour t
n Mifflin
(2x2 + 13x
+ 21)
pairs of 2:
1 and 2
© Houghto
42 = 2
the factor
Watch for the hardcover student edition page numbers for this lesson.
pairs of 21: List the factor 3 and 7 1 and 21,
of b that the value c, and find s of a and s of the factor of the factors. combination products listing the and inner inner Make a table the outer Outer + summing 21 results from (2)(1) = 23 Factors of (1)(21) + 2 Factors of (2)(3) = 13 1 and 21 (1)(7) + 1 and 2 3 and 7 ) = 17 7 ( ) 2 ( (1)(3) + 1 and 2 43 7 and 3 (2)(21) = (1)(1) + 1 and 2 21 and 1 1 and 2
Lesson 2
319
Module 8
8L2 319 30_U4M0
ESE3898
IN2_MNL
319
Lesson 8.2
4/9/14
6:46 PM
07/04/14 6:15 PM
E
F
Copy the pair of factors that resulted in an outer + inner sum of 13 into the binomial factors. Be careful to keep the inner and outer factors from the table as inner and outer coefficients in the binomials.
EXPLORE
2x 2 + 13x + 21 =
Factoring ax 2 + bx + c When c > 0
(1
x+ 3
)( 2 x + 7 )
Replace the common factor of the original coefficients to complete factorization of the original quadratic.
INTEGRATE TECHNOLOGY
4x 2 + 26x + 42 = 2 (x + 3)(2x + 7)
Students have the option of completing the factoring activity either in the book or online.
Reflect
1.
Critical Thinking Explain why you should use negative factors of c when factoring a quadratic with c > 0 and b < 0. A negative value of b tells you at least one of the factors of c must be negative, but since
QUESTIONING STRATEGIES
c > 0, in fact both must be negative. 2.
What If? If none of the factor pairs for a and c result in the correct value for b, what do you know about the quadratic? The quadratic cannot be factored into the product of two binomials.
3.
Discussion Why did you have to check each factor pair twice for the factors of c (3 and 7 versus 7 and 3) but only once for the factors of a (1 and 2, but not 2 and 1)? Hint: Compare the outer and inner sums of rows two and three in the table, and also check the outer and inner sums by switching the order of both pairs from row 2 (check 2 and 1 for a with 7 and 3 for c). It would be redundant to switch the order of both sets of factor pairs. Switching one
If there are three factor pairs for a and one factor pair for c, how many possible arrangements of factor pairs are there? Explain. 6; the factor pair for c can be paired with each factor pair for a in either order. After finding the combination of factors of a and c that give the correct value of b, how do you know which factor to put in each position in the binomial factors? The factors of a are the coefficients of x, and the factors of c are the constant terms. The factors you used to find the inner product should be the two inner values, and the factors you used to find the outer product should be the two outer values.
set represents a different set of possible binomial factors, while switching both pairs corresponds to exchanging the order of the two binomial factors, which is not a different answer. The second and third rows of the table (switching the factor order of c only)
of (2)(3) + (1)(7) = 13, which is the correct value. The corresponding factored equation
would be written as 4x 2 + 26x + 42 = 2(2x + 7)(x + 3), which is the same set of binomial factors as the original answer, just written in a different order.
Explain 1
Factoring ax 2 + bx + c When c < 0
Factoring x 2 + bx + c when c < 0 requires one negative and one positive factor of c. The same applies for equations of the form ax 2 + bx + c as long as a > 0. When checking factor pairs, remember to consider factors of c in both orders, and consider factor pairs with the negative sign on either member of the pair of c factors.
© Houghton Mifflin Harcourt Publishing Company
have different outer plus inner sums (17 versus 13). Switching both pairs results in a sum
EXPLAIN 1 Factoring ax 2 + bx + c When c < 0
When you find a combination of factors whose outer and inner product sum is equal to b, you have found the solution. Make sure you fill in the factor table systematically so that you do not skip any combinations. If a < 0, factor out -1 from all three coefficients, or use a negative common factor, so that the factors of a can be left as positive numbers.
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AVOID COMMON ERRORS
Lesson 2
PROFESSIONAL DEVELOPMENT Math Background
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Quadratic equations were probably first examined in Babylonia about 4000 years ago. Around 800 BCE in India, geometric methods were used to solve quadratic equations. These methods were similar to the method of completing the square, which was developed between 300 and 200 BCE. Both Euclid and Pythagoras grappled with finding a formula or procedure for solving quadratic equations. It was not until around 600 CE that the first explicit solution to the standard form of the equation was given, although it was not completely general. The practical notation using the symbols and methods currently in use was developed during the 15th century.
4/2/14 1:00 AM
Make sure students understand that the order of the coefficients in the binomials matters. Encourage students to check for errors by making sure that the product of the x-coefficients is a, the product of the constant terms is c, and the sum of the products of the inner and outer terms is b.
Solving Equations by Factoring ax 2 + bx + c
320
Example 1
QUESTIONING STRATEGIES
How many possible arrangements of factor pairs are there when a = 3 and c = -2? What are they? 4; Factors of a 1 and 3 1 and 3 1 and 3 1 and 3
Factor the quadratic by checking factor pairs.
6x 2 - 21x - 45 Find the largest common factor of 6, 21, and 45, and factor it out, keeping the coefficient of x 2 positive. 6x 2 - 21x - 45 = 3(2x 2 - 7x - 15)
Factors of c
Factors of c
1 and 2
1 and -15
(1)(-15) + (2)(1) = -13
1 and 2
3 and -5
(1)(-5) + (2)(3) = 1
1 and 2
5 and -3
(1)(-3) + (2)(5) = 7
1 and 2
15 and -1
(1)(-1) + (2)(15) = 29
1 and 2
-1 and 15
(1)(15) + (2)(-1) = 13
1 and 2
-3 and 5
(1)(5) + (2)(-3) = -1
1 and 2
-5 and 3
(1)(3) + (2)(-5) = -7
1 and 2
-15 and 1
(1)(1) + (2)(-15) = -29
1 and −2 2 and −1
−1 and 2 −2 and 1
What pairs of binomial factors result from these arrangements of factor pairs? (x + 1)(3x - 2), (x + 2)(3x - 1), (x - 1)(3x + 2), and (x - 2)(3x + 1)
Outer Product + Inner Product
Factors of a
Use the combination of factor pairs that results in a value of -7 for the b coefficient. 2x 2 - 7x - 15 = (x - 5)(2x + 3) Replace the common factor of the original coefficients to factor the original quadratic. 6x 2 - 21x - 45 = 3(x - 5)(2x + 3)
20x 2 - 40x - 25 Factor out common factors of the coefficients. 20x 2 - 40x - 25 = 5 (4x 2 - 8x - 5)
Factors of c
Outer Product + Inner Product
1 and 4
1 and - 5
(1)(-5) + (4)(1) = -1
1 and 4
5 and -1
(1)(-1) + (4)(5) = 19
1 and 4
-1 and 5
(1)(5) + (4)(-1) = 1
1 and 4
-5 and 1
(1)(1) + (4)(-5) = -19
2 and 2
1 and -5
(2)(-5) + (2)(1) = -8
2 and 2
-1 and 5
(2)(-1) + (2)(5) = 8
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Factors of a
Use the combination of factor pairs that results in a value of -8 for the b coefficient. 4x 2 - 8x - 5 =
( 2 x + 1 )(
2 x + -5
)
Replace the common factor of the original coefficients to factor the original quadratic. 20x 2 - 40x - 25 = 5 (2x + 1)(2x - 5) Module 8
Lesson 2
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Peer-to-Peer Activity Have student pairs discuss how to use the signs of b and c to decide whether binomial factors should contain + or - signs and then complete this chart:
b
c
+
+
-
-
+
321
Lesson 8.2
+ -
Example 2x 2 + 9x + 4 3n 2 - 11n + 6 4t 2 - t -10 6q 2 + q - 7
Sign in Factors
Factors
both +
(2x + 1)(x + 4)
product with > | | has +
(6q + 7)(q - 1)
(3n - 2)(n - 3) product with > | | has - (t - 2)(4t + 5) both −
Reflect
4.
EXPLAIN 2
What If? Suppose a is a negative number. What would be the first step in factoring ax 2 + bx + c? Factor out a negative common factor from all of the coefficients, even if the common
Solving Equations of the Form ax 2 + bx + c = 0 by Factoring
factor is -1. This will result in a new quadratic to factor with a positive value of a to which the previous methods can be applied. Your Turn
5.
INTEGRATE MATHEMATICAL PRACTICES Focus on Communication MP.3 Ask students to describe the general form of
Factor. -5x 2 + 8x + 4
-1(5x 2 - 8x - 4) = -1(x - 2)(5x + 2)
[# Explain 2
solutions to a factored quadratic equation of the form (ax + b)(cx + d) = 0. Students should find that the b d __ solutions are x = - __ a and x = - c . Therefore, when the coefficients of x in the factored form are not 1, the solutions are likely to be proper or improper fractions.
Solving Equations of the Form ax + bx + c = 0 by Factoring 2
For a quadratic equation in standard form, ax 2 + bx + c = 0, factoring the quadratic expression into binomials lets you use the Zero Product Property to solve the equation, as you have done previously. If the equation is not in standard form, convert it to standard form by moving all terms to one side of the equation and combining like terms. Example 2
Change the quadratic equation to standard form if necessary and then solve by factoring.
-2x 2 + 7x - 2 = 4x 2 + 4 Convert the equation to standard form: -2x 2 + 7x - 6 = 0
Subtract 4x 2 and 4 from both sides. Multiply both sides by -1.
2x 2 - 7x + 6 = 0
Use the combination pair that results in a sum of -7 and write the equation in factored form. Then solve it using the Zero Product Property.
(x - 2)(2x - 3) = 0 x-2=0
or
2x - 3 = 0
x=2
2x = 3
y
3 x=_ 2
3
3. The solution is x = 2 or x = _ 2 The solution can be checked by graphing the related function, ƒ (x) = 2x 2 - 7x + 6, and finding the x-intercepts.
2 1 x 0
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Consider factor pairs for 2 and 6. Use negative factors of 6 to get a negative value for b.
1 2 3 (2, 0) (1.5, 0) Lesson 2
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Multiple Representations
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Show students this alternate factoring method. Any trinomial can be written as a polynomial with four terms and then factored by grouping. To factor 6x 2 + 19x + 15, first find ac: 6 · 15 = 90. Then find factors of ac that sum to b: 9 and 10 are factors of 90 whose sum is 19. Rewrite the trinomial using those factors: 6x 2 + 10x + 9x + 15. Finally, factor by grouping: (6x 2 + 10x) + (9x + 15) = 2x(3x + 5) + 3(3x + 5) = (3x + 5)(2x + 3)
Solving Equations by Factoring ax 2 + bx + c
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QUESTIONING STRATEGIES
3(x 2 - 1) = -3x 2 + 2x + 5 Write the equation in standard form and factor so you can apply the Zero Product Property.
What is the first step for solving 7x 2 + 20x = x + 6? Convert the equation to standard form by subtracting x + 6 from both sides of the equation, which results in 7x 2 + 19x - 6 = 0.
3 x 2 + 3 = -3x 2 + 2x + 5 2 6 x - 2x - 8 = 0
3 x2 - x - 4 = 0
Use the combination pair that results in a sum of -1 .
Why is this step necessary for solving the equation by factoring? In order to use the Zero Product Property to solve the equation, the expression that you factor must be equal to zero.
(x + 1)
( 3 x - -4 ) = 0
x+1= 0
or
x = -1
3x - 4 = 0
3 x=4
_4
x= 3 4. The solutions are -1 and _ 3 Use a graphing calculator to check the solutions.
Reflect
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6.
In the two examples, a common factor was divided out at the beginning of the solution, and it was not used again. Why didn’t you include the common term again when solving x for the original quadratic equation? Solutions are arrived at by applying the Zero Product Property to the factors of the
quadratic expression. Although the common factor is a factor of the quadratic expression, it cannot equal 0 and so the 0-valued multiple predicted by the Zero Product Property must still be one of the binomial terms. Your Turn
7.
12x 2 + 48x + 45 = 0
3(4x 2 - 16x + 15) = 0
(2x + 3)(2x + 5) = 0
2x + 3 = 0
3 x = - _ 2
or
2x + 5 = 0
or
Module 8
5 x = - _ 2
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Lesson 2
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Connect Vocabulary English learners may need numerous exposures and structured practice with new vocabulary to truly acquire it for their own use. Before describing the new terminology for this lesson, review more basic terms such as sum, product, and factor. Students will need to rely on their understanding of these words to grasp phrases such as outer products, inner products, sum of inner and outer products, and common factors. It may be helpful for students to use diagrams and examples to record the meaning of new terminology.
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Explain 3
Solving Equation Models of the Form ax 2 + bx + c = 0 by Factoring
EXPLAIN 3
A projectile is an object moving through the air without any forces other than gravity acting on it. The height of a projectile at a time in seconds can be found by using the formula h = −16t 2 + vt + s, where v is in the initial upwards velocity in feet per second (and can be a negative number if the projectile is launched downwards) and s is starting height in feet. The a term of −16 accounts for the effect of gravity accelerating the projectile downwards and is the only appropriate value when measuring distance with feet and time in seconds.
Solving Equation Models of the Form ax 2 + bx + c = 0 by Factoring
To use the model to make predictions about the behavior of a projectile, you need to read the description of the situation carefully and identify the initial velocity, the initial height, and the height at time t. Example 3
QUESTIONING STRATEGIES
Read the real-world situation and substitute in values for the projectile motion formula. Then solve the resulting quadratic equation by factoring to answer the question.
To model the motion of a projectile when distance is measured in meters instead of feet, you use the formula h = -4.9t 2 + vt + s instead of h = -16t 2 + vt + s. By what factor does the coefficient of t 2 change? Why does that factor make sense? The coefficient is reduced by a factor of 3.3, because there are 3.3 feet in a meter.
When a baseball player hits a baseball into the air, the height of the ball at t seconds after the ball is hit can be modeled with the projectile motion formula. If the ball is hit at 3 feet off the ground with an upward velocity of 47 feet per second, how long will it take for the ball to hit the ground, assuming it is not caught? Find the parameters v and s from the description of the problem. v = 47
s=3
h=0 -16t + 47t + 3 = 0
Substitute parameter values.
2
Divide both sides by -1.
16t 2 - 47t - 3 = 0
Use the combination pair that results in a sum of -47.
(t - 3)(16t + 1) = 0 t-3=0
16t + 1 = 0
or
t=3
16t = -1 © Houghton Mifflin Harcourt Publishing Company
1 t = -_ 16
1. The solutions are 3 and -_ 16 The negative time answer can be rejected because it is not a reasonable value for time in this situation. The correct answer is 3 seconds.
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Solving Equations by Factoring ax 2 + bx + c
324
B
INTEGRATE MATHEMATICAL PRACTICES Focus on Modeling MP.4 When using an equation of the form
A child standing on a river bank ten feet above the river throws a rock toward the river at a speed of 12 feet per second. How long does it take before the rock splashes into the river? Find the parameters v and s from the description of the problem.
h = -16t 2 + vt + s to determine how many seconds it takes for an object to reach the ground, remind students that v represents the initial upward velocity. If a problem states that an object is thrown downward, students must use a negative value for v.
v = -12
s = 10
h= 0
-16 t 2 + -12 t + 10 = 0
Substitute parameter values. Divide both sides by -2 .
8t 2 + 6 t + -5 = 0
Use the combination pair that results in a sum of 6.
( 2 t - 1)( 4 t + 5) = 0
INTEGRATE MATHEMATICAL PRACTICES Focus on Technology MP.5 Have students use a graphing
ELABORATE QUESTIONING STRATEGIES How is factoring a trinomial in the form ax 2 + bx + c similar to factoring a trinomial in the form x 2 + bx + c? How is it different? For both forms, you find factors of the coefficient of x 2 and of c such that the sum of products of factors is equal to b. When the trinomial is in the form x 2 + bx + c, the coefficient of x 2 is 1, so the products of the factors are simply the factors of c.
325
Lesson 8.2
The solution are
__1 2
and
5 -__ .
or
4t + 5 = 0
2t = 1
4t = -5
t= 2
5 t = -4
__1
4
The only correct solution to the time it takes the rock to hit the water is
__
__1 2
seconds.
Your Turn © Houghton Mifflin Harcourt Publishing Company • Image Credites: ©Peter Guess/Shutterstock
calculator to graph an equation that represents the height of an object that is thrown upward, then falls. Discuss the shape of the graph, and ask students at what time the projectile reaches its maximum height. Students should recognize that the time of the object’s maximum height is the x-value for the vertex of the parabola. Because the graph is symmetric, they can also find this value by determining the x-value that is half way between the x-intercepts, or zeros, of the function.
2t - 1 = 0
8.
How long does it take a rock to hit the ground if thrown off the edge of a 72 foot tall building roof with an upward velocity of 24 feet per second?
v = 24; s = 72; h = 0 -16t 2 + 24t + 72 = 0
-8(2t 2 - 3t - 9) = 0
(x - 3)(2x + 3) = 0
x-3=0
x=3
or
2x + 3 = 0 x=-
_3
2 The rock lands 3 seconds after it is thrown.
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Lesson 2
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Elaborate 9.
SUMMARIZE THE LESSON
Discussion What happens if you do not remove the common factor from the coefficients before trying to factor the quadratic equation? If you do not remove the common factor from the coefficients, you can still find binomial
Create a flowchart like the one below for factoring trinomials in the form ax 2 + bx + c.
factors, but you may need to check many more factor pairs before you find the correct answer. Additionally, when you do find a set of factor pairs that work, one of the binomial factors will not be fully factored, such as (2x + 4), which should be written as 2(x + 2) if
Factor out the GCF, if any, of the terms.
you are asked to find the factors of a quadratic expression. 10. Explain how you can know there are never more than two solutions to a quadratic equation, based on what you know about the graph of a quadratic function. The graph of a quadratic equation is shaped like a parabola and can only cross the x-axis
Identify a, b, and c.
twice. The solutions of a quadratic equation in standard form are the x-intercepts of the
Find factor pairs for a.
corresponding quadratic function, and it can have up to two. 11. Essential Question Check-In Describe the steps it takes to solve a quadratic equation by factoring. Change the quadratic equation to standard form if it is not there already, and then find the
Find factor pairs for c.
factors of the quadratic expression by checking factor pairs of a and c. When the quadratic
Find the sum of the product of one set of factor pairs for a and c.
is factored, set each binomial equal to zero to find a possible solution.
Evaluate: Homework and Practice Factor the following quadratic expressions. 1.
6x 2 + 5x + 1
2.
3.
4x 2 - 8x + 3
4.
3x 2 - 2x - 5
6.
12x 2 + 22x - 14
8.
2(6x + 11x - 7) = 2(2x - 1)(3x + 7) 2
Module 8
-10x 2 + 3x + 4
-1(10x 2 - 3x - 4) = -1(2x + 1)(5x - 4)
(x + 1)(3x - 5) 7.
24x 2 - 44x + 12
4(6x 2 - 11x + 3) = 4(2x - 3)(3x - 1)
(2x - 1)(2x - 3)
5.
3(3x 2 + 11x + 10) = 3(x + 2) (3x + 5)
Exercise
Depth of Knowledge (D.O.K.)
EVALUATE
-3(5x 2 - 7x - 6) = -3(x - 2)(5x + 3)
Lesson 2
COMMON CORE
Mathematical Practices
1–16
2 Skills/Concepts
MP.4 Modeling
17–20
2 Skills/Concepts
MP.4 Modeling
21–22
2 Skills/Concepts
MP.4 Modeling
23
1 Recall of Information
MP.2 Reasoning
24
2 Skills/Concepts
MP.4 Modeling
3 Strategic Thinking
MP.4 Modeling
25–26
Write the factored form of ax2 + bx +c.
-15x 2 + 21x + 18
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No
Yes
© Houghton Mifflin Harcourt Publishing Company
(3x + 1)(2x + 1)
9x 2 + 33x + 30
Is the sum of the products equal to b?
• Online Homework • Hints and Help • Extra Practice
ASSIGNMENT GUIDE 4/2/14 12:59 AM
Concepts and Skills
Practice
Explore Factoring ax 2 + bx + c When c > 0
Exercises 1–4, 22
Example 1 Factoring ax 2 + bx + c When c < 0
Exercises 5–8, 21, 23, 25–26
Example 2 Solving Equations of the Form ax 2 + bx + c = 0 by Factoring
Exercises 9–16
Example 3 Solving Equation Models of the Form ax 2 + bx + c = 0 by Factoring
Exercises 17–20, 24
Solving Equations by Factoring ax 2 + bx + c
326
Solve the following quadratic equations.
INTEGRATE MATHEMATICAL PRACTICES Focus on Reasoning MP.2 Students can check their solutions by
9.
5x 2 + 18x + 9 = 0
10. 12x 2 - 36x + 15 = 0
(x + 3)(5x + 3) = 0
x+3=0
x = -3
substituting the values into the original equations and verifying that both solutions make the equation true. When checking solutions, remind students that the product of two negative numbers is positive and the product of two numbers with opposite signs is negative.
12x 2 - 36x + 15 = 0 5x + 3 = 0
or
4x 2 - 12x + 5 = 0
_
3 x=5
or
(2x - 1)(2x - 5) = 0
2x - 1 = 0 x= 1 2
11. 6x 2 + 28x - 2 = 2x - 10
6x 2 + 28x - 2 = 2x - 10
VISUAL CUES
Students sometimes forget that every number has 1 and itself as two of its factors. When students are making their lists of factors, remind them to try all possible factors.
-100x 2 + 55x + 3 = 50x 2 - 55x + 23 -150x 2 + 110x - 20 = 0
3x 2 + 13x + 4 = 0
15x 2 - 11x + 2 = 0
(x + 4)(3x + 1) = 0 x = -4
(3x - 1)(5x - 2) = 0
3x + 1 = 0
or
3x - 1 = 0 1 x= 3
or
3
_
or
4x + 1 = 0
or
x=-
-12x 2 - 34x + 28 = 0
_1
6x 2 + 17x - 14 = 0
4
(2x + 7)(3x - 2) = 0
15. (8x + 7)(x + 1) = 9
(8x + 7)(x + 1) = 9 8x 2 + 15x - 2 = 0
48x 2 + 24x - 9 = 48x
(x + 2) (8x - 1) = 0
IN2_MNLESE389830_U4M08L2.indd 327
_
3(16x 2 + 8x - 3) = 48x
8x + 8x + 7x + 7 = 9
Module 8
3x - 2 = 0 2 x= 3
3(4x - 1)(4x + 3) = 48x
2
or
2
or or
16. 3(4x - 1)(4x + 3) = 48x
48x 2 - 24x - 9 = 0
8x - 1 = 0 1 x= 8
16x 2 - 8x - 3 = 0
_
(4x + 1)(4x - 3) = 0
4x + 1 = 0
x=-
Lesson 8.2
_7
x=-
or
_
-12x 2 = 34x - 28
2x + 7 = 0
x+2=0
or
5x - 2 = 0 2 x= 5
14. -12x 2 = 34x - 28
(2x - 3)(4x + 1) = 0
2x - 3 = 0 3 x= 2
or
_
1 x = -_
13. 8x 2 - 10x - 3 = 0
x = -2
327
_
6x + 26x + 8 = 0
x+4=0
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Peter Guess/Shutterstock
AVOID COMMON ERRORS
or
2x - 5 = 0 5 x= 2
12. -100x 2 + 55x + 3 = 50x 2 - 55x + 23
2
When students are creating factor pairs, have them use two different colored pencils, one for positive numbers and one for negative numbers. This will allow students to more easily recognize a negative as they are multiplying. Additionally, using two colors helps them notice whether or not they have considered both positive and negative factor pairs, as needed.
or
_
327
_1 4
or or
4x - 3 = 0 3 x= 4
_ Lesson 2
4/2/14 12:59 AM
Read the real-world situation and substitute in values for the projectile motion formula. Then solve the resulting quadratic equation by factoring to answer the question.
AVOID COMMON ERRORS Students may forget how to determine the signs of the constant terms in the binomial factors. Remind students that the sign of c in a trinomial determines the signs of the constant terms in its binomial factors. If c > 0, then the signs of the constants must be the same. If c < 0, then the signs must be opposites.
17. A golfer takes a swing from a hill twenty feet above the cup with an initial upwards velocity of 32 feet per second. How long does it take the ball to land on the ground near the cup?
(2x + 1) (2x - 5) = 0
v = 32 s = 20
2x + 1 = 0
h=0
-16t 2 + 32t + 20 = 0
4t 2 - 8t - 5 = 0
1 x = -_ 2
2x - 5 = 0 5 or x = 2
or
_
1 seconds It takes the ball 2__ 2 to land.
18. An airplane pilot jumps out of an airplane and has an initial velocity of 60 feet per second downwards. How long does it take to fall from 1000 feet to 900 feet before the parachute opens?
v = -60
s = 1000 h = 900
-16t 2 - 60t + 1000 = 900
-16t 2 - 60t + 100 = 0 4t 2 + 15t - 25 = 0
(t + 5) (4t - 5) = 0
t+5=0
4t - 5 = 0 5 t = -5 or t= 4 1 The pilot falls for 1__ seconds before the 4 or
_
parachute opens.
A race car driving under the caution flag at 80 feet per second begins to accelerate at a constant rate after the warning flag. The distance traveled since the warning flag in feet is characterized by 30t 2 + 80t, where t is the time in seconds after the car starts accelerating again. 19. How long does it take the car to travel 30 feet after it begins accelerating?
30t + 80t - 30 = 0 2
3t 2 + 8t - 3 = 0
(t + 3)(3t - 1) = 0
t+3=0
© Houghton Mifflin Harcourt Publishing Company
30t 2 + 80t = 30
3t - 1 = 0 1 t = -3 or t= 3 1 __ It takes 3 second to travel 30 more feet. or
_
20. How long will the car take to travel 160 feet?
30t 2 + 80t = 160 30t + 80t - 160 = 0 2
3t 2 + 8t - 16 = 0
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(t + 4)(3t - 4) = 0
t+4=0
3t - 4 = 0 4 t= 3 It will take 1__13 seconds. t = -4
or or
328
_
Lesson 2
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Solving Equations by Factoring ax 2 + bx + c
328
Geometry For each rectangle with area given, determine the binomial factors that describe the dimensions.
INTEGRATE MATHEMATICAL PRACTICES Focus on Reasoning MP.2 When solving equations to determine how
21.
22.
area = 21x2 + 13x + 2
many seconds it takes for an object to reach the ground, discuss with students what the x- and y- intercepts of a graph of the equation represent. If time is represented on the x-axis and height is represented on the y-axis, the x-intercept represents the time it takes the object to reach the ground, and the y-intercept represents the original height of the object.
area =
6x2 + 17x -
3
21x 2 + 13x + 2 = (3x + 1)(7x + 2)
6x 2 + 17x - 3 = (x + 3)(6x - 1)
length = (6x - 1), width = (x + 3)
length = (7x + 2), width = (3x + 1)
23. Multiple Response Which of the following expressions in the list describes the complete factorization of the quadratic expression 15x 2 - 25x - 10? Circle all that apply. b. 5(3x + 1)(x - 2) c. 5(x + 2)(3x - 1) a. (3x + 1)(5x - 10) d. 5(x - 2)(3x + 1)
e. 5(3x - 1)(x + 2)
f.
(5x - 10)(3x + 1)
H.O.T. Focus on Higher Order Thinking
24. Multi-Part Response A basketball player shoots at the basket from a starting height of 6 feet and an upwards velocity of 20 feet per second. Determine how long it takes for the shot to drop through the basket, which is mounted at a height of 10 feet.
JOURNAL Have students explain how to factor a trinomial of the form ax 2 + bx + c when a and b are positive and c is negative.
a. Set up the equation for projectile motion to solve for time and convert it to standard form. v = 20 s=6 h = 10 -16t 2 + 20t + 6 = 10 -16t 2 + 20t - 4 = 0 b. Solve the equation by factoring. -16t 2 + 20t - 4 = 0 t-1=0
4t - 5t + 1 = 0 2
© Houghton Mifflin Harcourt Publishing Company
(t - 1)(4t - 1) = 0
or
t=1
4t - 1 = 0 1 t= 4
_
c. Explain why you got two positive solutions to the equation, and determine how you can rule one of them out to find the answer to the question. Hint: Solving the equation graphically may give you a hint. The solutions to the equation are the times at which the basketball is at the same height as the basket. The basketball must go up past the height of the basket before coming back down, and both times are positive (after the player shoots). The earlier time is when the ball passes the basket’s height traveling upward, and the later time is when it drops through the hoop. Therefore the correct answer is 1 second. 25. Critical Thinking Find the binomial factors of 4x 2 - 25. Rewrite the expression as 4x 2 + 0x - 25. The factor combination will have a sum of 0. 2 4x 2 + 0x - 25 = (2x + 5)(2x - 5) = (2x + 5) 26. Communicate Mathematical Ideas Find all the values of b that make the expression 3x 2 + bx - 4 factorable. -1, 4, 13, 1, -4, and -11 can be used for the value of b. Module 8
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Lesson Performance Task
QUESTIONING STRATEGIES What is the initial velocity of an object that is stopped? 0 m/s
The equation for the motion of an object with constant acceleration is d = d 0 + vt + __12 at 2, where d is distance from a given point in meters, d 0 is the initial distance from the starting point in meters, v is the starting velocity in meters per second, a is acceleration in meters per second squared, and t is time in seconds.
What is the acceleration of an object that does not change velocity? 0 m/s 2
A car is stopped at a traffic light. When the light turns green, the driver begins to drive, accelerating at a constant rate of 4 meters per second squared. A bus is traveling at a speed of 15 meters per second in another lane. The bus is 7 meters behind the car as it begins to accelerate.
If the car’s initial position is d 0 = 0, what value of d 0 describes the initial position of the bus? Explain. −7; since the bus is 7 meters behind the car’s starting point, the value for its position must be negative.
Find when the bus passes the car, when the car passes the bus, and how far each has traveled each time they pass one another.
The distance traveled from the intersection by the car is the following: 1 d c(t) = (0)t + (4)t 2 = 2t2 2 The distance traveled from the intersection by the bus is the following: 1 d b(t) = -7 + 15t + (0)t 2 = -7 + 15t 2 Set d c(t) equal to d b(t) to find the time when the bus passes the car and the
_
INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 Because the bus is moving at a constant
_
time when the car passes the bus. d c(t) = d b(t)
2t - 1 = 0 2t = 1 1 t= 2
_
0 = 2t 2 - 15t + 7
0 = (2t - 1)(t - 7)
The bus passes the car at t =
t-7=0 or
t=7
speed, its motion is represented by a linear equation. Discuss with students why the car’s motion is described by a quadratic equation. At the start, the car is stopped, so its initial velocity is 0 m/s. As the car accelerates, its speed increases steadily, so the distance traveled increases at an increasing rate. The increasing curve of a parabola models this type of motion.
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Aflo Foto Agency/Alamy
2t = -7 + 15t 2
_1 and the car passes the bus at t = 7. 2
1 1 The bus passes the car at t = _ seconds. Substitute t = _ into the
2 2 equation for d c(t). 2 1 d c(t) = 2t 2 = 2 = 0.5 2 Recall that the bus is 7 meters behind the car as it begins to accelerate. So, 7 + 0.5 = 7.5. Therefore, the bus travels a total of 7.5 meters before it passes the car, while the car travels 0.5 meter. The car passes the bus at t = 7 seconds. Substitute t = 7 into the equation for d c(t).
(_)
d c(t) = 2t 2 = 2(7) = 98 2
Recall that the bus is 7 meters behind the car as it begins to accelerate. So, 7 + 98 = 105. Therefore, the car travels a total of 98 meters before it passes the bus, while the bus travels 105 meters.
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Have students use graphing calculators to display the graphs of y = 2t 2 and y = -7 + 15t in the same window. Have them explain what information the graphs provide, and compare it to the information provided by the graph of y = 2t 2 - 15t + 7.
Students should recognize that the intersection points of the two graphs represent the times and locations at which the vehicles pass each other. The x-intercepts of the graph of y = 2t 2 - 15t + 7 indicate the times when the vehicles pass, but this graph does not provide any direct information about the distance traveled.
4/2/14 12:59 AM
Scoring Rubric 2 points: Student correctly solves the problem and explains his/her reasoning. 1 point: Student shows good understanding of the problem but does not fully solve or explain his/her reasoning. 0 points: Student does not demonstrate understanding of the problem.
Solving Equations by Factoring ax 2 + bx + c
330
LESSON
8.3
Name
Using Special Factors to Solve Equations
Class
8.3
Date
Using Special Factors to Solve Equations
Essential Question: How can you use special products to aid in solving quadratic equations by factoring?
Common Core Math Standards
Resource Locker
The student is expected to: COMMON CORE
Factor a quadratic expression to reveal the zeros of the function it defines. Also A-SSE.A.2, A-REI.B.4b
When you use algebra tiles to factor a polynomial, you must arrange the unit tiles on the grid in a rectangle. Sometimes, you can arrange the unit tiles to form a square. Trinomials of this type are called perfect-square trinomials.
Mathematical Practices COMMON CORE
Key
MP.1 Problem Solving
=1
Language Objective
= -1
Explain to a partner what a perfect-square trinomial is and how you can recognize one.
A
= -x
Arrange the algebra tiles on the grid. Place the 1 x 2-tile in the upper left corner, and arrange the 9 unit tiles in the lower right corner. ×
C
Fill in the empty spaces on the grid with x-tiles. ×
Module 8
be ges must EDIT--Chan DO NOT Key=NL-A;CA-A Correction
Lesson 3
331
gh “File info”
made throu
Date Class
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Using Speci Equations
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HARDCOVER PAGES 331348 Watch for the hardcover student edition page numbers for this lesson.
2
= -x
2
Key =1 = -1
=x
=x
= -x
x -tiles, 2 need 1 + 6x + 9. sion. You factor x a tiles to l the expres need to mode tiles you number of , and Identify the left corner unit tiles. , and 9 2 in the upper 6 x-tiles 1 x -tile Place the on the grid. algebra tiles corner. Arrange the lower right the in 9 unit tiles arrange the × 2
Use algebr
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empty spaces
on the grid
.
with x-tiles ×
Lesson 3 331 Module 8
8L3 331 30_U4M0
ESE3898
IN2_MNL
331
Lesson 8.3
= -x2
Use algebra tiles to factor x 2 + 6x + 9.
PREVIEW: LESSON PERFORMANCE TASK View the Engage section online. Discuss how the shape of the base of a fountain could affect the possible patterns made by the falling water. Then preview the Lesson Performance Task.
= x2
6 x-tiles, and 9 unit tiles.
© Houghton Mifflin Harcourt Publishing Company
By recognizing that a polynomial is a perfect-square trinomial or a difference of squares, you can use the appropriate special product rule to factor the polynomial, and then use the zero product property to solve the equation.
=x
Identify the number of tiles you need to model the expression. You need 1 x 2-tiles,
ENGAGE Essential Question: How can you use special products to aid in solving quadratic equations by factoring?
Exploring Factors of Perfect Square Trinomials
Explore
A-SSE.B.3a
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6:46 PM
07/04/14 6:19 PM
All 6 x-tiles were used, so all the tiles are accounted for and fit in the square with sides of length x + 3 . Read the length and width of the square to get the factors of the
(
trinomial x 2 + 6x + 9 = x + 3
EXPLORE
) (x + 3 ).
Exploring Factors of Perfect-Square Trinomials
Now, use algebra tiles to factor x 2 - 8x + 16. You need 1 x 2-tiles, 8 -x -tiles, and 16 unit tiles to model the expression.
INTEGRATE TECHNOLOGY
Arrange the algebra tiles on the grid. Place the 1 x 2-tile in the upper left corner, and
Students have the option of completing the algebra tiles activity either in the book or online.
arrange the 16 unit tiles in the lower right corner. ×
INTEGRATE MATHEMATICAL PRACTICES Focus on Modeling MP.4 Remind students that modeling the factors of
Fill in the empty spaces on the grid with -x-tiles.
trinomials with algebra tiles means the tiles must be arranged in a rectangle. When the factors of a perfect-square trinomial are modeled with algebra tiles, the tiles will be arranged in a square.
×
All 8 -x -tiles were used, so all the tiles are accounted for and fit in a square with sides of length
x - 4 . Read the length and width of the square to get the factors of the
trinomial x 2 - 8x + 16 =
(
x-4
)(
x-4
).
Reflect
1.
What If? Suppose that the middle term in x 2 + 6x + 9 was changed from 6x to 10x. How would this affect the way you factor the polynomial? The arrangement of unit tiles would have to be in a rectangle, not a square. The factored
© Houghton Mifflin Harcourt Publishing Company
form x 2 + 10x + 9 is (x + 1)(x + 9).
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Learning Progressions
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In this lesson, students expand their understanding of factoring trinomials of the form ax 2 + bx + c by learning to recognize and factor perfect-square trinomials, the difference of two squares, and polynomials that consist of one of these special products multiplied by a monomial factor. They also solve equations and real-world problems that involve these polynomials. As they work with polynomial expressions that follow certain patterns, students learn to recognize the patterns and apply them when appropriate. The ability to factor special products efficiently will be valuable in algebra as well as in future courses that require repeated reasoning with polynomials.
Using Special Factors to Solve Equations
332
2.
EXPLAIN 1 Factoring a 2x 2 + 2abx + b 2 and a 2x 2 - 2abx + b 2
If the positive unit squares are arranged in a square of unit tiles when factoring with algebra tiles, what will be true about the binomial factors? (The coefficient of the x 2 term is 1 as in the previous problems.) Both factors will be the same, as in (x + 3)(x + 3).
Explain 1
Recall that a perfect-square trinomial can be represented algebraically in either the form a 2 + 2ab + b 2 or the form a 2 - 2ab + b 2.
QUESTIONING STRATEGIES
Perfect-Square Trinomials
How can you tell whether a trinomial of the form ax 2 + bx + c is a perfect-square trinomial? If it is a perfect-square trinomial, both a and c will be perfect squares, and b will be equal to twice the product of the square roots of a and c, or twice the opposite of that product.
Perfect-Square Trinomials Perfect-Square Trinomial
Examples x + 6x + 9 = (x + 3)(x + 3) 2
= (x + 3)
2
a 2 + 2ab + b 2 = (a + b)(a + b) = (a + b)
c 2x 2 + 2cdx + d 2 = (cx) 2 + 2cdx + d 2
2
= (cx + d)(cx + d)
= (cx + d)
2
x 2 - 10x + 25 = (x - 5)(x - 5)
AVOID COMMON ERRORS
= (x - 5)
2
a 2 - 2ab + b 2 = (a - b)(a - b)
c 2x 2 - 2cdx + d 2 = (cx) 2 - 2cdx + d 2
= (a - b)
2
= (cx - d)(cx - d)
= (cx - d) Example 1
© Houghton Mifflin Harcourt Publishing Company
When asked to factor a trinomial that has a common monomial factor, such as 18x 2 - 60x + 50, students may see that the first and last terms are not perfect squares and therefore assume that they cannot use the rule for factoring perfect squares. Remind students to always begin by factoring out any common factors, and then examine the trinomial that remains to decide whether they can use a special product rule.
Factoring a 2x 2 + 2abx + b 2 and a 2x 2 - 2abx + b 2
2
Factor perfect-square trinomials.
4x 3 - 24x 2 + 36x 4x 3 - 24x 2 + 36x = 4x(x 2 - 6x + 9)
Factor out the common monomial factor 4x.
= 4x [x - 2(1 · 3)x + 3 ] 2
2
= 4x(x - 3)(x - 3)
Rewrite the perfect square trinomial in the form a 2x 2 - 2abx + b 2. Rewrite the perfect square trinomial in the form
(ax - b)(ax - b) to obtain factors.
The factored form of 4x 3 - 24x 2 + 36x is 4x(x - 3)(x - 3), or 4x(x - 3) . 2
x 2 + 16x + 64
(1
x 2 + 16x + 64 = x 2 + 2
(
= x+ 8
·
8
)x+
)(x + 8 )
2
Rewrite in the form a 2x 2 + 2abx + b2.
8
(
The factored form of x 2 + 16x + 64 is x + 8 Module 8
) (x + 333
Rewrite in the form (ax + b)(ax + b).
8
), or (x +
). 2
8
Lesson 3
COLLABORATIVE LEARNING IN2_MNLESE389830_U4M08L3.indd 333
Peer-to-Peer Activity Have students work in pairs. Ask each student to write five polynomials, each of which is a perfect-square trinomial or the difference of two squares. Students then trade polynomials and factor the ones they receive. Finally, ask students to explain to each other what steps they used to factor each polynomial.
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Lesson 8.3
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Your Turn
EXPLAIN 2
Factor perfect-square trinomials. 3.
2y 3 + 12y 2 + 18y
4.
2y 3 + 12y 2 + 18y = 2y(y 2 + 6y + 9)
100z 2 - 20z + 1
= (10z - 1)(10z - 1)
= 2y[y 2 + 2(1 · 3)y + 3 2]
= 2y(y + 3)(y + 3) = 2y(y + 3)
Factoring a 2x 2 - b 2 = 0
100z 2 - 20z + 1 = 10 2z 2 - 2(10 · 1)z + 1 2 = (10z - 1)
2
AVOID COMMON ERRORS
2
Students might think that y 6 is not a perfect square because the number 6 is not a perfect square. Remind them of the Product of Powers Property, which states that a m ∙ a n = a m + n. Thus, y 6 is a perfect square because it can be written as y 3 ∙ y 3. The Power of a Power Property can also be used to show this 2 as y 6 = (y 3) .
Factoring a 2x 2 - b 2 = 0
Explain 2
Recall that a difference of squares can be written algebraically as a 2 - b 2 and factored as (a + b)(a - b).
Difference of Squares
Difference of Two Squares Perfect-Square Trinomial
Examples
a - b = (a + b)(a - b)
x 2 - 9 = (x + 3)(x - 3)
2
2
4x 2 - 9 = (2x + 3)(2x - 3)
9x 2 - 1 = (3x + 1)(3x - 1)
QUESTIONING STRATEGIES
c 2x 2 - d 2 = (cx) 2 - d 2 = (cx + d)(cx - d) Example 2
Can you factor x 2 + 25 as a difference of two squares? Explain. No; it is a sum of two squares, not a difference. The operation sign for the difference of two squares must be −.
Factor each difference of squares.
x 2 - 49 x 2 - 49 = x 2 - 7 2
Rewrite in the form a 2x 2 - b 2. © Houghton Mifflin Harcourt Publishing Company
The factored form of x 2 - 49 is (x + 7)(x - 7).
49q 2 - 4p 2 49q 2 - 4p 2 = 7
2
2
q
-
(
2p
)
2
Rewrite in the form a 2x 2 - b 2.
Rewrite in the form (ax + b)(ax - b). ( 7q + 2p )( 7q - 2p ) The factored form of 49q - 4p is ( 7q + 2p ) ( 7q - 2p ). =
2
What are the values of a and b in the difference of squares 9 - x 4? a = 3 and b = x 2
Rewrite in the form (ax + b)(ax - b).
= (x + 7)(x - 7)
2
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334
INTEGRATE MATHEMATICAL PRACTICES Focus on Reasoning MP.2 Discuss with students what happens if you
Lesson 3
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Number Sense Show students a quick way to rule out the possibility that a trinomial is a perfect-square trinomial. Remind them that if ax 2 + bx + c is a perfect-square trinomial, then b must be an even number. This is because, in a perfect-square trinomial, b = 2 ∙ √a ∙ √b . If b is odd, then the trinomial is not a perfect square.
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forget to factor out a common factor of the terms in a difference of two squares. For example, to factor 4x 2 - 16y 4, you could first factor out a common factor of 4 to get 4(x 2 - 4y 4), then factor the difference of squares to get 4(x + 2y 2)(x - 2y 2). If you do not notice the common factor, you can still factor the expression as a difference of squares, to get (2x + 4y 2)(2x - 4y 2). Help students to see that this would be an equivalent expression, but it would not be a complete factorization. You would still need to factor a 2 out of each binomial to get the same final result as before.
― ―
Using Special Factors to Solve Equations
334
Reflect
EXPLAIN 3
5.
Discussion James was factoring a difference of squares but did not finish his work. What steps is he missing? 16x 4 - 1 = (4x 2) - 1 2
Solving Equations with Special Factors
= (4x 2 + 1)(4x 2 + 1)
James factored the first difference of squares, 16x 4 - 1, but forgot to factor the second one that came up, 4x 2 - 1. This is his completed work:
INTEGRATE MATHEMATICAL PRACTICES Focus on Patterns MP.8 When students factor a perfect-square
16x 4 - 1 = (4x 2) - 1 2
= (4x 2 + 1)(4x 2 - 1) 2 = (4x 2 + 1)⎡⎣(2x) - 1⎤⎦ = (4x 2 + 1)(2x + 1)(2x - 1)
trinomial, they should first look at the sign of the x-term, as this will tell them which pattern to use. If it 2 is +, they should use (a + b) ; if it is -, they should 2 use (a - b) .
Your Turn
Factor each difference of squares. 6.
x 2 - 144
81y 4 - 9y 2
81y 4 - 9y 2 = 9y 2(9y 2 - 1)
x 2 - 144 = x 2 - 12 2
= 9y 2(3 2y 2 - 1)
= (x + 12)(x - 12)
QUESTIONING STRATEGIES How are the equations 25x + 20x + 4 = 0 and 75x 2 + 60x + 12 = 0 related? How do their solutions compare? Explain. The second equation is equal to the first one multiplied by 3. 2 Both equations have the same solution, x = −__ . The 5 common factor of 3 in the second equation has no effect on the solution, because when you divide both sides of the equation by 3, it is identical to the first equation.
7.
= 9y 2(3y + 1)(3y - 1)
2
Explain 3
Solving Equations with Special Factors
© Houghton Mifflin Harcourt Publishing Company
Equations with special factors can be solved using the Zero Product Property. Remember, the Zero Product Property states that if the product of two factors is zero, then at least one of the factors must be zero. For example, if (x + 1)(x + 9) = 0 then x + 1 = 0 or x + 9 = 0. Consequently, the solutions for the equation are x = -1 or x = -9. Example 3
Solve the following equations with special factors. 4x 2 + 12x + 9 = 0
4x 2 + 12x + 9 = 0 2 2x 2 + 2(2 · 3)x + 3 2 = 0
Rewrite in the form a 2x 2 + 2abx + b 2.
(2x + 3)(2x + 3) = 0
Rewrite in the form (ax + b)(ax + b).
2x + 3 = 0
3 x = -_ 2
Module 8
Set factors equal to 0 using Zero Product Property. Solve equation.
335
Lesson 3
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Modeling Students may know and use the word perfect to mean flawless, but in mathematics the term perfect square has a specific meaning. A number that is a perfect square is the square of a whole number, and a perfect-square trinomial is the square of a binomial.
335
Lesson 8.3
Demonstrate how to model the factors of trinomials with algebra tiles. Point out that if a trinomial can be factored, the tiles form a rectangle. For a perfect-square trinomial, the tiles will be arranged in a square. Be sure to use and repeat the terminology as you work together so that students hear the words in context, which helps clarify their meaning.
02/04/14 3:55 PM
25x 2 - 1 = 0
EXPLAIN 4
25x 2 - 1 = 0 2
5 x2 - 1
2
=0
( 5x + 1 ) ( 5x - 1 ) = 0 5x - 1 = 0 or 5x + 1 = 0
_1
x=
5
_1 or x = 5
Rewrite in the form a 2x 2 + 2abx + b 2.
Solving Equation Models with Special Factors
Rewrite in the form (ax + b)(ax - b). Set factors equal to 0 using Zero Product Property.
AVOID COMMON ERRORS
Solve equation.
Remind students that the Zero Product Property can be applied only to quadratic equations in standard form, because it applies only to equations for which one side is zero. Students often try to apply it before setting one side equal to zero.
Your Turn
Solve the following equations with special factors. 8.
25x 2 - 10x + 1 = 0
9.
8x 4 - 2 x 2 = 0
25x 2 - 10x + 1 = 0
8x 4 - 2x 2 = 0
(5x - 1)(5x - 1) = 0
2x 2(2 2x 2 - 1) = 0
2x 2(4x 2 - 1) = 0
5 2x 2 - 2(5 · 1)x + 1 2 = 0
2x 2(2x + 1)(2x - 1) = 0
5x - 1 = 0
x2 = 0
5x = 1 1 x=_ 5
or
2x + 1 = 0
x=0
2x = -1
2
Explain 4
1 x=_
2x - 1 = 0 2x = 1
2
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Stocktrek Images, Inc./Getty Images
1 x = -_
or
Solving Equation Models with Special Factors
For each real-world scenario, solve the model which involves an equation with special factors. Example 4
Write the given information and manipulate into a familiar form. Solve the equation to answer a question about the situation.
As a satellite falls from outer space onto Mars, its distance in miles from the planet is given by the formula, where t is the number of hours it has fallen. Find when the satellite will be 200 miles away from Mars.
Analyze Information Identify the important information • The satellite’s distance in miles is given by the 2 formula d = -9t + 776 .
• The satellite distance at some time t is d = 200 .
Module 8
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Using Special Factors to Solve Equations
336
Formulate a Plan
QUESTIONING STRATEGIES
2 Substituting the value of the constant d = 200 into the equation d = -9t + 776 2 you get the equation 200 = -9t + 776 . Simplify the new equation into a
If the equation that models a real-world problem has the form x 2 - b 2 = 0 for some constant b, why might you need to discard one solution for x? Factoring x 2 - b 2 gives (x + b)(x - b), so the solutions to the equation are x = b and x = -b. One of these values is positive and one is negative. If x represents a quantity such as time or distance, a negative value may not make sense in the context of the problem.
familiar form and solve it.
Solve Rewrite the equation to be equal to 0. 200 = -9t 2 + 776
Subtract 200 from both sides.
0 = -9t 2 + 576
Divide both sides by -1.
0 = 9t 2 - 576
Factor out 9.
0= 9
(t
2
-
64
The equation contains a
difference of squares that you can factor.
(
) ( t - 8)
(
) (t -
0=9 t+8 Use the
)
Zero Product Property to solve.
0=9 t+ 8
t + 8 = 0 or t - 8 = 0
8
)
t = ±8 The answer is t = 8 because time must be
positive . So, the satellite has fallen
for 8 hours.
© Houghton Mifflin Harcourt Publishing Company
Justify and Evaluate t = 8 makes sense because time must be value of t into the original equation. -9 ·
+ 776 = - 9 · 64 + 776
= 200 This is what is expected from the given information.
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Lesson 8.3
2
8
= 776 - 576
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337
positive . Check by substituting this
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Your Turn
ELABORATE
Write the given information and manipulate it into a familiar form. Solve the equation to answer a question about the situation.
QUESTIONING STRATEGIES
10. A volleyball player sets the ball in the air, and the height of the ball after t seconds is given in feet by h = -16t 2 + 12t + 6. A teammate wants to wait until the ball is 8 feet in the air before she spikes it. When should the teammate spike the ball? How many reasonable solutions are there to this problem? Explain.
What types of quadratic equations have only one real solution? Quadratic equations that can be factored as perfect squares have only one real solution. For example, 4x 2 - 16x + 16 = 0, 2 which can be factored as 4(x - 2) = 0, has one real solution, x = 2.
-16t + 12t + 6 = 8 2
-16t 2 + 12t - 2 = 0 8t 2 - 6t + 1 = 0
(4t - 1)(2t - 1) = 0 4t - 1 = 0
or
1 t=_
2t - 1 = 0 1 t=_ 2
4
At 0.25 second or 0.5 second the ball will be 8 ft high. There are two solutions because both occur after t = 0, when the ball is set. 11. The height of a model rocket is given (in centimeters) by the formula h = -490t 2, where t is measured in seconds and h = 0 refers to its original height at the top of a mountain. It begins to fly down from the mountain-top at time t = 0. When has the rocket descended 490 centimeters?
-490 = -490t 2 490t 2 - 490 = 0
490(t 2 - 1) = 0
490(t + 1)(t - 1) = 0 t = ±1 After one second, the rocket will have descended a distance of 490 centimeters. The © Houghton Mifflin Harcourt Publishing Company
negative time cannot be used in this context.
Elaborate 12. Are the perfect square trinomials a 2 + 2ab + b 2 and a 2 - 2ab + b 2 very different? How can you get one from the other? Start with the first form a 2 + 2ab + b 2. Let b = -b new and simplify.
a 2 + 2ab + b 2 = a 2 + 2a(-b new) + (-b new) 2 = a 2 - 2ab new + ⎡⎣(-1)(b new)⎤⎦ 2 2 = a 2 - 2ab new + (-1) (b new) 2
= a 2 - 2ab new + b new 2
a 2 - 2ab new + b new 2 is the other form. They are not very different. Module 8
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13. How would you go about factoring a 2 - 2ab + b 2 - 1? There is a perfect-square trinomial inside the expression that you can factor. There is also a
SUMMARIZE THE LESSON
difference of squares.
What are two characteristics of a perfectsquare trinomial of the form ax 2 + bx + c? a and c are perfect squares, and b = 2 √ac .
a 2 - 2ab + b 2 - 1 = (a - b) - 1 2 2
―
= (a - b + 1 ) (a - b - 1 )
14. Setting a perfect-square trinomial equal to zero, a 2x 2 + 2abx + b 2 = 0, produces how many solutions? How many solutions are produced setting a difference of squares equal to zero, a 2x 2 - b 2 = 0? a 2x 2 + 2abx + b 2 = 0 a 2x 2 - b 2 = 0
What are two characteristics of a difference of squares? Both terms are perfect squares, and one term is subtracted from the other.
(ax + b)2 = 0
(ax + b)(ax - b) = 0
ax = -b
ax = ±b
b x = -_
b x = ±_ a
a
a 2x 2 + 2abx + b 2 = 0 produces
a 2x 2 - b 2 = 0 produces two
one solution for x.
solutions for x.
15. Physical problems involving projectile motion can be modeled using the general equation h = -16t 2 + v 0t. Here, h refers to the relative height of the projectile from its initial position, v 0 is its initial vertical velocity, and t is time elapsed from launch. If you are measuring the height of the projectile as it descends from a high place, and it was launched with v 0 = 0 (which means it was thrown horizontally or dropped), how would you find the solution using special factors? (Assume that the height the projectile has descended is a square number in this question, although this is not a requirement in real life). Set v 0 = 0 and h = -s 2 for some value of s. h is negative because the projectile is
descending from its original position. Substitute these values into the equation and simplify. h = -16t 2 + v 0t h = -16t 2 © Houghton Mifflin Harcourt Publishing Company
-s 2 = -16t 2 16t 2 - s 2 = 0 At this point you can use the difference of squares to solve.
(4t + s)(4t - s) = 0 4t = ±s
s t = ±_ 4
s Only the positive answer applies to the model, so the solution is t = _ . The difference of 4
squares was used to find this. 16. Essential Question Check-In How can you use special products to solve quadratic equations? Once you recognize that a polynomial is a perfect-square trinomial or a difference of
squares, you can factor the polynomial. Then, use the Zero Product Property to solve the equation. Module 8
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Evaluate: Homework and Practice
EVALUATE • Online Homework • Hints and Help • Extra Practice
For each trinomial, draw algebra tiles to show the factored form. Then, write the factored form. 1.
x 2 - 10x + 25
2.
x 2 + 8x + 16
×
×
ASSIGNMENT GUIDE
x 2 + 8x + 16 = (x + 4)(x + 4)
x 2 - 10x + 25 = (x - 5)(x - 5) Factor. 3.
4x 2 + 4x + 1
4.
9x 2 - 18x + 9
9x 2 - 18x + 9 = 9(x 2 - 2x + 1)
4x 2 + 4x + 1 = 2 2x 2 + 2(2 · 1)x + 1 = (2x + 1)(2x + 1)
= 9(x - 1)(x - 1)
= (2x + 1)
= 9 (x - 1 )
2
5.
16x 3 + 8x 2 + x
16x + 8x + x = x (16x + 8x + 1) 3
2
6.
2
= x(4x + 1)(4x + 1)
7.
= 2x(4x - 1) (4x - 1) = 2x(4x - 1)
8.
x - 169 = x - 13 2
32x 4 - 8x 2
2
4p 2 - 9q 4
4p 2 - 9q 4 = (2p) - (3q 2) 2
2
2
= (2p + 3q 2)(2p - 3q 2)
= (x + 13)(x - 13)
9.
32x 3 - 16x 2 + 2x = 2x (16x 2 - 8x + 1)
2
x 2 - 169 2
32x 3 - 16x 2 + 2x
10. 2y 5 - 32z 4y
2y 5 - 32z 4 y = 2y(y 4 - 16z 4)
32x 4 - 8x 2 = 8x 2 (4x 2 - 1)
2 2 = 2y⎡⎣(y 2) - (4z 2) ⎤⎦
= 8x 2 (2 2x 2 - 1)
= 2y(y 2 + 4z 2)(y 2 - 4z 2)
= 8x 2(2x + 1) (2x - 1)
Practice
Explore Exploring Factors of Perfect-Square Trinomials
Exercises 1–2
Example 1 Factoring a 2x 2 + 2abx + b 2 and a 2x 2 - 2abx + b 2
Exercises 3–6
Example 2 Factoring a 2x 2 - b 2 = 0
Exercises 7–10, 25, 28, 30
Example 3 Solving Equations with Special Factors
Exercises 11–18, 26, 29
Example 4 Solving Equation Models with Special Factors
Exercises 19–24, 27
© Houghton Mifflin Harcourt Publishing Company
= x(4x + 1)
2
Concepts and Skills
2 = 2y(y 2 + 4z 2)⎡⎣y 2 - (2z) ⎤⎦
= 2y(y 2 +4z 2)(y + 2z)(y- 2z)
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Depth of Knowledge (D.O.K.)
COMMON CORE
Mathematical Practices
1–10
1 Recall of Information
MP.2 Reasoning
11–18
2 Skills/Concepts
MP.2 Reasoning
19–24
2 Skills/Concepts
MP.4 Modeling
25
3 Strategic Thinking
MP.3 Logic
26
2 Skills/Concepts
MP.2 Reasoning
27
2 Skills/Concepts
MP.4 Modeling
3 Strategic Thinking
MP.3 Logic
28–30
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Solve the following equations with special factors.
QUESTIONING STRATEGIES
11. 25x 2 + 20x + 4 = 0
If the values of a and c in a trinomial in the form ax 2 + bx + c are both perfect squares, is the trinomial always a perfect-square trinomial? Explain. Provide an example to defend your answer. No; if the first and last terms are both perfect squares, the coefficient of the middle term must be two times the square root of the first coefficient, a, times the square root of the constant, c. Possible example: 16x 2 + 90x + 36; both 16 and 36 are perfect squares, but the middle term would need to be 48x for this to be a perfect-square trinomial.
12. x 3 - 10x 2 + 25x = 0
25x + 20x + 4 = 0
x 3 - 10x 2 + 25x = 0
2
x(x 2 - 10x + 25) = 0
5 2 x 2 + 2(5⋅2)x + 2 2 = 0
x ⎡⎣x 2 - 2(1⋅5)x+ 5 2⎤⎦ = 0
(5x + 2)(5x + 2) = 0 (5x + 2) = 0
x(x - 5)(x - 5) = 0
2
5x + 2 = 0
x(x - 5) = 0
5x = -2 x = -2 5
x = 0 or x - 5 = 0
2
_
x=5
13. 4x 4 + 8x 3 + 4x 2 = 0
4x + 8x + 4x = 0 4
3
14. 4x 2 - 8x + 4 = 0
4x 2 - 8x + 4 = 0
2
4x 2(x2 + 2x + 1) = 0
4(x 2 - 2x + 1) = 0
4x 2 ⎡⎣x 2 + 2(1⋅1)x + 1 2⎤⎦ = 0
4 ⎡⎣x 2 - 2(1⋅1) + 1 2⎤⎦ = 0
4x (x + 1)(x + 1) = 0
4 (x - 1 ) (x - 1 ) = 0
2
4 x (x + 1 ) = 0
4 (x - 1 ) = 0
2
2
2
x-1 =0
4x 2 = 0 or x + 1 = 0 x=0
x=1
x = -1
15. x 2 - 81 = 0
16. 2x 3 - 2x = 0
x 2 - 81 = 0
2x - 2x = 0 3
2 2x (x 2 - 1) = 0
x2 - 92 = 0
© Houghton Mifflin Harcourt Publishing Company
(x + 9 ) (x - 9 ) = 0
x + 9 = 0 or x - 9 = 0 x = -9 17. 16q - 81 = 0
x+1=0
x =0
x = -1 x = 1
x-1=0
(4q + 9)(4q - 9) = 0 or
4
4p 4 - 25p 2 + 16p 2 = 0
(4q) 2 - (9) 2 = 0
4q = -9 9 q = -_ 4
0r
18. 4p - 25p = -16p 2
2
4q + 9 = 0
2x = 0
4
16q - 81 = 0
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2
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2x(x + 1)(x - 1) = 0
4p 4 - 9p = 0 2
p 2(4p 2 - 9) = 0
2 p 2⎡⎣(2p) - 3 2⎤⎦ = 0
4q - 9 = 0
p 2(2p + 3)(2p - 3) = 0
4q = 9 9 q=_ 4
p 2 = 0 or p=0
341
2p + 3 = 0
or 2p - 3 = 0
2p = -3 3 p=2
_
2p = 3 3 p= 2
_
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Jivesh is analyzing the flight of a few of his model rockets that he assembled with various equations. In each equation, h is the height of the rocket in centimeters, and the rocket was fired from the ground at time t = 0 , where t is measured in seconds.
MODELING Students may wish to check their answers to factoring problems by using algebra tiles. Remind them that the algebra tiles that model a perfect-square trinomial will form a square.
19. For Jivesh’s Model A rocket, he uses the equation h = -490t 2 + 1120t. When is the height of the Model A rocket 640 centimeters? h = -490t2 + 1120t
640 = -490t 2 + 1120t 490t 2 + 1120t 2 + 640 = 0 49t 2 + 112t + 64 = 0
(7t - 8)(7t - 8) = 0 (7t - 8) 2 = 0
7t = 8 8 t=_ 7 8 ≈ 1.14 seconds, the rocket will have reached a height of 640 centimeters. After t = _ 7 20. Jivesh also has a more powerful Model B rocket. For this rocket, he uses the equation. When is the height of the Model B rocket 810 centimeters? h = -490t2 + 1260t 810 = -490t 2 + 1260t 490t 2 - 1260t + 810 = 0 49t 2 - 126t + 81 = 0
(7t - 9)(7t - 9) = 0 (7t - 9) 2 = 0
© Houghton Mifflin Harcourt Publishing Company
7t = 9 9 t=_ 7 9 After t = _ ≈ 1.29 seconds, the rocket will have reached a height of 810 centimeters. 7 21. Jivesh brought his Model B rocket on a camping trip near the top of a mountain. He wants to model how it descends down the mountain. Here, he uses the equation h = 490t 2. When has the rocket descended 1000 centimeters? h = -490t2 -1000 = -490t 2 490t 2 -1000 = 0 49t 2 - 100 = 0
(7t + 10)(7t - 10) = ±0
7t = ±10 10 t = ±_ 7 10 ≈ 1.43 seconds, the rocket will have descended a distance of 1000 After t = _ 7 centimeters. The negative value of t cannot be used in this context. Module 8
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AVOID COMMON ERRORS When checking the answer for a perfect-square 2 trinomial, such as (x + 6) , students might write the product as x 2 + 6 2 = x 2 + 36. Encourage students to 2 write (x + 6) as (x + 6)(x + 6) and then use the FOIL method.
22. Geometry Claire is cutting a square out of a bigger square for an art project. She cuts out a square with an area of 9 cm 2. The leftover area is 16 cm 2. What is the length of one of the sides of the bigger square? The area of a square is A = / 2 where / is the length of one of its sides.
Solve for l. I 2 - 9 = 16 I 2 - 9 - 16 = 0 I 2 - 25 = 0
(I + 5)(I - 5) = 0
I = ±5
QUESTIONING STRATEGIES
Disregard the negative value of l because squares cannot have negative lengths.
Explain why the product of binomials (r + s)(r - s) is not a trinomial. When you multiply these binomials together by FOIL, the inner and outer products cancel each other out, so the result is a difference of two squares: (r + s)(r - s) = r 2 - s 2.
The square has side lengths of 5 centimeters. 23. The height of a diver during a dive can be modeled by h = -16t 2 , where h is height in feet relative to the diving platform and t is time in seconds. Find the time it takes for the diver to reach the water if the platform is 49 feet high.
-49 = -16t 2 16t 2 - 49 = 0
(4t + 7)(4t - 7) = 0 4t = ±7
7 t = ±_ 4 7 After t = _ = 1.75 seconds, the diver will have reached the water. The negative 4 value of t cannot be used in this context.
24. Physics Consider a particular baseball player at bat. The height of the ball at time t can be modeled by h = -16t 2 + v 0t + h 0. Here, v 0 is the initial upward velocity of the ball, and h 0 is the height at which the ball is hit. If a ball is 4 feet off the ground when it is hit with a negligible upward velocity close to 0 feet per second, when will the ball hit the ground?
h = -16t 2 + v 0t + h 0 0 = -16t 2 + 0t + 4 0 = 16t 2 - 4
0 = 4 (4 t 2 - 1 )
0 = 4(2t + 1)(2t - 1) 1 t = ±_ 2
1 The answer should be positive, so t = _ second. 2
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25. Explain the Error Jeremy factored 144x 2 - 100 as follows:
QUESTIONING STRATEGIES
144x 2 - 100 = (12x + 10)(12x - 10) = 2(6x + 5)(6x - 5)
Is 4x 2 + 20x - 25 a perfect-square trinomial? How can you tell? No; the last term is negative. This means that the operation sign in one binomial factor is + and in the other is -. In a perfect-square trinomial, the constant term is always positive.
What was his error? Correct his work.
He factored out a 2 from both (12x + 10) and (12x + 10), which is the same as factoring out a 4 from the whole expression. It should be 4(6x + 5)(6x - 5).
144x 2 - 100 = (12x + 10)(12x - 10) = 2(6x + 5) ⋅ 2(6x - 5) = 4(6x + 5)(6x - 5)
Is 16x 4 - 1 the difference of two squares? Explain. Yes; 16x 4 - 1 is equivalent 2 2 to (4x ) - 1 2.
26. Which of the following are solutions to the equation x 5 - 2x 3 + x = 0? Select all that apply. a. x = -1 b. x = 2 c.
x=1
d. x = 0.5 e. x = 0
x 5 - 2x 3 + x = 0
x(x 4 - 2x 2 + 1) = 0
x(x 2 - 1)(x 2 - 1) = 0
AVOID COMMON ERRORS
x(x 2 - 1) = 0 2
2 x⎡⎣(x + 1)(x - 1)⎤⎦ = 0
When the constant term of the difference of two squares is even, such as in x 2 - 16, some students may divide by 2 rather than taking the square root, producing an answer of (x + 8)(x - 8). Emphasize to students that they are taking the square root of each term, so the correct factorization is (x + 4)(x - 4).
2 2 x (x + 1 ) (x - 1 ) = 0
x=0
or
x+1=0
or
x-1=0
x = -1
x=1
The letters a, c, and e correspond to solutions of the equation x 5 - 2x 3 + x = 0. H.O.T. Focus on Higher Order Thinking 4x
a. Find and completely factor the expression for the area of the frame.
The area of the frame only is the difference of
4x
2y
the large square number and the small square number.
2y
16x 2 - 4y 2 = 4(4x 2 - y 2)
= 4(2x + y)(2x - y)
© Houghton Mifflin Harcourt Publishing Company
27. Multi-Step An artist framed a picture. The picture is a square with a side length of 2y. It is surrounded by a square frame with a side length of 4x.
The area of the frame is 4(2x + y)(2x - y).
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INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 Some students may speculate that because a
Solve for the width of the frame, which is the width of the frame minus the width of the square picture, divided by two: 4x - 2y _____ = 2x - y 2
Set the area of the picture, 4y 2, to 25 and solve.
difference of squares in the form a 2 - b 2 can be factored as (a + b)(a - b), a difference of cubes can be factored similarly. Inform students that there are different patterns for the factors of the sum and difference of cubes: a 3 + b 3 = (a + b)(a 2 - ab + b 2) a 3 - b 3 = (a - b)(a 2 + ab + b 2).
25 = 4y 2 0 = 4y 2 - 25
0 = (2y + 5)(2y - 5)
±5 = 2y 5 ±_ =y 2
In the context of lengths and area, you can only take the positive y as a possible 5 solution, so y = _ . 2
Have them multiply each pair of factors together to confirm these patterns.
5 into this Next, solve for x. The area of the frame is 4(4x 2 - y 2) = 11. Substitute y = _ 2
equation and solve for x. 4(4x 2 - y 2) = 11 5 2⎤ 4 ⎡⎣4x 2 - _ = 11 2 ⎦ 25 ⎤ ⎡ __ 2 4 ⎣4x - 4 ⎦ = 11
() ( )
QUESTIONING STRATEGIES
(16x 2 - 25) = 11
16x 2 - 25 - 11 = 0 16x 2 - 36 = 0
(4x + 6)(4x - 6) = 0 © Houghton Mifflin Harcourt Publishing Company
When factoring a trinomial, do you think it is easier to first list all the factor pairs or to first check whether it fits the pattern for a perfect-square trinomial? Explain your reasoning. Possible answer: It is easier to check for the pattern, because once you see that a and c are perfect squares, you have only to check one product of factors to confirm whether it gives the correct value of b for a perfect-square trinomial. If it does, you can write the answer, and if it does not, you can list more factor pairs to determine how to factor the trinomial.
4x = ±6 3 x = ±_ 2
3 Use the positive value of x = _ . Solve for the width of the frame using the values for x and y. 2
5 5 1 3 2x - y = 2 ⋅ _ -_ =3-_ =_ 2 2 2 2 1 The frame is _ inch wide. 2
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28. Critical Thinking Sinea thinks that the fully factored form of the expression x 4 - 1 is (x 2 - 1)(x 2 + 1). Is she correct? Explain.
JOURNAL
No, the binomial (x 2 - 1) is also a difference of squares, so the fully factored form is
Have students write a journal entry explaining how a perfect-square trinomial and a difference of squares are alike and how they are different.
(x + 1)(x -1)(x 2 + 1).
29. Persevere in Problem Solving Samantha has the equation x 3 + 2x 2 + x = x 3 - x. Explain how she can find the solutions of the equation. Then solve the equation.
She can factor each side of the equation, and then subtract the right side from both sides so one side of the equation is 0. She can then simplify the equation and set the factors equal to 0 to solve for x. x 3 + 2x 2 + x = x 3 - x
x(x 2 + 2x + 1) = x(x 2 - 1)
x(x + 1)(x + 1) = x(x + 1)(x - 1)
x(x + 1)(x + 1) - x(x + 1)(x - 1) = 0 x(x + 1)⎡⎣(x + 1) - (x - 1)⎤⎦ = 0 x(x + 1)[2] = 0 2x(x + 1) = 0
2x = 0
or
x=0
x+1=0 x = -1
The solutions are 0 or −1. 30. Communicate Mathematical Ideas Explain how to fully factor the expression 4 x - 2x 2y 2 + y 4.
Use the rule for perfect square trinomials to factor x 4 - 2x 2y 2 + y 4 as (x 2 - y 2). The
© Houghton Mifflin Harcourt Publishing Company
binomial (x 2 - y 2) is a difference of squares that can be factored as (x + y)(x - y), so
the fully factored form is (x + y) 2(x - y) 2.
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Lesson Performance Task
AVOID COMMON ERRORS Students may see that the first two fountain equations are the difference of squares and surmise that they each have a square shape. Remind students that a square’s area is the product of two equal lengths. The difference of squares does not factor into the product of two identical expressions.
A designer is planning to place a fountain in the lobby of an art museum. Four artists have each designed a fountain to fit the space. Some have designed rectangular fountains and the others designed square fountains. Given a quadratic equation representing the area of the fountain and the actual area of the fountain, find the dimensions of each fountain.
Artist
Artemis
Beatrice
Geoffrey
Daniel
Area equation
A A = 9x 2 - 25
A B = 4x 2 - 25
A G = 25x 2 + 80x + 64
A D = 81x 2 + 198x + 121
Fountain area
39 square feet
28x - 76 square feet
160x square feet
198x + 242 square feet
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Karen Huntt/Corbis
Artemis:
Beatrice:
A A = 9x 2 - 25
A B = 4x 2 - 25
39 = 9x - 25
28x - 76 = 4x 2 - 25
= (3x + 5)(3x - 5) 2
= (2x + 5)(2x - 5)
0 = 9x 2 - 64
0 = 4x 2 - 28x + 49
0 = (3x - 8)(3x + 8) 8 8 or x = - x= 3 3 8 Disregard x = - _ . 3
_
0 = (2x - 7)
2
_
2x - 7 = 0 2x = 7 7 x= 2 A B = 2x + 52x - 5 7 7 = 2 +5 2 -5 2 2 = (7 + 5)(7 - 5)
_
A A = (3x + 5)(3x - 5)
( (_) )( (_) )
= 3
8 8 +5 3 -5 3 3
( (_) )( (_) )
= (8 + 5)(8 - 5)
= (13)(3)
= (12)(2)
The fountain by Artemis is a rectangle measuring 13 feet by 3 feet.
The fountain by Beatrice is a rectangle measuring 12 feet by 2 feet.
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Have students consider a square fountain with side length x feet. Ask them to explore how the area of the fountain changes if they make each side d feet longer. Have them draw a diagram to show the area of each fountain and explain how the diagram corresponds to the algebraic expressions representing the area.
Students should find that while the original fountain has area x 2, the enlarged fountain has area x 2 + 2dx + d 2. A diagram may show that the area of the enlarged fountain can be divided into a square with area x 2, a square with area d 2, and two rectangles with area dx.
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Geoffrey:
Daniel:
A G = 25x 2 + 80x + 64 = (5x + 8)
= (9x + 11)
2
160x = 25x + 80x + 64 2
198x + 242 = 81x + 198x + 121
0 = 25x - 80x + 64
0 = 81x 2 - 121
0 = (9x - 11)(9x + 11)
2
5x - 8 = 0
9x - 11 = 0
5x = 8 8 x= 5
9x = -11 x = - 11 9
_ _
A G = (5x + 8)
2
A D = (9x + 11)
( (_ ) )
8 +8 = 5 5 2 = (8 + 8) = 16
9x + 11 = 0
9x = 11 or 11 x= 9 Disregard x = - 11 . 9
_
How many possible values of x are there for the fountains designed by Beatrice and Geoffrey? Explain. For each of these fountains, there is only one solution for x, because when the appropriate equation is written in standard form, the expression set equal to zero is a perfect-square trinomial. Both factors are the same, so setting each factor equal to zero yields the same solution.
2
2
2
0 = (5x - 8)
QUESTIONING STRATEGIES
A D = 81x 2 + 198x + 121
2
( (_) )
2
= 9
_
11 + 11 9
2
= (11 + 11)
2
2
= 22 2
The fountain by Geoffrey is a square with
The fountain by Daniel is a square with
16-foot sides.
22-foot sides.
© Houghton Mifflin Harcourt Publishing Company
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Scoring Rubric 2 points: Student correctly solves the problem and explains his/her reasoning. 1 point: Student shows good understanding of the problem but does not fully solve or explain his/her reasoning. 0 points: Student does not demonstrate understanding of the problem.
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MODULE
8
MODULE
STUDY GUIDE REVIEW
8
Using Factors to Solve Quadratic Equations
Study Guide Review
Essential Question: How can you use factoring a quadratic equation to solve real-world problems?
ASSESSMENT AND INTERVENTION
KEY EXAMPLE
Key Vocabulary
difference of two squares (diferencia de dos cuadrados) perfect-square trinomial (trinomio cuadrado perfecto)
(Lesson 8.1)
Factor x 2 - 2x - 8. Find the factor pair of -8 whose sum is -2. The factor pair is -4 and 2.
x - 2x - 8 = (x - 4)(x + 2)
Factors of -8
Assign or customize module reviews.
Sum of Factors
-1 and 8
2
7
1 and -8
-7
-2 and 4
2
2 and -4
MODULE PERFORMANCE TASK
KEY EXAMPLE
Mathematical Practices: MP.1, MP.2, MP.4, MP.6 A-SSE.A.2, A-SSE.B.3a, F-IF.B.4
• Can the crate be turned on its side? Students should assume that the crate has a top and bottom and that turning the crate might damage its contents. • Can I decide on a crate that has less depth than width and can therefore be turned? Students should assume that the crate’s depth is greater than its width.
Find the factor pairs of 4 and 3 that result in a sum of 8.
Factors of 4
Factors of 3
The factor pairs are 2 and 2 and 1 and 3.
1 and 4
1 and 3
2
1 and 4
3 and 1
(2x + 1)(2x + 3) = 0
2 and 2
1 and 3
4x + 8x + 3 = 0
SUPPORTING STUDENT REASONING
2x + 1 = 0 © Houghton Mifflin Harcourt Publishing Company
• How much clearance should I allow? The crate is sliding and will not need much “wiggle room”. However, the crate should not scrape against the archway.
(Lesson 8.2)
Solve 4x 2 + 8x + 3 = 0.
COMMON CORE
Students should begin this problem by first finding the width of the arched doorway. Then they should focus on what additional information they will need. Here is some of the information they may ask for.
-2
1 x = -_ 2
and
and
(2)(3) + (2)(1) = 8
3 x = -_ 2 (Lesson 8.3)
Solve 16x - 25 = 0. 2
4 ⋅x -5 =0 2
(1)(1) + (4)(3) = 13
2x + 3 = 0
KEY EXAMPLE
2
Outer Product + Inner Product (1)(3) + (4)(1) = 7
2
(4x + 5)(4x - 5) = 0
Rewrite in the form a 2 x 2 - b 2.
Rewrite in the form (ax + b)(ax - b). Set factors equal to 0 using Zero Product Property.
4x + 5 = 0
and 4x - 5 = 0 5 5 and x=_ x = -_ 4 4
Module 8
349
Study Guide Review
SCAFFOLDING SUPPORT
IN2_MNLESE389830_U4M08MC 349
• Students may initially find that the arch is 4 feet wide. Point out that 4 feet is only half the width of the base of the arch, since the width includes 4 feet on both sides of the origin. • Some students may need help in understanding that the taller the crate is, the narrower the base needs to be. • Students who need more structure will benefit from being given a specific height, such as 5 feet.
349
Module 8
• Challenge students by asking them in how many different ways a crate having dimensions of 5 ft × 6 ft × 7 ft could fit through the archway, assuming it is packed well enough that it can be turned on any of its faces.
4/12/14 9:08 AM
EXERCISES Solve each equation. (Lessons 8.1, 8.2, 8.3) 1.
SAMPLE SOLUTION
x - 81 = 0
2.
2x - 8x - 10 = 0 −1 and 5
The width of the arched doorway at its base (y = 0) is found by solving -x 2 + 16 = 0. Factor to solve: (4 + x)(4 - x) = 0.
x 2 + 7x + 12 = 0
4.
x 2 - 14x = -49
Therefore, x = 4, -4. The width of the doorway at its base is 4 -(-4) = 8, or 8 ft.
2
-9 and 9
3.
−4 and −3
5.
7.
16 - 4x 2 = 0 −2 and 2
2
7
6.
If the height of the crate is 5 ft, its maximum width can be found by solving 5 = -x 2 + 16. Students may wish to sketch the parabola and draw a box of height 5 to better visualize the situation:
6x 2 + 5x + 1 = 0 1 1 and -_ -_ 2 3
The area of a rectangular pool is (x 2 + 17x + 72) square meters. There is a 3-meter-wide concrete walkway around the pool. Write expressions to represent the dimensions of the outside border of the walkway. (Lesson 8.1)
y 14
(x + 12) meters and (x + 11) meters
12 10
ESSENTIAL QUESTION
(Lesson 15.2)
Solve each equation. (Lessons 18.1, 18.2, 18.3)
each expression. (Lesson 18.1, 18.2, 18.3) Identify each expression as a perfect-square trinomial, a difference of squares, or neither. Factor
Height (ft)
Polynomials 18.1–18.3 Multiplying and Dividing
MODULE PERFORMANCE TASK
Fitting Through the Arch
The Ship-Shape Shipping Company just unloaded several crates outside the arch ranging in height from 2 feet to 6 feet. Choose a particular crate height. Then, find the maximum width the crate could have and still fit through the arched doorway. Start by listing in the space below how you will tackle this problem. Then use your own paper to complete the task. Be sure to write down all your data and assumptions. Then use graphs, tables, or algebra to explain how you reached your conclusion.
Module 8
350
6 4
© Houghton Mifflin Harcourt Publishing Company
The Ship-Shape Shipping Company ships items in rectangular crates. At one shipping destination, each crate must be able to fit through an arched doorway. The shape of this arched doorway can be modeled by the quadratic equation y = -x 2 + 16, where x is the distance in feet from the center of the arch and y is the height of the arch. Find the width of the archway at its base.
8
2 x -3
-2
―
-1
0 Width (ft)
1
2
3
―
So, x = ± √ 11 and the maximum width is 2 √11 or about 6.6 feet.
Study Guide Review
DISCUSSION OPPORTUNITIES
IN2_MNLESE389830_U4M08MC 350
4/12/14 9:08 AM
• What happens if the crate is light enough to be tilted up on one of its edges? Would a larger crate then be able to fit through the arch? (Students will not be able to model this mathematically, but if they are interested, encourage them to explore the issue by manipulating various rectangular cutouts on a graph of the parabola.) Assessment Rubric 2 points: Student correctly solves the problem and explains his/her reasoning. 1 point: Student shows good understanding of the problem but does not fully solve or explain. 0 points: Student does not demonstrate understanding of the problem.
Study Guide Review 350
Ready to Go On?
Ready to Go On?
ASSESS MASTERY
8.1–8.3 Using Factors to Solve Quadratic Equations
Use the assessment on this page to determine if students have mastered the concepts and standards covered in this module.
• Online Homework • Hints and Help • Extra Practice
Identify each expression as a perfect-square trinomial, a difference of squares, or neither. Factor each expression. (Lessons 8.1, 8.2, 8.3) 4p 2 + 12p + 9
1.
ASSESSMENT AND INTERVENTION
2.
perfect-square trinomial; (2p + 3)
a 2 - 9a - 36
neither; (a - 12)(a + 3)
2
Solve each equation. (Lessons 8.1, 8.2, 8.3) x 2 - 4x - 21= 0
3.
4.
10 10 and __ -__ 7 7
-3 and 7
Access Ready to Go On? assessment online, and receive instant scoring, feedback, and customized intervention or enrichment.
5x 2 - 33x - 14 = 0
5.
ESSENTIAL QUESTION
(Lesson 15.2)
Solve each equation. (Lessons 18.1, 18.2, 18.3)
each expression. (Lesson 18.1, 18.2, 18.3) Identify each expression as a perfect-square trinomial, a difference of squares, or neither. Factor
Polynomials 18.1–18.3 Multiplying and Dividing
• Reading Strategies • Success for English Learners • Challenge Worksheets Assessment Resources
© Houghton Mifflin Harcourt Publishing Company
Differentiated Instruction Resources
x 2 + 16x + 64 = 0
-8
A golfer hits a ball from a starting elevation of 4 feet with a vertical velocity of 70 feet per second down to a green with an elevation of -5 feet. The number of seconds t it takes the ball to hit the green can be represented by the equation -16t 2 + 70t + 4 = -5. How long does it take the ball to land on the green? (Lesson 8.2)
7.
• Reteach Worksheets
6.
2 and 7 -_ 5
ADDITIONAL RESOURCES Response to Intervention Resources
49x 2 - 100 = 0
4.5 seconds
ESSENTIAL QUESTION How can you use factoring to solve quadratic equations in standard form?
8.
Possible Answer: You can factor the right side of the quadratic equation. Then set each linear factor equal to 0. Solve each linear equation. These are the solutions of the original quadratic equation.
• Leveled Module Quizzes
Module 8
COMMON CORE IN2_MNLESE389830_U4M08MC 351
351
Module 8
Study Guide Review
351
Common Core Standards
4/12/14 9:08 AM
Content Standards
Mathematical Practices
A-SSE.B.3, A-SSE.A.2
MP.2
3
A-SSE.B.3, A-SSE.A.2, A-REI.B.4
MP.2
8.3
4
A-SSE.B.3, A-SSE.A.2, A-REI.B.4
MP.2
8.2
5
A-SSE.B.3, A-SSE.A.2, A-REI.B.4
MP.2
8.1
6
A-SSE.B.3, A-SSE.A.2, A-REI.B.4
MP.2
8.2
7
A-SSE.B.3, A-SSE.A.2, A-REI.B.4
MP.1
Lesson
Items
8.2
1–2
8.1
MODULE MODULE 8 MIXED REVIEW
MIXED REVIEW
Assessment Readiness
Assessment Readiness
1. Consider the equation 5x(2x + 1) - 3(2x + 1) = 0. Choose True or False for each statement about the equation. A. It is equivalent to (5x - 3)(2x + 1) = 0. 1. B. A solution of the equation is x = _ 2 3 C. A zero of the equation is _. 5
8
True True True
ASSESSMENT AND INTERVENTION
False False False
2. Factor to solve each equation. Does the equation have a solution of x = 2? Select Yes or No for each. Yes No A. 4x 2 - 16 = 0 B. x 2 - 4x + 4 = 0 C. 4x 2 + 16x + 16 = 0
Yes Yes
No No
Assign ready-made or customized practice tests to prepare students for high-stakes tests.
4x 2 + 7x - 15
is 4x - 5. Explain how you can check his answer 3. Larry thinks the quotient of _________ x+3 using multiplication. Then, check his answer. Is Larry correct? Possible answer: You can check the quotient by multiplying the quotient by the divisor, or (4x - 5)(x + 3). The product is 4x 2 + 7x - 15, so Larry is correct.
ADDITIONAL RESOURCES
4. Marcello is replacing a rectangular sliding glass door with dimensions of (x + 7) and (x + 3) feet. The area of the glass door is 45 square feet. What are the length and width of the door? Explain how you got your answer.
Assessment Resources • Leveled Module Quizzes: Modified, B
I found the area of the glass door by multiplying the length by the width, which is x 2 + 10x + 21. I set this product, which represents the area in terms of x, equal to
AVOID COMMON ERRORS
45. Then, I set the equation equal to 0 and solved for x and got x = -12 and x = 2. 2 + 7, or 9 feet, and the width is 2 + 3, or 5 feet.
Module 8
COMMON CORE
Item 2 If students have trouble factoring the expressions quickly, point out that an alternate solution method would be to substitute the value into each equation and directly check whether it is a solution.
© Houghton Mifflin Harcourt Publishing Company
A dimension of the door cannot be negative, so x = 2. The length of the door is
Study Guide Review
352
Common Core Standards
IN2_MNLESE389830_U4M08MC 352
4/12/14 9:08 AM
Content Standards Mathematical Practices
Lesson
Items
7.3
1*
A-APR.B.3
MP.1
8.3
2
A-SSE.A.2
MP.2
5.2
3*
A-SSE.A.1
MP.3
5.2, 8.1
4*
A-SSE.A.1, A-REI.B.4
MP.4
* Item integrates mixed review concepts from previous modules or a previous course.
Study Guide Review 352
MODULE
9
Using Square Roots to Solve Quadratic Equations
Essential Question: How can you use quadratic
equations to solve real-world problems?
ESSENTIAL QUESTION:
LESSON 9.2
LESSON 9.3
Using the Quadratic Formula to Solve Equations LESSON 9.4
Choosing a Method for Solving Quadratic Equations
This version is for PROFESSIONAL DEVELOPMENT Algebra 1 and Geometry only VIDEO
Author Juli Dixon models successful teaching practices in an actual high-school classroom.
LESSON 9.5
Solving Nonlinear Systems © Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Getty Images/PhotoDisc
Professional Development Video
my.hrw.com
REAL WORLD VIDEO The designers of a fireworks display need to make precise timing calculations. An explosion too soon or too late could spell disaster!
MODULE PERFORMANCE TASK PREVIEW
Fireworks Display As with any other projectile, the relationship between the time since a firework was launched and its height is quadratic. Fireworks must be carefully timed in order to ignite at the most impressive height. In this task, you will figure out how you can use math to launch fireworks that are safe and achieve the maximum possible effect.
Module 9
DIGITAL TEACHER EDITION IN2_MNLESE389830_U4M09MO 353
Access a full suite of teaching resources when and where you need them: • Access content online or offline • Customize lessons to share with your class • Communicate with your students in real-time • View student grades and data instantly to target your instruction where it is needed most
353
Module 9
LESSON 9.1
Solving Equations by Taking Square Roots
Solving Equations by Completing the Square
Answer: You can choose a method depending on the form of the quadratic equation that is written from the real-world problem. If you can’t solve it using square roots, factoring, or completing the square, then you can use the quadratic formula to solve the equation.
Professional Development
9
MODULE
Using Square Roots to Solve Quadratic Equations
353
PERSONAL MATH TRAINER Assessment and Intervention Assign automatically graded homework, quizzes, tests, and intervention activities. Prepare your students with updated, Common Core-aligned practice tests.
4/12/14 2:56 PM
Are YOU Ready?
Are You Ready?
Complete these exercises to review skills you will need for this chapter.
Exponents
ASSESS READINESS
1 _
Simplify 25 2 .
Example 1
―
1 _
1 power is A number raised to the _ 2 equal to the square root of the number.
25 2 = √25 = 5 Simplify. 1.
1 _
100 2
2. 10
1 _
50 2
5 √― 2
3.
(_3681 )
Use the assessment on this page to determine if students need strategic or intensive intervention for the module’s prerequisite skills.
• Online Homework • Hints and Help • Extra Practice
1 _ 2
2 __ 3
ASSESSMENT AND INTERVENTION
Algebraic Expressions
()
2
b when b = 18. Evaluate _ 2 2 b _ 2 2 18 = 9 2 = 81 _ 2
Example 2
()
() ( )
Substitute 18 for b and evaluate the expression.
2
b for the given value of b. Evaluate _ 2 4. b = 24 5. b = -10 144 25
Example 3
Factor x 2 + 14x + 49. x 2 + 14x + 49
2
(x + 7) (x + 7) (x + 7)
2
2 1
6.
b=3
x2 + 14x + 49 is a perfect square. Rewrite in the form a 2 + 2ab + b 2. Rewrite in the form (a + b) (a + b).
Factor each perfect square trinomial. 7. x 2 - 12x + 36 (x - 6) 2
8.
x 2 + 22x + 121 (x + 11)2
4x 2 + 12x + 9 (2x + 3)2
10.
16x 2 - 40x + 25 (4x - 5)2
9.
Module 9
IN2_MNLESE389830_U4M09MO 354
Tier 1 Lesson Intervention Worksheets Reteach 9.1 Reteach 9.2 Reteach 9.3 Reteach 9.4 Reteach 9.5
TIER 1, TIER 2, TIER 3 SKILLS
Personal Math Trainer will automatically create a standards-based, personalized intervention assignment for your students, targeting each student’s individual needs!
9 _ 4
© Houghton Mifflin Harcourt Publishing Company
x + 2(x)(7) + 7 2
3
ADDITIONAL RESOURCES See the table below for a full list of intervention resources available for this module. Response to Intervention Resources also includes: • Tier 2 Skill Pre-Tests for each Module • Tier 2 Skill Post-Tests for each skill
354
Response to Intervention Tier 2 Strategic Intervention Skills Intervention Worksheets 5 Algebraic Expressions 10 Exponents 16 Multi-Step Equations
Differentiated Instruction
4/12/14 2:55 PM
Tier 3 Intensive Intervention Worksheets available online Building Block Skills 19, 22, 23, 24, 27, 29, 30, 40, 59, 69, 76, 81, 98, 100
Challenge worksheets Extend the Math Lesson Activities in TE
Module 9 354
LESSON
9.1
Name
Solving Equations by Taking Square Roots
Class
9.1
Date
Solving Equations by Taking Square Roots
Essential Question: How can you solve quadratic equations using square roots? Resource Locker
Common Core Math Standards COMMON CORE
Recall that the square root of a non-negative number a is the real number b such that b 2 = a. Since 4 2 = 16 2 and (-4) = 16, the square roots of 16 are 4 and -4. Thus, every positive real number has two square roots, _ _ one positive and one negative. The positive square root is given by √ a and the negative square root by -√ a . _ √ These can be combined as ± a .
A-REI.B.4b
Solve quadratic equations by... taking square roots …
Mathematical Practices COMMON CORE
Exploring Square Roots
Explore
The student is expected to:
Properties of Radicals
MP.8 Patterns
Property
Language Objective
Symbols
Explain to a partner how to solve ax = c by taking square roots, and how to tell if there are one, two, or no real solutions.
_
_
_ _ √ 36 = √ 9 ⋅ 4 _ _ = √9 ⋅ √4
For a ≥ 0 and b ≥ 0, √ab = √ a ⋅ √b .
Product Property of Radicals
2
Example
_
=3⋅2 =6
_
_
√
ENGAGE
√
View the Engage section online. Discuss the photo, explaining that the motion of a clock’s pendulum is accelerated by gravity. The pendulum swings back and forth with periods that vary with the square of its length. Then preview the Lesson Performance Task.
© Houghton Mifflin Harcourt Publishing Company
PREVIEW: LESSON PERFORMANCE TASK
_
√ 16 _ = -_ √ 100 4 = -_ 10 = -0.4
Essential Question: How can you solve quadratic equations using square roots? Isolate the square root on one side of the equation by adding, subtracting, multiplying, or dividing. Then take the square root. Finally, solve for both the positive and negative square root.
_
_ 16 -√ 0.16 = - _ 100
√a a _. For a ≥ 0 and b > 0, _ = _ b √b
Quotient Property of Radicals
Find each square root. _
A
±√49 = + 7
C
_ ±√12
= ±
√
=±
2
⋅
and - 7
__
_
√
4 ⋅ 3 = ±
_
√
_
±√25 = + 5
B 4
_
16 √_ 16 = ±_ D ±√_ 9 √9
_
⋅
√
and - 5
_
3
3
4 = ±_ 3
_
E
_ ±√0.27
Module 9
√
__
_
_
√
√
_
√
√
9 ⋅3 9 ⋅ √_ 3 ⋅ 27 27 3 3 _ = ± _ = ± __ = ± __ = ± _ = ±_ 100 √ 100 10 10 10
be ges must EDIT--Chan DO NOT Key=NL-A;CA-A Correction
Lesson 1
355
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Date Class
9.1
IN2_MNLESE389830_U4M09L1 355
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√
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_
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_
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√_b = √b
_
√ 16 _ = - √_ 100
= -0.4
__
and b > 0,
4 _ = - 10
- 7 _ + 7 and ±√ 49 = _ ±√ 12
Watch for the hardcover student edition page numbers for this lesson.
_
9 ⋅ 4_ = √_ √4 = √9 ⋅ =3⋅2 =6
Radic
Quotient Radicals
HARDCOVER PAGES 355366
2
⋅
4
⋅
√
_ + 5 ±√ 25 =
16
3
√
16 _ _ 9
4 _ =± 3
_
√
and - 5
_
_
_=± ±√ 9 √
_
_
√
⋅ 3 = ±
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_
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_ 3 3 ⋅_ _ __ 9 ⋅ √3 _ ±_ __ = _ 9 ⋅3 10 27 _=± 10 27 _=± = ± √_ _ 10 100 ± _ 100 ±√ 0.27 =
√
√
√
√
Lesson 1
355 Module 9
9L1 355 30_U4M0
ESE3898
IN2_MNL
355
Lesson 9.1
09/04/14
6:51 PM
07/04/14 6:48 PM
Reflect _
EXPLORE
_
1.
Discussion Explain why √6 2 and √(-6_) 2 have the same value. 2 6 2 = (-6) = 36 and the symbol √ represents the positive square root of 36, or 6.
2.
Discussion Explain why a must be non-negative when you find √a . By the definition of a square root of a, a = b 2 for some b, so a is equal to the square
Exploring Square Roots
_
of a real number, and the square of a real number is always non-negative. 3.
INTEGRATE TECHNOLOGY
Does 0 have any square roots? Why or why not? Yes, because 0 2 = 0, 0 is its own (and only) square root.
Explain 1
Students have the option of completing the Engage activity either in the book or online.
Solving ax 2 - c = 0 by Using Square Roots
Solving a quadratic equation by using square roots may involve either finding square roots of perfect squares or finding square roots of numbers that are not perfect squares. In the latter case, the solution is irrational and can be approximated.
Solve the equation. Give the answer in radical form, and then use a calculator to approximate the solution to two decimal places, if necessary. Use a graphing calculator to graph the related function and compare the roots of the equation to the zeros of the related function.
Example 1
INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 Make certain that students understand the
―
4x 2 - 5 = 2 Solve the equation for x. 4x 2 - 5 = 2
―
Original equation
4x - 5 + 5 = 2 + 5 2
Add 5 to both sides.
4x 2 = 7
Simplify.
4x 2 = _ 7 _ 4 4 x 2 = 1.75
Divide both sides by 4.
x = ±√1.75
Definition of a square root
x ≈ ±1.32
Use a calculator to approximate the square roots.
The approximate solutions of the equation are x ≈ 1.32 and x ≈ -1.32. Use a graphing calculator to graph the related function, ƒ(x) = 4x 2 - 7, and find the zeros of the function. The graph intersects the x-axis at approximately (1.32, 0) and (-1.32, 0). So, the roots of the equation are the zeros of the related function.
Module 9
EXPLAIN 1
© Houghton Mifflin Harcourt Publishing Company
Simplify.
_
―
difference between √a , - √a , and the phrase take the square roots of a. The first is the positive square root of a. The second is the negative square root of a. The third is both the positive and negative square roots of a and can be represented by ± √a .
Solving ax 2 - c = 0 by Using Square Roots AVOID COMMON ERRORS Remind students to first add or subtract to isolate the x 2 term before taking square roots.
Lesson 1
356
PROFESSIONAL DEVELOPMENT IN2_MNLESE389830_U4M09L1.indd 356
4/2/14 1:21 AM
Math Background
_
√ For a positive number a, we write the positive square root of a as _ a . Likewise, for a positive number b, we write the positive square root of b as √ b . By the
_
definition of square root, (√ a )2 = a and (√ b ) = b. Therefore, _
_
ab = (√a )2 (√b ) = _
2
_
(√_a · √b )
2
2
by the Power of a Product Property. This _
_
_
means, again by the definition of square root, that √ a · √ b = √ab . A similar argument can be made for the Quotient Property of Radicals.
Solving Equations by Taking Square Roots
356
QUESTIONING STRATEGIES
2x 2 - 8 = 0 Solve the equation for x.
When does a quadratic equation of the form x 2 = c have one solution? When c = 0 the equation has one solution, because √0 = 0.
2x 2 - 8 = 0
―
Original equation
2x 2 - 8 + 8 = 0 + 8 2x 2 = 8
Simplify.
2x 2 = _ 8 _
2
Add 8 to both sides.
Divide both sides by 2 .
2
x2 = 4
Simplify. _
√
x=±
4
x=± 2
Definition of a square root Evaluate the square roots.
The solutions of the equation are x = 2 and x = -2 . Use a graphing calculator to graph the related function, ƒ(x) = 2x 2 - 8, and find the zeros of the function. The graph intersects the x-axis at
( 2 , 0 ) and ( -2 , 0 ). So,
the
© Houghton Mifflin Harcourt Publishing Company
the
roots
of the equation are
zeros
of the related function.
Your Turn
Solve the equation. Give the answer in radical form, and then use a calculator to approximate the solution to two decimal places, if necessary. Use a graphing calculator to graph the related function to check your answer. 4.
3x 2 + 6 = 33
5.
3x + 6 = 33
3x = 27
5x 2 = 11
2
x =9
x 2 = 2.2
2
x=
_ ±√ 9
_
x = ±√ 2.2
x = ±3
x ≈ ±1.48
The solutions are 3 and -3. The graph intersects the x-axis at (3, 0) and (-3, 0).
Module 9
5x 2 - 9 = 2
5x 2 - 9 = 2
2
The approximate solutions are 1.48 and -1.48. The graph intersects the x-axis at approximately (1.48, 0) and (-1.48, 0).
357
Lesson 1
COLLABORATIVE LEARNING IN2_MNLESE389830_U4M09L1.indd 357
Peer-to-Peer Activity Have students work in pairs to write a real-world problem and solution that can be solved by using square roots. Some common themes in real-world problems are finding the area of a rectangle, square, or triangle. Encourage students to draw a model to go along with the problem. Have each pair share their problem with the class. Discuss any extraneous solutions.
357
Lesson 9.1
4/2/14 1:21 AM
Solving a (x + b) = c by Using Square Roots 2
Explain 2
EXPLAIN 2
Solving a quadratic equation may involve isolating the squared part of a quadratic expression on one side of the equation first. Example 2
Solving a(x + b) = c by Using Square Roots 2
Solve the equation. Give the answer in radical form, and then use a calculator to approximate the solution to two decimal places, if necessary.
(x + 5)2 = 36 (x + 5)2 = 36 x+5=
Original equation
_ ±√ 36
x + 5 = ±6
Simplify the square root.
x = ±6 - 5 x = -6 - 5
INTEGRATE MATHEMATICAL PRACTICES Focus on Reasoning MP.2 When students come across equations with
Take the square root of both sides.
or
x = -11
Subtract 5 from both sides. x=6-5
Solve for both cases.
no real solution, explain that although it is possible to create a square root of a negative number, it is not a real number. No real number squared is equal to a negative number since a negative times a negative is a positive, and a positive times a positive is a positive.
x=1
The solutions are x = -11 and x = 1.
2 3 (x - 5) = 18
3 (x - 5) = 18 2
(x - 5) = 6
Original equation
2
Divide both sides by 3 .
_
√ √6 + x = ±_ x=√ 6 +5
5
x ≈ 7.45
or x ≈ 2.55
6 x-5=± _
Take the square roots of both sides. Add 5 to both sides. _
or x = -√6 + 5
Solve for both cases.
Reflect
6.
Find the solution(s), if any, of 2(x - 3) = -32. Explain your reasoning. 2 2(x - 3) = -32 2
(x - 3)2 = -16
The square of a number is never negative, so this equation does not have any real-number solutions.
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The approximate solutions are x ≈ 7.45 and x ≈ 2.55 .
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Introduce the idea that squaring and finding square roots are inverse operations. This means that squaring undoes the effect of taking the square root, and taking the square root undoes the effect of squaring. Other examples of inverse operations are addition and subtraction, and multiplication and division. Stress to students that before taking a square root of a squared term, they must isolate the squared term on one side of the equation.
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358
Your Turn
QUESTIONING STRATEGIES
Solve the equation. Give the answer in radical form, and then use a calculator to approximate the solution to two decimal places, if necessary.
When are there two solutions to an equation in the form ax 2 - c = 0? When you apply the definition of a square root, you get two square c roots, a positive and a negative. For this to be true, __ a must be greater than or equal to 0.
7.
4(x + 10) = 24 2
(x + 10)2 = 6
8.
― ―
x + 10 = ± √6 x = ± √6 - 10
―
x = - √6 -10
or
x ≈ -12.45
EXPLAIN 3
Explain 3
―
x - 9 = ± √64 x - 9 = ±8
x = ±8 + 9
―
x = √6 - 10
x = -8 + 9
x ≈ -7.55
x=1
or x = 8 + 9 x = 17
Solving Equation Models by Using Square Roots
Real-world situations can sometimes be analyzed by solving a quadratic equation using square roots.
Solving Equation Models by Using Square Roots
Example 3
AVOID COMMON ERRORS
Solve the problem.
A contractor is building a fenced-in playground at a daycare. The playground will be rectangular with its width equal to half its length. The total area will be 5000 square feet. Determine how many feet of fencing the contractor will use. First, find the dimensions. Let A = 5000, ℓ = x, and w = __12 x.
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Nancy Hixson/Shutterstock
Some students may incorrectly interpret the solution to a real-world problem and include an extraneous solution. Remind students to use the context of the word problem to interpret the solution. Often, the negative solution will not make sense, so it should be disregarded.
(x - 9)2 = 64
A = ℓw 1x 5000 = x · _ 2 1 x2 5000 = _ 2 10,000 = x 2 _
±√10,000 = x ±100 = x
Take the square root of both sides. Evaluate the square root.
Since the width of a rectangle cannot be negative, the length of the playground is 100 feet. The width is half the length, or 50 feet. Find the amount of fencing. P = 2ℓ + 2w = 2(100) + 2(50) = 200 + 100
Multiply.
= 300
Add.
So, the contractor will use 300 feet of fencing.
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Connect Vocabulary Preview this lesson with English learners to review new terms such as square root, real number, non-negative, and property. Students need to review and revisit this key vocabulary in order to understand and acquire its use. After explaining the vocabulary, have students try to rephrase it in their own words to a partner. Listen for students who may need support.
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B
A person standing on a second-floor balcony drops keys to a friend standing below the balcony. The keys are dropped from a height of 10 feet. The height in feet of the keys as they fall is given by the function h(t) = -16t 2 + 10, where t is the time in seconds since the keys were dropped. The friend catches the keys at a height of 4 feet. Find the elapsed time before the keys are caught.
INTEGRATE MATHEMATICAL PRACTICES Focus on Modeling MP.4 Make sure that students understand the
Let h(t) = 4 . Substitute the value into the equation and solve for t. h(t) = -16t 2 + 10
Original equation
4 = - 16t + 10
Substitute.
2
4 - 10 = -16t 2 + 10 - 10
-6 = -16t
2
-6 = _ -16t 2 _
-16
-16
±√
Subtract 10 from both sides. Simplify.
= t2
Simplify.
0.375
=t
Take the square root of both sides.
± 0.61 ≈ t
QUESTIONING STRATEGIES
Divide both sides by -16.
0.375
__
connections between the real-world situation and the equations they create. They should always make a note of what the variables and solutions represent.
When solving a quadratic equation using square roots, what must you do before you can take the square root of both sides of the equation? isolate the square of the variable on one side of the equation
Use a calculator to approximate the square roots.
Since time cannot be negative, the elapsed time before the keys are caught is approximately 0.61 second(s). Your Turn
9.
A zookeeper is buying fencing to enclose a pen at the zoo. The pen is an isosceles right triangle. There is already a fence along the hypotenuse, which borders a path. The area of the pen will be 4500 square feet. The zookeeper can buy the fencing in whole feet only. How many feet of fencing should he buy?
_ _ _
――
± √9000 = x
±94.87 ≈ x Since the length of the base and the height of a triangle cannot be negative, the length of the base and height of the pen is approximately 94.87. Since the zookeeper can only buy fencing in whole feet, use 95 as the length. There are two sides of the same length that need fencing, so the zookeeper should buy 95 + 95, or 190, feet of fencing. Module 9
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1 bh. The length of the base and the height of The equation for the area of a triangle is A = __ 2 an isosceles triangle are equal, so let b = h = x. Let A = 4500. Substitute the values into the equation for the area and solve for x. 1 A = bh 2 1 4500 = x · x 2 1 4500 = x 2 2 9000 = x 2
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Elaborate
ELABORATE
10. How many real solutions does x 2 = -25 have? Explain. x 2 = -25 has no real solutions, because the square of a real number is always
non-negative.
QUESTIONING STRATEGIES
11. Suppose the function h(t) = -16t 2 + 20 models the height in feet of an object after t seconds. If the final height is given as 2 feet, explain why there is only one reasonable solution for the time it takes the object to fall. The equation 2 = -16t 2 + 20 has two real solutions for t. One solution is positive, and the
When solving equations of the form (x + b) 2 = c, is it possible to get two positive or two negative answers? If so, give an example of each. It is possible to get two positive answers; for 2 example, solving (x - 8) = 25 gives solutions of 13 and 3. It is also possible to get two negative 2 answers. For example, solving (x + 5) = 4 gives solutions of -7 and -3.
other is negative. Since time cannot be negative, the negative solution is rejected. So, the positive solution is the only reasonable solution for the time it takes the object to fall. 12. Essential Question Check-In What steps would you take to solve 6x 2 - 54 = 42? Isolate the squared term by adding 54 to both sides, and then divide both sides by 6.
Finally, take the square root of both sides and solve for both the negative and positive square root.
SUMMARIZE THE LESSON
Evaluate: Homework and Practice
Copy and complete the graphic organizer to review the lesson. In each box, write an example of a quadratic equation with the given number of solutions. Solve each equation.
No real solutions: x2 + 81 = 0 no real solutions One solution: x2 + 81 = 81 x=0 Two solutions: x2 = 81 x = ±9
1.
_
±√0.0081
――― √― 81 = ± _ 10,000 √――― 9 = ± _ ±
© Houghton Mifflin Harcourt Publishing Company
Solving Quadratic Equations by Using Square Roots When the Equation Has ...
100 = ± 0.09
√ √― 8 ±_ √― 25 √―― 4·2 =± _ ―5 · √―2 √4 _ =± 5 2 √― 2 _
8 ± _ 25
=±
3.
_
±√96
―― ― ― ―
± √16 ∙ 6 = ± √16 ∙ √6 = ±4 √6
5
Solve each equation. Give the answer in radical form, and then use a calculator to approximate the solution to two decimal places, if necessary. Use a graphing calculator to graph the related function to check your answer. 4.
5x 2 - 21 = 39
5.
5x = 60
0.1x 2 - 1.2 = 8.8
0.1x 2 = 10
2
x = 12
x 2 = 100 x = ± √100
x = ± √― 12 2
x ≈ ±3.46
x = ±10
Module 9
――
The graph intersects the x-axis at (10, 0) and (-10, 0).
The graph intersects the x-axis at approximately (3.46, 0) and (-3.46, 0).
Exercise
Lesson 9.1
_
2.
81 √_ 10,000
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• Online Homework • Hints and Help • Extra Practice
Use the Product Property of Radicals, the Quotient Property of Radicals, or both to simplify each expression.
Lesson 1
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Depth of Knowledge (D.O.K.)
COMMON CORE
Mathematical Practices
1–3
1 Recall of Information
MP.2 Reasoning
4–9
1 Recall of Information
MP.5 Using Tools
10–15
1 Recall of Information
MP.2 Reasoning
16–21
2 Skills/Concepts
MP.4 Modeling
22
2 Skills/Concepts
MP.2 Reasoning
23
3 Strategic Thinking
MP.3 Logic
24
3 Strategic Thinking
MP.4 Modeling
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6.
6x 2 - 21 = 33
1 x 2 = -20 6-_ 3 1 - x 2 = -26 3 x 2 = 78
7.
6x = 54
_
2
x2 = 9
―
x = ± √9
x = ±3
The graph intersects the x-axis at approximately (8.83, 0) and (-8.83, 0).
5 - 2x 2 = 3
7x 2 + 10 = 18
9.
- 2x 2 = - 8 x2 = 4
7x 2 = 8 8 x2 = _ 7 x=±
―
x = ± √4 x = ±2
―
x = ± √78
x ≈ ±8.83
The graph intersects the x-axis at (3, 0) and (-3, 0). 8.
EVALUATE
ASSIGNMENT GUIDE
―
√_87
x ≈ ±1.07
The graph intersects the x-axis at (2, 0) and (-2, 0).
The graph intersects the x-axis at approximately (1.07, 0) and (-1.07, 0).
Solve each equation. Give the answer in radical form, and then use a calculator to approximate the solution to two decimal places, if necessary. 11. (x + 15) = 81
2 10. 5(x - 9) = 15
2
(x - 9) = 3 2
― ―
x - 9 = ± √3 x = ± √3 + 9
―
x = - √3 + 9
or
x ≈ 7.27
x + 15 = ±9
x = ±9 - 15
―
x = √3 + 9
x = -9 - 15 or x = 9 - 15
x = -24
x ≈ 10.73
12. 3(x + 1) = 27
_
x - 40 = ±6
x = ±3 - 1
x = -6 + 40 or x = 6 + 40
x=2
14. (x - 12) = 54
― ―― x - 12 = ± √― 9 ∙ √― 6 x - 12 = ±3 √― 6 x = ±3 √― 6 + 12 x = -3 √― 6 + 12 or x = 3 √― 6 + 12 2
x - 12 = ± √54 x - 12 = ± √9 ∙ 6
x ≈ 4.65
x = ±6 + 40
or x = 3 - 1
x = 34
x = 46
15. (x + 5.4) = 1.75 2
――
x + 5.4 = ± √1.75
―
7 x + 5.4 = ± _ 4
―7 x + 5.4 = ± ___ ―4
― √― 7 ___
Example 1 Solving ax 2 - c = 0 by Using Square Roots
Exercises 4–9, 22
Example 2 2 Solving a(x + b) = c by Using Square Roots
Exercises 10–15, 22
Example 3 Solving Equation Models by Using Square Roots
Exercises 16–21, 23–26
A graphing calculator can be used to check irrational solutions to an equation. Write the equation solved for 0, for example, in the form 2 2 0 = a(x + b) - c. Graph y = a(x + b) - c. Compare the graph’s zeros, or x–intercepts, to the approximate decimal equivalent of the solutions.
x = ± 2 - 5.4 √― 7 or x = ___ - 5.4 2 2
―7 x = - ___ - 5.4
x ≈ -6.72
Exercise
Exercises 1–3
√7
x ≈ -4.08
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Explore Exploring Square Roots
x + 5.4 = ± ___ 2
x ≈ 19.35
Module 9
Practice
INTEGRATE TECHNOLOGY © Houghton Mifflin Harcourt Publishing Company
x + 1 = ±3 x = -3 - 1
―
x - 40 = ± √36
x + 1 = ±√ 9
x = -4
x = -6
2 (x - 40)2 = 24 13. _ 3 (x - 40)2 = 36
2
(x + 1) 2 = 9
―
x + 15 = ± √81
Concepts and Skills
Depth of Knowledge (D.O.K.)
COMMON CORE
Mathematical Practices
25
3 Strategic Thinking
MP.3 Logic
26
3 Strategic Thinking
MP.3 Logic
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16. The area on a wall covered by a rectangular poster is 320 square inches. The length of the poster is 1.25 times longer than the width of the poster. What are the dimensions of the poster?
QUESTIONING STRATEGIES
A = ℓw
How do you take the square root of a fraction? Take the square root of the numerator and denominator separately. For ― ― 4 __4 = ± __2 . example, √__ = √ 9 9 3
320 = 1.25x ⋅ x
320 = 1.25x 2
Since the width of a rectangle cannot be negative, the width of the poster is 16 inches. So, the length of the poster is 1.25 ⋅ 16, or 20, inches.
256 = x 2
± √―― 256 = x ±16 = x
What is wrong with solving the equation x 2 = 3x by dividing both sides by x? When you divide by x, you lose the solution x = 0. If x = 0, division of both sides by x is undefined.
17. A circle is graphed with its center on the origin. The area of the circle is 144 square units. What are the coordinates of the x-intercepts of the graph? Round to the nearest tenth.
A = πr 2 144 = πr
2
144 ___ = r2 π
VISUAL CUES
― 144 ___
Since the radius of a circle cannot be negative, the radius of the graphed circle is approximately 6.8 units. The x-intercepts are 6.8 units from the origin. So, they are located at (-6.8, 0) and (6.8, 0).
± π = r
Encourage students to circle on their papers what is to be isolated in each step of solving the equations in this lesson.
12 ±___ ― =r
√π
±6.8 ≈ r
18. The equation d = 16t 2 gives the distance d in feet that a golf ball falls in t seconds. How many seconds will it take a golf ball to drop to the ground from a height of 4 feet? 64 feet?
1. Circle the entire x 2 –term and think of how it can be isolated.
4 = 16t 2
2. Circle only x 2 and think of how it can be isolated.
64 = 16t 2
_1 = t 2
4 = t2
―1 = t ± √_ 4
―
4
© Houghton Mifflin Harcourt Publishing Company
3. Circle the x in x 2, showing that the last operation to be undone is the square.
± √4 = t ±2 = t
1 =t ±_ 2
Since time cannot be negative, it will take 1 second to drop to the ground a golf ball __ 2 from a height of 4 feet.
Since time cannot be negative, it will take a golf ball 2 seconds to drop to the ground from a height of 64 feet.
19. Entertainment For a scene in a movie, a sack of money is dropped from the roof of a 600-foot skyscraper. The height of the sack above the ground in feet is given by h = −16t 2 + 600, where t is the time in seconds. How long will it take the sack to reach the ground? Round to the nearest tenth of a second.
0 = 16t 2 + 600 -600 = 16t 2 37.5 = t 2
――
Since time cannot be negative, the solution is t = 6.12. So, it will take the sack approximately 6.12 seconds to reach the ground.
± √37.5 = t ±6.12 ≈ t
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20. A lot for sale is shaped like a trapezoid. The bases of the trapezoid represent the widths of the front and back yards. The width of the back yard is twice the width of the front yard. The distance from the front yard to the backyard, or the height of the trapezoid, is equal to the width of the back yard. Find the width of the front and back yards, given that the area is 6000 square feet. Round to the nearest foot. 1 A=_ h b + b 2). 2 ( 1
1( ) 6000 = _ 2x (x + 2x) 2 _ 6000 = 1(2x)(3x) 2
6000 = 3x 2
AVOID COMMON ERRORS When solving equations in which the number isolated on one side is the opposite of a perfect square, students may take the square root as if the number were positive. Remind students that you cannot take the square root of the opposite of a perfect square, such as -25. It has no square roots; 5 and -5 are the square roots of 25, not -25.
Since the length of the base of a trapezoid cannot be negative, the width of the front yard is approximately 45 feet. So, the width of the back yard is 2 · 45, or 90, feet.
2000 = x 2
――
± √2000 = x
±45 ≈ x
21. To study how high a ball bounces, students drop the ball from various heights. The function h(t) = −16t 2 + h 0 gives the height (in feet) of the ball at time t measured in seconds since the ball was dropped from a height of h 0. If the ball is dropped from a height of 8 feet, find the elapsed time until the ball hits the floor. Round to the nearest tenth.
0 = -16t 2 + 8
-8 = -16t 2
_1 = t 2
―12 = t ± √_ 2
Since time cannot be negative, the ball will hit the ground after approximately 0.7 second.
±0.7 ≈ t
_
a. 2x 2 − 2 = 16
C
2√33 =±_ 3
2 b. 2(x - 2) = 16
A
x = ±3
B
x = 2 ± 2√2
D
x = −8 and x = 0
x2 = 9
(x - 2) = 8
© Houghton Mifflin Harcourt Publishing Company • Image Credits:©MarcelClemens/Shutterstock
22. Match each equation with its solutions.
2
c.
3x 2 + 4 = 48 44 x 2 = __ 3
d. 3(x + 4) = 48 2
(x + 4) = 16
_
2
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JOURNAL
H.O.T. Focus on Higher Order Thinking
23. Explain the Error Trent and Lisa solve the same equation, but they disagree on the solution of the equation. Their work is shown. Which solution is correct? Explain.
Have students explain how to solve an equation of the form ax 2 - c = 0 by using square roots.
Trent:
Lisa:
5x 2 + 1000 = -125
5x 2 + 1000 = -125
5x 2 = -1125
5x 2 = -1125
x 2 = -225
x 2 = -225
x=
_ ±√-225
no real solutions
x = ±15 Lisa’s solution is correct. By the definition of a square root of a x 2 = a, so the square of a real number is always non-negative. So, a must be non-negative. In the equation x 2 = -225 , a is negative. So, there are no real solutions of the equation. 24. Multi-Step Construction workers are installing a rectangular, in-ground pool. To start, they dig a rectangular hole in the ground where the pool will be. The area of the ground that they will be digging up is 252 square feet. The length of the pool is twice the width of the pool. a. What are the dimensions of the pool? Round to the nearest tenth.
A = ℓw
252 = 2x ∙ x
126 = x 2
±11.2 ≈ x
The width is approximately 11.2 feet. So, the length is approximately 2 ∙ 11.2, or 22.4, feet. The pool is approximately 11.2 feet by 22.4 feet.
© Houghton Mifflin Harcourt Publishing Company
b. Once the pool is installed, the workers will build a fence, that encloses a rectangular region, around the perimeter of it. The fence will be 10 feet from the edges of the pool, except at the corners. How many feet of fencing will the workers need?
11.2 + 20 = 31.2
22.4 + 20 = 42.4
31.2 + 31.2 + 42.4 + 42.4 = 147. 2
So, the workers will need approximately 147.2 feet of fencing. 25. Communicate Mathematical Ideas Explain why the quadratic equation x 2 + b = 0 where b > 0, has no real solutions, but the quadratic equation x 2 - b = 0 where b > 0, has two real solutions.
When b > 0, the solution of x 2 + b = 0 is the square root of a negative number, which is not real; but the solution of x 2 - b = 0 is the square root of a positive number, which has 2 possible values. 26. Justify Reasoning For the equation x 2 = a, describe the values of a that will result in two real solutions, one real solution, and no real solution. Explain your reasoning.
― ―
―
Two real solutions: If a is positive, x will have the values √a and - √a . One real solution: If a is equal to 0, x will have the value √0 = 0. If a is negative, there are no real values of x because no real number squared is negative. Module 9
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Lesson Performance Task
CONNECT VOCABULARY Some students may not understand some of the terms used in this Lesson Performance Task. Have a volunteer draw a pendulum clock on the board and locate the pendulum and clock face. Explain that the period of a pendulum’s motion is how long it takes to swing back (the tick) and forward (the tock), in seconds. The longer the pendulum, the longer the period.
You have been asked to create a pendulum clock for your classroom. The clock will be placed on one wall of the classroom and go the entire height of the wall. Choose how large you want the face and hands on your clock to be and provide measurements for the body of the clock. The pendulum will start halfway between the center of the clock face and its bottom edge and will initially end 1 foot above the floor. Calculate the period of the pendulum using the formula L = 9.78t 2, where L is the length of the pendulum in inches and t is the length of the period in seconds. Now, adjust the length of your pendulum so the number of cycles in 1 minute or 60 seconds is an integer value. How long is your pendulum and how many periods equal one minute?
Answers will vary. If the ceiling height is 10 feet and the clock face has a radius
AVOID COMMON ERRORS
of 2, the pendulum goes from about 7 feet off the ground to 1 foot off the
Some students may fail to convert the length of the pendulum to inches. Have students begin writing their solutions by rewriting the formula, then stating what each variable stands for.
ground, so it has a length of 6 feet or 72 inches. L = 9.78t 2
L _ =t 9.78 t=
2
――
L √_ 9.78
When L = 72,
――
72 √_ 9.78
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Design Pics/Don Hammond/Corbis
t=
≈ 2.71
There are 60 seconds in one minute. Divide 60 by 2.71 to find the number of cycles in 1 minute. 60 ≈ 22.14 2.71 The number of periods is not an integer value, so try a period of 2.5 seconds since 60 divided by 2.5 is 24, which is an integer value.
_
L = 9.78(2.5)
2
= 9.78(6.25) = 61.125
So, make the pendulum 61.125 inches or 5 feet 1
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Have students look at the first eleven perfect squares, and ask them to look for patterns. Some patterns students may find include the following.
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• The difference between two consecutive perfect squares goes up by consecutive odd integers (0, 1, 4, 9, 16, …, the difference is 1, 3, 5, 6, …). • All perfect squares end in 0, 1, 4, 5, 6, or 9. • The ones digits of the first eleven squares are symmetrical (0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100).
Scoring Rubric 2 points: Student correctly solves the problem and explains his/her reasoning. 1 point: Student shows good understanding of the problem but does not fully solve or explain his/her reasoning. 0 points: Student does not demonstrate understanding of the problem.
Solving Equations by Taking Square Roots
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LESSON
9.2
Name
Solving Equations by Completing the Square
Class
9.2
Date
Solving Equations by Completing the Square
Essential Question: How can you use completing the square to solve a quadratic equation? Resource Locker
Common Core Math Standards The student is expected to: COMMON CORE
You can use algebra tiles to model a perfect square trinomial.
Complete the square ... to reveal the maximum or minimum value of the function .... Also A-SSE.A.2, A-SSE.B.3a, A-REI.B.4b, A-REI.B.4a, F-IF.C.8a
Key
Mathematical Practices COMMON CORE
Modeling Completing the Square
Explore
A-SSE.B.3b
=1 = -1
MP.4 Modeling
=x
= -x
= x2
= -x2
Language Objective Explain to a partner how to complete the square of a quadratic equation in the form ax 2 + bx + c = 0.
ENGAGE
You can transform one side of the equation into a perfect-square trinomial, and then solve the equation by taking the square root of both sides.
PREVIEW: LESSON PERFORMANCE TASK View the Engage section online. Discuss the photo, and explain that the spacing between the tiles depends on the design the tiles make when installed. Then preview the Lesson Performance Task.
The algebra tiles shown represent the expression x 2 + 6x. The expression does not have a constant term, which would be represented with unit tiles. Create a square diagram of algebra tiles by adding the correct number of unit tiles to form a square.
How many unit tiles were added to the expression? 9
Write the trinomial represented by the algebra tiles for the complete square.
×
1 x2 + 6 x + 9 © Houghton Mifflin Harcourt Publishing Company
Essential Question: How can you use completing the square to solve a quadratic equation?
It should be easily recognized that the trinomial 1 x 2 + 6 x + 9 is an example of
the special case (a + b 2) = a 2 + 2ab + b 2. Recall that trinomials of this form are called perfect-square trinomials. Since the trinomial is a perfect square, it can be factored into two identical binomials.
1 x2 + 6 x + 9 =
( 1 x+ 3 )
2
Refer to the algebra tiles in the diagram. What expression is represented by the tiles?
×
4 x2 + 8 x
Complete the square in Step G by filling the bottom right corner with unit tiles. How many unit tiles were added to the diagram? 4
Module 9
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Lesson 2
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Lesson 2
367 Module 9
9L2 367 30_U4M0
ESE3898
IN2_MNL
367
Lesson 9.2
09/04/14
6:52 PM
07/04/14 6:46 PM
Write the trinomial represented by the algebra tiles for the complete square.
EXPLORE
4 x2 + 8 x + 4
The trinomial is a square of a binomial. Use the algebra tiles to write the trinomial in factored form.
4 x2 + 8 x + 4 =
(2
x+ 2
)
Modeling Completing the Square
2
Reflect
1.
INTEGRATE TECHNOLOGY
Discussion When using algebra tiles to model the expression x 2 + 6x + c, the x-tiles are divided equally, with 3 tiles on the right and bottom sides of the x 2-tile. How does the number 3 relate to the total number of x-tiles? How does the number 3 relate to the total number of unit tiles that were added? The 3 x-tiles are half of the total number of x-tiles. The number of unit tiles added to the
Students have the option of completing the Explore activity either in the book or online.
grid, 9, is equivalent to 3 2, or the square of half of the total number of x-tiles. 2.
QUESTIONING STRATEGIES
In order to form a perfect square trinomial with the expression x 2 + 8x + c, how would the algebra tiles be arranged? How many unit tiles must be added? How is the number of unit tiles added related to the total number of x-tiles? 4 x-tiles would be arranged on the right and bottom sides of the x 2-tile. 16 unit tiles
What shape do the algebra tiles make in the Explore activity? They make a square.
would be added to complete the square. The number of unit tiles added to the grid, 16, is
Why do you think adding the unit tiles is called completing the square? Without the unit tiles, the diagram is not a complete square. Adding the tiles completes the square.
equivalent to 4 2, or the square of half of the total number of x-tiles.
Completing the Square When a = 1
Explain 1
Completing the square is a process of rewriting a quadratic expression as a perfect square trinomial so that it can be solved by taking square roots. In the Explore, the method for completing the square when a = 1 was modeled with algebra tiles. First, place half of the x-tiles along the right side of the x 2-tile and half underneath the tile. Then add unit tiles to fill in the rectangle started with the x 2 - and x-tiles. The number of unit tiles equals the square of the number of x-tiles on either side of the x 2-tile.
()
(
2
)
2
2
then be x 2 + bx + _b2 and its factored form will be x + __b2 . Example 1
Complete the square to form a perfect-square trinomial.
x 2 + 4x x + 4x 2
b=4
()
2
4 = 22 = 4 _ 2
x + 4x + 4 2
Module 9
Identify b.
() b to the expression. Add (_ 2) 2
b . Find _ 2
© Houghton Mifflin Harcourt Publishing Company
()
In other words, to complete the square for the expression x 2 + bx + c, add _b2 . The perfect-square trinomial will
Can you complete a square using unit tiles if you start with an x 2-tile and four x-tiles? If so, how? Yes; add four unit tiles.
EXPLAIN 1 Completing the Square when a = 1 QUESTIONING STRATEGIES Can the number added to complete the square ever be negative? Explain. No, it 2 b cannot be negative. Add __ to both sides of the 2 equation, and any number squared is positive.
2
368
()
Lesson 2
AVOID COMMON ERRORS
PROFESSIONAL DEVELOPMENT IN2_MNLESE389830_U4M09L2.indd 368
Math Background The relationships between sets of numbers extend to include complex numbers, such as 3 - 5i, pure imaginary numbers, such as 10i, and transcendental numbers, √― 2 such as π and 3 . A nonzero rational number with 0 as a denominator is undefined and thus not a real number. However, 0 divided by 0 is indeterminate because any number multiplied by 0 is 0, so the quotient can be any number.
4/2/14 1:26 AM
Some students may not pay attention to whether b is positive or negative, since c is positive, regardless of the sign of b. Have students change the sign of b in a problem and compare the factored forms of the two expressions.
Solving Equations by Completing the Square
368
EXPLAIN 2
x 2 - 8x x 2 - 8x b = -8
Solving x + bx + c = 0 by Completing the Square
( )
2
Identify b.
2
( )=
-8 _ = -4 2
() b to the expression. Add (_ 2)
2
2
b . Find _ 2
16
2
x 2 - 8x + 16
QUESTIONING STRATEGIES
Reflect
In what form should the equation be before 2 you find __b2 ? The equation should be in the form x 2 + bx = c.
()
3.
When b is negative, why is the result added to the expression still positive? When b is negative, the result added to the expression is still positive because the squared
numbers are always positive.
When will solutions be irrational? Give an example. Solutions will be irrational when 2 b __ c + 2 is not a perfect square. For example, the solutions of x 2 + 6x + 3 = 0 will be irrational, since , 6 2 -3 + __ = 6 which is not a perfect square. 2
Your Turn
()
4.
Complete the square: x 2 + 12x
12 = 6 = 36 (_ 2) 2
()
2
x 2 + 12x + 36
2
Explain 2
Solving x 2 + bx + c = 0 by Completing the Square
Completing the square can also be used to solve equations in the forms x 2 + bx + c = 0 or x 2 + bx = c.
Solve each equation by completing the square. Check the answers.
Example 2
x 2 - 4x = 3
© Houghton Mifflin Harcourt Publishing Company
x 2 - 4x = 3
() ( )
b = _ -4 = 4 to both sides. Add _ 2 2 Factor and simplify. 2
x 2 - 4x + 4 = 3 + 4
(x - 2) 2 = 7
_
x - 2 = ± √7
Take the square root of both sides.
_
x - 2 = √7
2
x-2=
or _
x = 2 + √7
_ -√7
x=2-
Write and solve two equations.
_ √7
Add 2 to both sides.
Check the answers. _
(2 + √ 7 )
2
_
- 4(2 + √7 )
_
_
_
(2 - √7 ) 2 - 4(2 - √7 )
_
_
_
= 4 + 4√7 + 7 - 8 - 4√7
= 4 - 4√7 + 7 - 8 + 4√7
= 4 + 7 - 8 + 4√7 - 4√7
= 4 + 7 - 8 - 4√7 + 4√7
=3
=3
_
_
_
_
_
_
2 + √7 and 2 - √7 are both solutions of the equation x 2 - 4x = 3. Module 9
Lesson 2
369
COLLABORATIVE LEARNING IN2_MNLESE389830_U4M09L2.indd 369
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Peer-to-Peer Activity
Have pairs of students create a content table to show the number of unit tiles needed to complete the square for trinomials in the form x 2 + bx + c and ax 2 + bx + c. Have students use algebra tiles, if necessary, to write trinomials and complete the table. Sample:
Expression x 2 - 4x
369
Lesson 9.2
Number of Unit Tiles Needed to Complete Square 4
Square Form of Expression (x - 2)2
B
x 2 + 16x = 36
INTEGRATE TECHNOLOGY
x 2 + 16x = 36
()
2
b = 64 Add _ 2
1 x 2 + 16 x + 64 = 36 + 64
( 1 x + 8 ) = 100 2
A graphing calculator can be used to check if an equation has one, two, or no real solutions. Students can graph the equation as y = x 2 + bx + c and see if the graph of the resulting parabola crosses the x-axis once, twice, or never.
to both sides.
Factor and simplify.
1 x + 8 = ± 10
Take the square root of both sides.
1 x + 8 = 10 or 1 x + 8 = - 10 x = -18
x= 2 Check the answers. x 2 + 16x = 36
(-18) 2 + 16 · -18 = 36
324 - 288 = 36 36 = 36 x = -18 is/is not a solution to the equation x 2 + 16x = 36. x 2 + 16x = 36
2 2 + 16 · 2 = 36
4 + 32 = 36
© Houghton Mifflin Harcourt Publishing Company
36 = 36 x = 2 is/is not a solution to the equation x 2 + 16x = 36 Your Turn
Solve each equation by completing the square. Check the answers. 5.
x 2 - 10x = 11
6.
x 2 - 10x + 25 = 36
x 2 + 6x + 9 = 11
(x + 3) = 11 2
(x - 5) = 36 2
― ― x + 3 = √11
_
x - 5 = ±√ 36 = ±6 x-5=6
x = 11
Module 9
x 2 + 6x = 2
or
x + 3 = ± √11
x - 5 = -6
x = -3 + √― 11
x = -1
370
or
―
x + 3 = - √11
―
x = -3 - √11
Lesson 2
DIFFERENTIATE INSTRUCTION IN2_MNLESE389830_U4M09L2.indd 370
Modeling
4/2/14 1:26 AM
When solving an equation such as x 2 + 4x = 5, students can use algebra tiles to complete the square on x 2 + 4x. If they need further visualization, they can use a large = and construct the entire equation using algebra tiles. When they add 4 to one side of the equation to complete the square, remind them to add 4 to the other side to maintain the equality. Have them form a square with nine unit tiles so that they can compare the side lengths of the two squares. On the left, the side length is x + 2 , while on the right, the side length is 3, so x + 2 = 3. Because length is not negative, x + 2 = -3 is discarded.
Solving Equations by Completing the Square
370
EXPLAIN 3
When a is a perfect square, completing the square is easier than2 in other cases. Recall that the number of unit tiles b needed is equal to the square of b divided by four times a, or __ . This is always the case when a is a perfect square. 4a
Solving ax + bx + c = 0 by Completing the Square when a is a Perfect Square 2
AVOID COMMON ERRORS
()
Example 3
(
__1 2
)
2
( ) -__38
2
Solve each equation by completing the square.
4x 2 - 8x = 21
4x 2 - 8x = 21 (-8)2 _ _ = 64 = 4 4∙4 16 4x 2 - 8x + 4 = 21 + 4
2
Students may have difficulty finding __b2 when b is a fraction. Remind them that dividing by 2 is the same 2 as multiplying by __12 . For example, if b = -__34 , __b2 -__34
Solving ax 2 + bx + c = 0 by Completing the Square When a Is a Perfect Square
Explain 3
()
b Find __ . 4a 2
b Add __ to both sides. 4a 2
(2x - 2)2 = 25
Factor and simplify.
(2x - 2) =
Take the square root of the both sides.
2
_ ±√25
2x - 2 = ±5
· , or . Let students practice becomes with a few examples and lead them to see that dividing a fraction by 2 is the same as doubling the denominator.
QUESTIONING STRATEGIES
Simplify.
2x - 2 = 5
or
2x - 2 = -5
2x = 7
or
2x = -3
7 x=_ 2
or
3 x = -_ 2
Write and solve 2 equations. Add to both sides. Divide both sides by 2.
9x 2 + 6x = 10 9x 2 + 6x = 10
What are the first ten perfect squares, starting with 1? 1, 4, 6, 16, 25, 36, 49, 64, 81, 100
2
© Houghton Mifflin Harcourt Publishing Company
6 36 _=_= 1 36 4∙ 9
b . Find _ 4a 2
b Add __ to both sides. 4a 2
9 x 2 + 6 x + 1 = 10 + 1
( 3 x+ 1 ) = 2
3 x+ 1 =± 3
x+
1
=
11
Factor and simplify.
―――
√
11
―――
√
11
3 x=- 1 +
Take the square root of the both sides.
3
or
x+
―――
√
11
―――
√
1 11 + x = __ 3
Module 9
1
=±
―――
√ -11
3 x=- 1 -
―――
√
11
―――
√
1 11 x = __
Write and solve two equations. Subtract 1 from both sides. Divide both sides by 3 .
3
371
Lesson 2
LANGUAGE SUPPORT IN2_MNLESE389830_U4M09L2.indd 371
Connect Vocabulary Help students to understand how the phrase completing the square is used in the visual representation of the concept (using algebra tiles) and how the phrase is used in the symbolic representation (the algebraic solution). Point out the geometric and algebraic uses of the word square and explain how they are related in this context.
371
Lesson 9.2
4/2/14 1:26 AM
Reflect
EXPLAIN 4
In order for the procedure used in this section to work, why does a have to be a perfect square? If a binomial is squared, by definition, the leading coefficient will always be a perfect square.
7.
Solving ax 2 + bx + c = 0 by Completing the Square when a is Not a Perfect Square
Your Turn
Solve each equation by completing the square. 16x 2 - 16x = 5
8.
(-16)2 _____ 4 ∙ 16
9.
256 = ___ =4 65
12 144 ____ = ___ =9 2
4 ∙ 16
16x 2 - 16x + 4 = 9
QUESTIONING STRATEGIES
16
4x 2 + 12x + 9 = 14
(4x - 2) = 9 2
4x - 2 = 3
Is the process of completing the square the same for quadratic equations with leading coefficients of 1 and those with leading coefficients not equal to 1? Explain. Mostly; the only difference is that the constant added to each side changes 2 b b2 from __ to ___ when a is a perfect square that is not 4a 2 equal to 1.
(2x + 3) = 14 2
4x - 2 = ± √― 9 = ±3 or
4x = 5
― ― 2x + 3 = √14
2x + 3 = ± √14
4x - 2 = -3
2x = -3 + √― 14
4x = -1
5 x=_ 4
Explain 4
4x 2 + 12x = 5
1 x = -_ 4
x=
or
―
2x + 3 = - √14
-3 + √― 14 ________ 2
―
2x = -3 - √14 x=
-3 - √― 14 ________
()
2
Solving ax 2 + bx + c = 0 by Completing the Square When a Is Not a Perfect Square
When the leading coefficient a is not a perfect square, the equation can be transformed by multiplying both sides by a value such that a becomes a perfect square. Example 4
Solve each equation by completing the square.
2x 2 - 6x = 5 © Houghton Mifflin Harcourt Publishing Company
Since the coefficient of x 2 is 2, which is not a perfect square, multiply both sides by a value so the coefficient will have a perfect square. In this case, use 2. 2x 2 - 6x = 5 2(2x 2 - 6x) = 2(5)
Multiply both sides by 2.
4x - 12x = 10 (-12)2 _ _ = 144 = 9 4∙4 16 2x 2 - 6x + 9 = 9 + 10 2
Simplify. b Find __ . 4a 2
b Add __ to both sides. 4a 2
(2x - 3)2 = 19
― ― or 2x - 3 = √19 ― 2x = 3 + √19 ― √ 19 3 + _ 2x - 3 = ± √19
x=
Module 9
IN2_MNLESE389830_U4M09L2.indd 372
2
Factor and simplify. Take the square root of the both sides.
― ― 2x = 3 - √19 ― √ 3 19 _
2x - 3 = - √19 x=
Write and solve 2 equations. Add to both sides. Divide both sides by 2.
2
372
Lesson 2
4/2/14 1:26 AM
Solving Equations by Completing the Square
372
B
AVOID COMMON ERRORS
3x 2 + 3x = 16
Emphasize that when multiplying both sides of an equation so that the coefficient of x 2 is a perfect b2 square, and then adding __ to both sides, b and a are 4a the new coefficients that result from the multiplication. Some students will mistakenly revert 12 144 original equation to find the coefficients. ____to the = ___ =9
3 (3x 2 + 2x) = 3 (16)
2
(2x + 3) = 14 2
―
2x = -3 + √14
―
x = ________ 2 -3 + √14
―
2x + 3 = - √14
Simplify.
6 36 _=_= 1 4∙ 9 36
4x + 12x + 9 = 14
or
.
Multiply both sides by 3 .
9 x 2 - 6 x = 48
16
― ― 2x + 3 = √14
3
3x 2 + 3x = 16
2
2x + 3 = ± √14
, which is not a perfect square, multiply both sides by a
value so the coefficient will have a perfect square. In this case, use
2
4 ∙ 16
3
Since the coefficient of x 2 is
b Find __ . 4a 2
b Add __ to both sides. 4a 2
9 x 2 + 6 x + 1 = 48 + 1
( 3 x+ 1 ) = 2
―
2x = -3 - √14
―
x = ________ 2 -3 - √14
49
3 x+ 1 =±
3 x+ 1 = 7
Factor and simplify.
―――
√
49 = ± 7
Take the square root of the both sides.
3 x+ 1 =- 7
Write and solve two equations.
3 x= 6
3 x=- 8
Subtract 1 from both sides.
x= 2
x=- 3
or
_8
Divide both sides by 3 .
Reflect
© Houghton Mifflin Harcourt Publishing Company
10. Consider the equation 2x 2 + 11x = 12. Why is 2 the best value by which to multiply both sides of the equation before completing the square? Since 2 is the least value by which to multiply so that a will be a perfect square, it is the
best value by which to multiply both sides of the equation.
Your Turn
Solve each equation by completing the square. 1 x 2 + 3x = 14 11. _ 2
x 2 + 6x = 28 x 2 + 6x + 9 = 37
6 36 ___ = __ =9 2
4∙1
―
x + 3 = √37
x = -3 + √― 37
Module 9
IN2_MNLESE389830_U4M09L2.indd 373
373
Lesson 9.2
(-8) ____ 2
4x 2 - 8x = 32
4
4∙ 4
4x 2 - 8x + 4 = 36
(x + 3) = 37 2
12. 2x 2 - 4x = 16
or
―
x + 3 = - √37
(2x - 2) 2 = 36
―
2x = -3 - √37 373
64 = __ =4 16
2x -2 = 6 x=4
or
2x -2= -6 x = -2 Lesson 2
4/2/14 1:26 AM
Explain 5
Modeling Completing the Square for Quadratic Equations
EXPLAIN 5
Completing the square can be useful when solving problems involving quadratic functions, especially if the function 2 cannot be factored. In these cases, complete the square to rewrite the function in vertex form: ƒ(x) = a(x - h) + k. Completing the square in this situation is similar to solving equations by completing the square, but instead of adding a term to both sides of the equation, you will both add and subtract it from the function’s rule.
Modeling Completing the Square for Quadratic Functions
Recall that the height of an object moving under the force of gravity, with no other forces acting on it, can be modeled by the quadratic function h = -16t 2 + vt + s, where t is the time in seconds, v is the initial vertical velocity, and s is the initial height in feet. Example 5
QUESTIONING STRATEGIES
Write a function in standard form for each model. Then, rewrite the equation in vertex form and solve the problem. Graph the function on a graphing calculator and find the x-intercepts and maximum value of the graph.
Describe in words what happens to an object thrown upwards. The object moves upward until it reaches its maximum height; then, it falls to the ground.
Sports A baseball is thrown by someone who is 5 feet tall. If the person throws the baseball at a velocity of 30 feet per second, what will be the maximum height of the baseball? How long will it take the baseball to hit the ground?
Describe in words how the vertex of the graph relating time and height of an object thrown upwards relates to the motion of the object. The value of h(t) at the vertex is the object’s greatest height. The value of t at the vertex is the amount of time it takes the object to reach its greatest height.
The function for this situation is h = -16t 2 + 30t + 5. Complete the square to find the vertex of the function’s graph. h = -16t 2 + 30t + 5 h = -1(16t 2 - 30t) + 5
Factor out -1.
( ) 15 - _ 225 + 5 h = -1((4t - _ 4) 16 ) 15 + _ 225 + 5 h = -(4t - _ 4) 16 15 305 _ _ h = -4(t - ) + 4 16
225 + 5 225 - _ h = -1 16t 2 - 30t + _ 16 16
Complete the square. Factor the perfect-square trinomial.
2
Distribute the -1.
2
Combine the last two terms.
15 ___ The coordinates of the vertex are ( __ , 305 , or about (3.75, 19.06). The maximum height will be about 4 16 ) 19 feet.
The graph of the function confirms the vertex at about (3.75, 19.06). The x-intercept at about 2.03 indicates that the boll will hit the ground after about 2 seconds.
Module 9
IN2_MNLESE389830_U4M09L2.indd 374
374
You know that the graph of a quadratic function is a parabola. Why isn’t the complete parabola drawn when graphing h(t)? Time and distance are both non-negative, and this restricts the graph to first-quadrant values.
© Houghton Mifflin Harcourt Publishing Company · Image Credits: ©Elaine Willcock/Shutterstock
2
AVOID COMMON ERRORS Some students may be less comfortable working with functions than with equations, or may not know the difference. As a result, they may try to solve a function as they would an equation, dividing both sides or adding to both sides. Explain that f(x) is read f of x, and that it is not treated the same way as a variable like y. Tell students that to complete the square given a function, they must add and subtract the same thing to the same side to preserve equality. So that the addition does not undo the subtraction, keep terms separate using parentheses.
Lesson 2
4/2/14 1:26 AM
Solving Equations by Completing the Square
374
ELABORATE
Sports A person kicks a soccer ball with an initial upward velocity of 16 feet per second. What is the maximum height of the soccer ball? When will the soccer ball hit the ground? An equation for this situation is h = -16t 2 + 16 t + 0 .
AVOID COMMON ERRORS
Complete the square to find the vertex of the function’s graph.
Make sure students understand that they must keep b2 to both sides of the equation balanced by adding __ 4a the equation, not to one side only.
h = -16t 2 + 16 t + 0 h = -16 h = -16
SUMMARIZE THE LESSON Copy and complete the graphic organizer shown below. In each box, write and solve an example of the given type of quadratic equation.
h = -16
h = -16
Solving Quadratic Equations by Completing the Square
h = -16
x2 + bx + c = 0 x2 - 3 x - 1 = 0 2 x2 - 3 x = 1 2 3 2 x - x + 9 = 1+ 9 2 16 16 x- 3 4
2
= 25 16
x- 3 = ±5 4 4 1 x = - or x = 2 2 ax2 + bx + c = 0
2(2x2 + 4x -20) = 0 4x2 + 8x - 40 = 0 4x2 + 8x = 40 4x2 + 8x + 4 = 40 + 4 (2x + 2)2 = 44 2x + 2 = ± 44
2x + 2 = ± 2 11
2x + 2 = 2 11 or 2x + 2 = -2 11 2x = -2 + 2 11 or 2x = - 2 -2 11 x = -1 + 11 or x = -1 - 11
375
Lesson 9.2
(
( ( (
2
)
1 t + 0
)
1 1 1 t2 - 1 t + _ - _ + 0 4 4
) ) 2
1 1 1 t-_ -_ +0 4 2
) )
Factor out -16 . Complete the square.
Factor the perfect-square trinomial.
2
1 16 1 t-_ +_+0 2 4
Distribute the -16.
2
1 1 t-_ + 4 2
( )
1 The coordinates of the vertex are _, 4 2 © Houghton Mifflin Harcourt Publishing Company
x2 + bx = c x2 - 4x = 5 x2 - 4x + 4 = 5 + 4 (x - 2)2 = 9 x - 2 = ±3 x = -1 or x = 5
(1t-
Combine the last two terms.
.
The soccer ball will be at its highest when it is at its vertex, or at 4 feet. The graph of the function confirms the vertex at (0.5, 4). The x-intercept at 1 indicates that the ball will hit the ground after 1 second.
Your Turn
13. Physics A person standing at the edge of a cliff 48 feet tall throws a ball up and just off the cliff with an initial upward velocity of 8 feet per second. What is the maximum height of the ball? When will the ball hit the ground? 49 feet; after 2 seconds
Module 9
IN2_MNLESE389830_U4M09L2.indd 375
375
Lesson 2
4/2/14 1:25 AM
Elaborate
()
EVALUATE
2
14. When b > 0, the perfect square-trinomial of the expression x + bx is ax 2 + bx + __b2 . What is the perfectsquare trinomial when b < 0? Does the sign of the constant change? Why or why not?
( _b2 ) . The sign of the constant 2
When b < 0, the perfect-square trinomial is ax 2 - bx + -
does not change since the squaring function will get rid of the negative value.
15. Essential Question Check-In What is the first step in completing the square to solve a quadratic equation of the form ax 2 + bx = c? The first step in completing the square to solve a quadratic equation in the form
ASSIGNMENT GUIDE
ax 2 + bx = c is to transform the left side of the equation into a perfect-square trinomial.
Evaluate: Homework and Practice Complete the square to form a perfect-square trinomial. 1.
x + 26x x 2 + 26x + 169
2.
x - 18x x 2 - 18x + 81
3.
x 2 - 18x + 81
4.
x - 24x
2
x 2 - 2x + 1
• Online Homework • Hints and Help • Extra Practice
2
2
x 2 - 24x + 144
Solve each equation by completing the square. Check the answers. 5.
x 2 + 8x = 33 x 2 + 8x + 16 = 49
6.
(x + 4) 2 = 49
x 2 - 6x = 8 x 2 - 6x + 9 = 17
(x - 3)2 = 17
x + 4 = ± √― 49 = ± 7
x+4=7
or
7.
x + 12x = 5 x 2 + 12x + 36 = 41
x = -11
x + 6 2 = 41
―
8.
―
or
2
x - 7 2 = 144
―
x + 6 = ± √41
x + 6 = √41 x = -6 + √41
x - 14x = 95 x 2 - 14x + 49 = 144
x - 3 = - √― 17 x = 3 - √― 17
―
――
x - 7 = ± √144 = ±12 x - 7 = 12
x + 6 = - √41 x = -6 - √41
―
or
x = 19
x - 7 = -12 x = -5
Solve each equation by completing the square. 9.
9x + 12x = 32
10. 4x 2 + 20x = 2
2
9x + 12x + 4 = 36
(2x + 5)2 = 27
2
3x + 2 = 6
4 x=_
or
― __―
3x + 2 = -6
3
2x + 5 = 3 √3 -5 + 3 √3 x= 2
8 x = -_
Module 9
3
Exercise
Explore Modeling Completing the Square
Exercise 22
Example 1 Completing the Square when a = 1
Exercises 1–4
Example 2 Solving x 2 + bx + c = 0 by Completing the Square
Exercises 5–8, 23–24
Example 3 2 Solving ax + bx + c = 0 by Completing the Square when a Is a Perfect Square
Exercises 9–12, 21
Example 4 Solving ax 2 + bx + c = 0 by Completing the Square when a Is Not a Perfect Square
Exercises 13–16, 21
Example 5 Modeling Completing the Square for Quadratic Functions
Exercises 17–20, 25
Depth of Knowledge (D.O.K.)
― __―
or 2x + 5 = -3 √3 -5 - 3 √3 x= 2 Lesson 2
376
IN2_MNLESE389830_U4M09L2.indd 376
Practice
4x 2 + 20x + 25 = 27
2
(3x + 2) = 36
© Houghton Mifflin Harcourt Publishing Company
x=3
x + 4 = -7
2
― x - 3 = √― 17 or x = 3 + √― 17
x - 3 = ± √17
Concepts and Skills
COMMON CORE
Mathematical Practices
1–16
1 Recall of Information
MP.5 Using Tools
17–20
2 Skills/Concepts
MP.4 Modeling
21
2 Skills/Concepts
MP.5 Using Tools
22
2 Skills/Concepts
MP.4 Modeling
23
3 Strategic Thinking
MP.3 Logic
24
3 Strategic Thinking
MP.3 Logic
25
3 Strategic Thinking
MP.3 Logic
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Solving Equations by Completing the Square
376
11. 16x 2 - 32x = 65
CONNECT VOCABULARY
12. 9x 2 - 24x = 1
9x 2 - 24x + 16 = 17
16x 2 - 32x + 16 = 81
(3x - 4)2 = 17
(4x - 4) = 81 2
Relate completing the square to arranging algebra tiles in a shape that is part of a square, and then adding 1-tiles to complete the square.
― 4 + √― 17 x=_
4x = 4 + 9 or 4x = 4 - 9 13 x=_
3x - 4 = √17
5 x = -_
4
4
3
1 x 2 + 4x = 10 13. _ 2 x 2 + 8x = 20 2 x + 8x + 16 = 36
QUESTIONING STRATEGIES
(x + 4)2 = 36
In a perfect square trinomial, is the last term always positive? Explain. Yes; a perfect square trinomial can take either the form (a + b)2 = a 2 + 2ab + b 2 or the form (a - b)2 = a 2 - 2ab + b 2. In both forms, the last term is positive.
x+4=6
9x 2 - 12x = 60
(3x - 2)2 = 64
or
3x - 2 = 8
x + 4 = -6
or 3x - 2 = -8
_
10 x= 3
x = -10
x = -2
1 x 2 - 5x = 18 16. _ 2 x 2 - 10x = 36
15. 2x 2 - 14x = 4
4x 2 + 28x = 8 4x 2 + 28x + 49 = 57
x 2 - 10x + 25 = 61
(2x + 7) 2 = 57
― ― _
3
14. 3x 2 - 4x = 20
x=2
or 2x + 7 = √57 -7 + √57 x= 2
― 4 - √― 17 x = -_
or 3x - 4 = - √17
(x - 5)2 = 61
―
― ― _
x - 5 = √61 x = 5 + √61
2x + 7 = - √57 -7 - √57 x= 2
―
―
or x - 5 = √61 x = 5 - √61
―
Projectile Motion Write an equation for each model, rewrite the equation into vertex form, and solve the problem. Then graph the function on a graphing calculator and state the x-intercepts of the graph. 17. Sports A person kicks a ball into the air with an initial upward velocity of 8 feet per second. What is the maximum height of the ball? When will the ball hit the ground?
© Houghton Mifflin Harcourt Publishing Company
model: h = -16t 2 + 8t
)
(
1 t+ h = -16 t 2 - _ 2
(
1 1 _ - _) 16
16
)
( )
1 1 1 t + __ + (-16) -__ h = -16 t 2 - _ 16 2 16
(
)
1 +1 h = -16 t - _ 4 2
The ball will be at its highest when it is at its vertex, or at 1 foot. The graph of the function confirms the vertex at (0.25, 1). The x-intercept at 0.5 indicates that the ball will hit the ground after 0.5 second.
IN2_MNLESE389830_U4M09L2.indd 377
Lesson 9.2
(
1 h = -16 t 2 - _ t 2
Module 9
377
h = -16t 2 + 8t
377
Lesson 2
4/2/14 1:25 AM
18. Physics A person reaching out to the edge of a building ledge 85 feet off the ground flicks a twig up and off the ledge with an initial upward velocity of 11 feet per second. What is the maximum height of the twig? When will the twig hit the ground?
VISUAL CUES
h = -16t 2 + 11t + 85
(
)
Suggest that students circle the leading coefficient in each step of the solution process. This visual cue can help them to remember to take this coefficient into account when completing the square.
11 h = -16 t 2 - __ t + 85 16
( ) ( )
⎡ 11 11 2 11 2⎤ h = -16⎢t 2 - __ t + __ - __ ⎥ + 85 16 32 32 ⎦ ⎣
11 11 + 85 ⎥ + (-16) (- ___ (__ 32 ) ⎦ 32 )
⎡ 11 t+ h = -16⎢t 2 - __ 16 ⎣ 2 11 h = -16 t + 32
(
2
⎤
2 2
COLLABORATIVE LEARNING
_) _
The vertex is
5561 64
5561 11 _ , ). (_ 64
32
The twig will be at its highest when it is at its vertex, or at
Have students work in small groups to make a poster showing how to apply the steps for solving quadratic equations by completing the square. Give each group a different equation to solve. Then have each group present its poster to the rest of the class, explaining each step.
5561 _ ≈ 86.89 feet.
64 The graph of the function confirms the vertex at about (0.34, 86.89). The x-intercept at about 2.7 indicates that the twig will hit the ground after about 2.7 seconds. 19. Construction An elevated bike ramp’s area can be represented be the expression __12 x 2 + 4x + c. Find the value of c that makes the equation __12 x 2 + 4x + c = 0 true. What does c represent?
_1 x
2 + 2x + c = 0 2 2 x + 4x + c = 0
This value represents the maximum height of the bike ramp. 20. Gardening The area of a square patch of soil can be represented by the expression x 2 - 12x + c. Find the value of c that makes this equation x 2 - 12x + c = 0 true. What are the width and length of the garden?
x 2 - 12x + c = 0 The value of c that makes the equation true is 36 feet. x 2 - 12x + 36 = 0
(x - 6)2 = 0 x=6
The garden is 6 feet long and 6 feet wide. 21. Identify the value of a in each equation of the form ax 2 + bx + c = 0. a. 11x 2 + 2x = 4 b. 4x 2 + 5 = 0 c.
3x 3 = 7
d. 5x 2 + 11x = 1 e. 3x = 5 2
Module 9
IN2_MNLESE389830_U4M09L2.indd 378
a = 11
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Wave Royalty Free/Design Pics Inc/Alamy
The value of c that makes this equation true is 4 feet.
a=4
a=0 a=5 a=3 378
Lesson 2
4/2/14 1:25 AM
Solving Equations by Completing the Square
378
22. The diagram represents the expression x 2 + 8x. Use algebra tiles to model completing the square. Then write the perfect square trinomial expression.
QUESTIONING STRATEGIES
×
x 2 + 8x + 16
How can you confirm that you have factored a trinomial correctly? Use FOIL to square the binomial to make sure that the result is equal to the trinomial being factored.
H.O.T. Focus on Higher Order Thinking
23. Explain the Error A student was instructed to solve the equation x 2 + 4x = 77 and produced the following work. Explain the student’s error. What is the correct solution? x 2 + 4x = 77 x 2 + 4x + 4 = 77 + 4
INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 A perfect square trinomial can be used to
(x + 2)2 = 81 x+2=9
x=7
The student forgot that the square root of 81 has two solutions: −9 and 9. The correct solution is x + 2 = ±9, or x = 7 and x = -11.
show that a quadratic equation can have only one solution, even when the leading coefficient is not 1. For example, 4x 2 - 12x + 9 can be factored 2 as (2x - 3) , so the equation 4x 2 - 12x + 9 = 0 has a single solution, x = __32 .
24. Justify Reasoning Will the equation x 2 + 6x = -10 produce an answer that is a real number after the square is completed? Explain.
The equation x 2 + 6x = -10 will not produce an answer that is a real number after the square is completed, since completing the square will produce the equation x 2 + 6x + 9 = -1, and there are no real solutions whose square is −1.
JOURNAL
© Houghton Mifflin Harcourt Publishing Company
Have students explain what it means to complete the square. 25. Draw Conclusions When solving a quadratic model, why are some solutions considered extraneous? Is this always the case, or can some quadratic models have two solutions?
When solving a quadratic model, some solutions are considered extraneous because they have a negative value, which is not useful in a real-world context. However, this is not always the case, as some quadratic models will have two valid solutions.
Module 9
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Lesson 9.2
379
Lesson 2
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Lesson Performance Task
INTEGRATE MATHEMATICAL PRACTICES Focus on Modeling MP.4 Regardless of the types of tiles used, students
An architect is designing the lobby of a new office building. The company that hired her has reclaimed a large quantity of stone floor tiles from the building previously on the site and wishes to use them to tile the lobby. The lobby needs to be 18 feet longer than it is wide to incorporate an information desk. The table below shows the types of tile available, the general color of the tile, and the area that can be covered by the tile.
Stone
Color
should recognize that they should all begin their solutions with a length of (x + 18), a width of x, and an area of x 2 + 18x. Discuss with students why 2 b 18 2 everyone will find that __ = __ = 81 when they 2a 2 complete the square.
( ) ( )
Area in Square Feet
Cream
175
Marble
Cream with gold flecks
115
Marble
Black
648
Marble
White with black flecks
360
Slate
Gray
280
Slate
Gray with blue gray regions
243
Travertine
Caramel
208
Travertine
Latte
760
Adoquin
Dark gray with black regions
319
Adoquin
Light gray with darker gray regions
403
Limestone
Pewter
448
Limestone
Beige
544
INTEGRATE MATHEMATICAL PRACTICES Focus on Communication MP.3 Have students explain how they would use
Design the lobby using at least all of one type of tile. You can add additional types of tiles to create patterns in the floor. For this exercise, you can decide on the dimensions of the tiles in order to make any pattern you wish.
Complete the square and solve.
Answers will vary but the procedure will be as follows.
x 2 + 18x + 81 = 360 + 81
(x + 9) 2 = 441
Use the marble tiles that are white with black flecks.
x + 9 = ±21
length ⋅ width = area
x + 9 = 21
(x + 18)x = 360 x 2 + 18x = 360
(_b2 ) . 18 (_b2 ) = (_ 2 )
x = 12
2
2
length = x + 18
= 12 + 18
= 92
= 30
= 81
Module 9
x = -30
In the context of this problem, a negative dimension does not make sense. So, use x = 12.
Find
2
x + 9 = -21
or
algebra tiles to represent the situation in this task. Students should report that there is 1 x 2-tile with 9 x-tiles to its right and 9 x-tiles below it. It takes 81 unit tiles to complete the diagram.
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Kzenon/ Shutterstock
Marble
The lobby will be 30 feet long by 12 feet wide.
380
Lesson 2
EXTENSION ACTIVITY IN2_MNLESE389830_U4M09L2.indd 380
Have students investigate “tiling” a floor with pennies. Have students choose how much money they will spend in pennies; determine how many square inches their pennies will cover; and then create and solve a real-world problem similar to the Lesson Performance Task. Students may determine that, for example, an 18” by 16” penny tiling requires an array of 22 by 20 pennies or 440 pennies ($4.40). If the area to be covered, in square inches, is 1152 and the length is 4 inches longer than the width, then (x + 4) x = 1152 and (x + 2)2 = 1156, so x = 32 and (x + 4) = 36. This would describe a floor covered with four of the 18” by 16” penny tiles at a cost of $17.60.
4/2/14 1:25 AM
Scoring Rubric 2 points: Student correctly solves the problem and explains his/her reasoning. 1 point: Student shows good understanding of the problem but does not fully solve or explain his/her reasoning. 0 points: Student does not demonstrate understanding of the problem.
Solving Equations by Completing the Square
380
LESSON
9.3
Name
Using the Quadratic Formula to Solve Equations
9.3
Use ... completing the square to ... Derive the quadratic formula … Also A-REI.B.4b
A B
MP.5 Using Tools
Explain to a partner how to solve an equation in the form ax2 + bx + c = 0 using the quadratic formula.
C
0
Subtract c from both sides.
Multiply both sides by 4a to make the coefficient of x 2 a perfect square. 4a 2 x 2 +
D
ENGAGE
=
4abx
-4ac
Add b 2 to both sides of the equation to complete the square.
© Houghton Mifflin Harcourt Publishing Company
4a 2x 2 + 4abx + b 2 = -4ac + b 2
E
Factor the left side to write the trinomial as the square of a binomial.
( 2ax + b ) = b - 4ac 2
2
F
Take the square roots of both sides. __
2ax + b
G
√
=±
b 2 - 4ac
Subtract b from both sides. __ 2ax = -b ±
H
View the Engage section online. Discuss the photo, asking students to speculate on how math could be used to place an outfielder in exactly the right spot to make a catch. Then preview the Lesson Performance Task.
Write the standard form of a quadratic equation.
ax 2 + bx = -c
Language Objective
PREVIEW: LESSON PERFORMANCE TASK
Deriving the quadratic formula
ax 2 + bx + c =
Mathematical Practices
The quadratic formula is a formula derived from the general quadratic equation by completing the square. It gives the real solutions (zero, one, or two) of any quadratic equation.
√
b 2 - 4ac
Divide both sides by 2a to solve for x. __
√
b 2 - 4ac -b ± x = __ 2a
Module 9
Lesson 3
381
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Lesson 3 381
Module 9
9L3.indd 30_U4M0
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381
Lesson 9.3
Resource Locker
You can complete the square on the general form of a quadratic equation to derive a formula that can be used to solve any quadratic equation.
A-REI.B.4a
Essential Question: What is the quadratic formula, and how can you use it to solve quadratic equations?
Using the quadratic formula to Solve Equations
Explore
The student is expected to:
COMMON CORE
Date
Essential Question: What is the quadratic formula, and how can you use it to solve quadratic equations?
Common Core Math Standards COMMON CORE
Class
381
09/04/14
6:53 PM
07/04/14 6:43 PM
__
-b ± √b 2 - 4ac The formula you just derived, x = __, is called the quadratic formula. 2a It gives you the values of x that solve any quadratic equation where a ≠ 0.
EXPLORE
Reflect
1.
Deriving the Quadratic Formula
What If? If the derivation had begun by dividing each term by a, what would the resulting binomial of x have been after completing the square? Does one derivation method appear to be simpler than the other? Explain.
INTEGRATE TECHNOLOGY
ax 2 + bx + c = 0 b c _ x2 + _ ax + a = 0 b c _ x2 + _ ax = -a 2 2 b b b c _ _ _ x2 + _ a x +2 2a = - a +2 2a b b c _ x+ = -_ a + 2a 2a b The resulting binomial is x + . The previous derivation appears simpler because no 2a fractions are involved in the derivation.
(
( )
_)
( )
Students have the option of completing the Explore activity either in the book or online.
( )
(_)
QUESTIONING STRATEGIES What is it called when you add b 2 to both sides of the equation? What does this step allow you to do? Completing the square; it allows you to rewrite the left side of the equation as the square of a binomial, so that you can then take the square roots of both sides.
Using the Discriminant to Determine the Number of Real Solutions
Explain 1
Recall that a quadratic equation, ax 2 + bx + c, can have two, one, or no real solutions. By evaluating the part of the quadratic formula under the radical sign, b 2 - 4ac, called the discriminant, you can determine the number of real solutions.
Determine how many real solutions each quadratic equation has.
Example 1
EXPLAIN 1
x 2 - 4x + 3 = 0 Identify a, b, and c.
b 2 - 4ac
Use the discriminant.
(-4) 2 - 4(1)(3)
Substitute the identified values into the discriminant.
16 - 12 = 4
Simplify.
Since b 2 - 4ac > 0, the equation has two real solutions.
x 2 - 2x + 2 = 0 a = 1 , b = -2 , c = 2
Identify a, b, and c.
b - 4ac
Use the discriminant.
2
( -2 ) - 4( 2
4
- 8
1
)(
= -4
2
)
Using the Discriminant to Determine the Number of Real Solutions
© Houghton Mifflin Harcourt Publishing Company
a = 1, b = -4, c = 3
QUESTIONING STRATEGIES Does the discriminant include the radical sign? No, the discriminant is b2 -4ac , the expression under the radical sign.
Substitute the identified values into the discriminant.
INTEGRATE MATHEMATICAL PRACTICES Focus on Reasoning MP.2 Point out the benefit of finding the
Simplify.
Since b 2 - 4ac < 0, the equation has no real solution(s).
Module 9
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Lesson 3
PROFESSIONAL DEVELOPMENT IN2_MNLESE389830_U4M09L3.indd 382
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Learning Progressions In this lesson, students derive the quadratic formula and use it to solve equations. They also learn to recognize when the quadratic formula gives complex-number solutions. The quadratic formula is a generalization of the method of completing the square. Students should choose the method that best fits the situation at hand when solving quadratic equations. Students can use the quadratic formula to find the zeros of and to graph non-factorable quadratic equations, an important skill for college readiness.
discriminant first, so that students will be aware of the number of real solutions. Knowing this will help students later verify whether their solutions are correct.
Using the Quadratic Formula to Solve Equations
382
Reflect
EXPLAIN 2
2.
Solving Equations by Using the Quadratic Formula
When the discriminant is positive, the quadratic equation has two real solutions. When the discriminant is negative, there are no real solutions. How many real solutions does a quadratic equation have if its discriminant equals 0? Explain. One real solution; if the discriminant is 0, then you are adding or subtracting the square
root of 0 in the quadratic formula. Since the only square root of 0 is 0, the answer will be the same whether it is added or subtracted.
AVOID COMMON ERRORS
Your Turn
Students might make sign errors in finding -b or b2 - 4ac when any of a, b, or c are negative. Encourage students to use parentheses and to write down every step.
Use the discriminant to determine the number of real solutions for each quadratic equation. 3.
x 2 + 4x + 1 = 0
4.
2x 2 - 6x + 15 = 0
a = 1, b = 4, c = 1
a = 2, b = -6, c = 15
(4) - 4(1)(1)
(-6) - 4(2)(15)
b 2 - 4ac
b 2 - 4ac 2
2
5.
x 2 + 6x + 9 = 0
a = 1, b = 6, c = 9 b 2 - 4ac
(6) 2 - 4(1)(9)
16 - 4 = 12
36 - 120 = -84
36 - 36 = 0
two real solutions
no real solutions
one real solution
Explain 2
Solving Equations by Using the Quadratic Formula
To use the quadratic formula to solve a quadratic equation, check that the equation is in standard form. If not, rewrite it in standard form. Then substitute the values of a, b, and c into the formula. Example 2
© Houghton Mifflin Harcourt Publishing Company
Solve using the quadratic formula.
2x 2 + 3x - 5 = 0 a = 2, b = 3, c = -5
Identify a, b, and c.
-b ± √b 2 - 4ac x = __ 2a
Use the quadratic formula.
__
___
2 -3 ± √(3) - 4(2)(-5) x = ___ 2(2)
Substitute the identified values into the quadratic formula.
x=
Simplify the radicand and the denominator.
_ -3 ± √49 _
4 -3 ± 7 x=_ 4 -3 + 7 -3 - 7 x = _ or x = _ 4 4 5 x = 1 or x = -_ 2 5. The solutions are 1 and -_ 2
Module 9
Evaluate the square root. Write as two equations. Simplify both equations.
383
Lesson 3
COLLABORATIVE LEARNING IN2_MNLESE389830_U4M09L3.indd 383
Peer-to-Peer Activity Have students work in pairs to make a poster showing how to apply the steps for solving quadratic equations by completing the square. Give each pair a different equation to solve. Then have each pair of students present its poster to another pair, explaining each step.
383
Lesson 9.3
4/2/14 1:29 AM
Graph y = 2x 2 + 3x - 5 to verify your answers.
QUESTIONING STRATEGIES Why is the symbol ± in the quadratic formula? The definition of square root is the positive or negative number that, when multiplied by itself, equals the number inside the radical. Where does the Quadratic Formula come from? The quadratic formula comes from completing the square and simplifying the general form of a quadratic equation.
The graph does verify the solutions.
2x = x 2 - 4 x 2 - 2x - 4 = 0
Write in standard form.
a = 1 , b = -2 , c = -4
Identify a, b, and c.
x=
――― -b ± √b - 4ac __ 2
2a
Use the quadratic formula.
―――――――――
( ) √( -2 ) - 4( 1 )( -4 ) x = ―――――――――――――― 2( 1 ) ―― - -2 ±
2
Substitute the identified values into the quadratic formula.
√
2± 20 x = __ 2
―――
√
―
Simplify the radicand and the denominator.
―
2± 2 ± 2 √5 4 ∙5 x = __ = _= 1 ± √5 2 2
Simplify.
x = 1 + √5 or x = 1 - √5
Write as two equations.
―
or x ≈ -1.236
―
―
Use a calculator to find approximate solutions to three decimal places.
The exact solutions are 1 + √5 and 1 - √5 . The approximate solutions are
3.236
and -1.236 .
Graph y = x 2 - 2x - 4 and find the zeros using the graphing calculator. The calculator will give approximate values. The graph
does
confirm the solutions.
© Houghton Mifflin Harcourt Publishing Company
x ≈ 3.236
―
Reflect
6.
Discussion How can you use substitution to check your solutions? Substitute each value into the given quadratic equation to see if it leads to a true equality.
Module 9
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Lesson 3
DIFFERENTIATE INSTRUCTION Visual Cues
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To help prevent computational errors, some students may find it helpful to circle everything after b2, calculate that, and then add the answer to b2. For example:
√――――― 72 - 4(2)(-3) = √――― 49 + 24
―
= √73
Using the Quadratic Formula to Solve Equations
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Your Turn
EXPLAIN 3
Solve using the quadratic formula. 7.
Using the Discriminant with Real-World Models
x 2 - 6x - 7 = 0
―――――― -(-6)± √(-6) - 4(1)(-7) x = ___ 2(1) 6 ± √― 64 x=_ 2 6±8 x=_ 2 6+8 6-8 x = _ or x = _
8.
2
QUESTIONING STRATEGIES
2 x = 7 or
How can you use the discriminant to determine the number of real solutions of a quadratic equation? How is the process different when using a quadratic equation to model situations in the real world? Evaluate the discriminant. If it is negative, there are no real solutions; if it is positive, there are two real solutions; if it is zero, there is one real solution. In some cases, when using a quadratic equation to model a real-world situation, negative solutions will not make sense; in that case, even if real solutions exist they may be disregarded.
2x 2 = 8x - 7
―――――― -(-8)± √(-8) - 4(2)(7) ___ 2(2) 8 ± √― 8 x=_ 4 8 ± 2 √― 2 x=_ 4 ― 8 - 2 √― 8 + 2 √2 2 x = _ or x = _ 4 4 ― ― √2 √2 x = 2 + _ or x = 2 - _ 2 2 √― √― 2 2 The solutions are 2 + _ and 2 - _ . 2
x=
2 x = -1
The solutions are 7 and -1.
2
Explain 3
2
Using the Discriminant with Real-World Models
Given a real-world situation that can be modeled by a quadratic equation, you can find the number of real solutions to the problem using the discriminant, and then apply the quadratic formula to obtain the solutions. After finding the solutions, check to see if they make sense in the context of the problem. In projectile motion problems where the projectile height h is modeled by the equation h = −16t 2 + vt + s, where t is the time in seconds the object has been in the air, v is the initial vertical velocity in feet per second, and s is the initial height in feet. The -16 coefficient in front of the t 2 term refers to the effect of gravity on the object. This equation can be written using metric units as h = −4.9t 2 + vt + s, where the units are converted from feet to meters. Time remains in units of seconds.
© Houghton Mifflin Harcourt Publishing Company
Example 3
For each problem, use the discriminant to determine the number of real solutions for the equation. Then, find the solutions and check to see if they make sense in the context of the problem.
A diver jumps from a platform 10 meters above the surface of the water. The diver’s height is given by the equation h = −4.9t 2 + 3.5t + 10, where t is the time in seconds after the diver jumps. For what time t is the diver’s height 1 meter? Substitute h = 1 into the height equation. Then, write the resulting quadratic equation in standard form to solve for t. 1 = −4.9t 2 + 3.5t + 10
0 = −4.9t 2 + 3.5t + 9
First, use the discriminant to find the number of real solutions of the equation. b 2 - 4ac
Use the discriminant.
(3.5) - 4(-4.9)(9) = 188.65 2
Since b 2 − 4ac > 0, the equation has two real solutions.
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Connect Vocabulary In the introduction, students learn about the discriminant of the quadratic formula. Students may know the word discriminate, which means to differentiate or make a distinction. A discriminant is something used to help make a distinction.
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Next, use the quadratic formula to find the real number solutions. a = −4.9, b = 3.5, c = 10
AVOID COMMON ERRORS
Identify a, b, and c.
――― -b ± √b - 4ac 2
t = __ 2a -3.5 ± √188.65 __ t= 2(-4.9)
Students may forget to write the equation in standard form before finding the values of a, b, and c. Remind them that the quadratic equation must be in the form ax2 + b + c = 0 before they can begin to use the Quadratic Formula.
Use the quadratic formula.
―――
Substitute the identified values into the quadratic formula and the value of the discriminant.
-3.5 ± 13.73 t ≈ __ -9.8 -3.5 - 13.73 -3.5 + 13.73 or t ≈ __ __ t≈ -9.8 -9.8 t ≈ -1.04 or t ≈ 1.76
Simplify. Write as two equations. Solutions
Disregard the negative solution because t represents the seconds after the diver jumps and a negative value has no meaning in this context. So, the diver is at height 1 meter after a time of t ≈ 1.76 seconds.
B
The height in meters of a model rocket on a particular launch can be modeled by the equation h = −4.9t 2 + 102t + 100, where t is the time in seconds after its engine burns out 100 meters above the ground. When will the rocket reach a height of 600 meters? Substitute h = 600 into the height equation. Then, write the resulting quadratic equation in standard form to solve for t. h = −4.9t 2 + 102t + 100
600 = −4.9t 2 + 102t + 100 0 = −4.9t 2 + 102t - 500
First, use the discriminant to find the number of real solutions of the equation. a = −4.9, b = 102 , c = -500
Identify a, b, and c.
b - 4ac
Use the discriminant.
2
10404
-
9800
-500
)
= 604
Substitute the identified values into the discriminant. Simplify.
Since b 2 - 4ac > 0, the equation has 2 real solutions.
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© Houghton Mifflin Harcourt Publishing Company
( 102 ) - 4(-4.9)( 2
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Next, use the quadratic formula to find the real number solutions.
―――
√
102 604 - ± t = __ 2(-4.9)
Substitute the identified values into the quadratic formula and the value of the discriminant.
-102 24.58 ± t = ___ -9.8
Simplify.
24.58 24.58 - 102 + - 102 t ≈ __ or t ≈ __ -9.8 -9.8
Write as two equations.
t ≈ -7.90
Solutions
Disregard the launched and a
or
negative negative
t ≈ 12.92
solution because t represents the seconds after the rocket has value has no meaning in this context. So, the rocket is at
height 600 meters after a time of t ≈
12.92
seconds.
Your Turn
For each problem, use the discriminant to determine the number of real solutions for the equation. Then, find the solutions and check to see if they make sense in the context of the problem. 9.
A soccer player uses her head to hit a ball up in the air from a height of 2 meters with an initial vertical velocity of 5 meters per second. The height h in meters of the ball is given by h = −4.9t 2 + 5t + 2, where t is the time elapsed in seconds. How long will it take the ball to hit the ground if no other players touch it?
h = −4.9t 2 + 5t + 2 0 = −4.9t 2 + 5t + 2 © Houghton Mifflin Harcourt Publishing Company
Find the discriminant.
(5)2 − 4(-4.9)(2) = 64.2 Since b 2 - 4ac > 0, the equation has two real solutions. Use the quadratic formula to find the solutions of the quadratic equation.
―― -5 ± √64.2 __ -9.8 -5 ± 8.01 t≈_
t=
-9.8
t ≈ -0.31 or t ≈ 1.33 Disregard the negative solution because there is no negative time in this problem context. The soccer ball reached the ground after about t ≈ 1.33 seconds.
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10. The quarterback of a football team throws a pass to the team’s receiver. The height h in meters of the football can be modeled by h = −4.9t 2 + 3t + 1.75, where t is the time elapsed in seconds. The receiver catches the football at a height of 2 meters. How long does the ball remain in the air until it is caught by the receiver?
ELABORATE QUESTIONING STRATEGIES
The ball is caught by the receiver when h = 2.
Explain why the value of the discriminant indicates the number of solutions. When the number inside the radical sign is positive, there are two solutions, because you can take the positive and negative square root of a positive number. When the number is negative, there are no real solutions because you cannot take the square root of a negative number. When the number inside the radical sign is 0, there is one solution, because the square root of 0 is 0 and adding and subtracting 0 will not change the value of -b.
h = −4.9t 2 + 3t + 1.75 0.25 = −4.9t 2 + 3t + 1.75 0 = −4.9t 2 + 3t + 1.5
(3)2 - 4(−4.9)(1.5) = 38.4 Since b 2 - 4ac > 0, the equation has two real solutions.
―― __ _
Use the quadratic formula to find the solutions of the quadratic equation. -3 ± √38.4 t= -9.8 -3 ± 6.20 t≈ -9.8 t ≈ -0.33 or t ≈ 0.94 Disregard the negative solution because there is no negative time in this problem context.The ball was in the air for t ≈ 0.94 second.
SUMMARIZE THE LESSON
Elaborate
at one point. If the discriminant is positive, the equation will have two solutions: it will intersect the x-axis at two points. If the discriminant is negative, the equation will have no solutions: the graph will not intersect the x-axis at all. 12. What advantage does using the quadratic formula have over other methods of solving quadratic equations? The quadratic formula works for all quadratic equations. Other methods only work in certain situations.
13. Essential Question Check-In How can you derive the quadratic formula?
The quadratic formula is derived by completing the square for the standard form of a quadratic equation.
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Copy and complete the graphic organizer shown below with students to summarize lesson concepts. In each box, write the number of real solutions.
© Houghton Mifflin Harcourt Publishing Company · Image Credits: © Corbis Royalty Free
11. How can the discriminant of a quadratic equation be used to determine the number of zeros (x-intercepts) that the graph of the equation will have? If the discriminant is zero, the equation will have one solution: it will intersect the x-axis
The Number of real solutions of ax2 + bx + c = 0 when...
b2 - 4ac > 0 Two real solutions b2 - 4ac < 0 No real solutions b2 - 4ac = 0 One real solution
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Evaluate: Homework and Practice
EVALUATE
Determine how many real solutions each quadratic equation has. 1.
4x 2 + 4x + 1 = 0
2.
a = 1, b = -1, c = 3
a = 4, b = 4, c = 1
b 2 - 4ac
b - 4ac 2
(-1)2 - 4(1)(3)
(4)2 - 4(4)(1)
Concepts and Skills
1 - 12 = -11
16 - 16 = 0
ASSIGNMENT GUIDE
Since b 2 - 4ac < 0, the equation has no real solutions.
Since b - 4ac = 0, the equation has one real solution. 2
Practice
3.
Explore Deriving the Quadratic Formula
x 2 - 8x 2 - 9 = 0
4.
Exercises 1–8, 20, 22
Explore 2 Solving Equation by Using the Quadratic Formula
Exercises 9–14
Explore 3 Using the Discriminant with Real-World Models
Exercises 15–19
b 2 - 4ac
64 + 36 = 100
5 - 16 = -11
(- √―5 )
Since b 2 - 4ac > 0, the equation has two real solutions. 6.
_1 , b = -1, c = _1
2 b - 4ac 2
(_)(_1 ) 1 _ 12 4 _ 1- =
(- √―7 )
© Houghton Mifflin Harcourt Publishing Company
Exercise
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-4
(_14 )(7)
Since b 2 - 4ac = 0, the equation has one real solution. 8.
―
1 =0 x 2 √2 - x + _ 2
―
a = √2 , b = -1, c =
b - 4ac
― (_ ) 2 ― 1 - 2 √2 < 0 2
(_ )
_1 2
(-1)2 - 4( √2 ) 1
Since b 2 - 4ac < 0, the equation has no real solutions.
Since b 2 - 4ac = 0, the equation has one real solution.
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2
7-7=0
―
Module 9
_1 , b = - √―7 , c = 7
4 b - 4ac
x 2 - x √2 + 1 = 0 _ 2 1 a = , b = - √2 , c = 1 2 2 b - 4ac 2 (- √2 ) - 4 12 (1) 2-2=0
―
―
2
2 2 Since b 2 - 4ac > 0, the equation has two real solutions.
_
- 4(2)(2)
x 2 - x √7 + 7 = 0 _ 4
a=
4
(-1)2 - 4 1
―
2
Since b 2 - 4ac < 0, the equation has no real solutions.
x2 - x + _ 1 =0 _ 2 4
a=
7.
―
b 2 - 4ac
(-8)2 - 4(1)(-9)
5.
―
2x 2 - x √5 + 2 = 0
a = 2, b = - √5 , c = 2
a = 1, b = -8, c = -9
Explore 1 Using the Discriminant to Determine the Number of Real Solutions
• Online Homework • Hints and Help • Extra Practice
x2 - x + 3 = 0
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Depth of Knowledge (D.O.K.)
COMMON CORE
Mathematical Practices
1–16
2 Skills/Concepts
MP.2 Reasoning
17
2 Skills/Concepts
MP.4 Modeling
18–19
3 Strategic Thinking
MP.4 Modeling
20
3 Strategic Thinking
MP.3 Logic
21
3 Strategic Thinking
MP.3 Logic
22
3 Strategic Thinking
MP.4 Modeling
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Solve using the quadratic formula. Leave answers that are not perfect squares in radical form. 9.
10x + 4 = 6x 2
INTEGRATE MATHEMATICAL PRACTICES Focus on Reasoning MP.2 Students can check their solutions for
10. x 2 + x - 20 = 0
10x + 4 = 6x 2
x=
5x + 2 = 3x 2 0 = 3x 2 - 5x - 2
―――――― -1 ± √1 - 4(1)(-20) ___ 2 ――― √ -1 ± 1 + 80 __ 2
―――――― x = ___ 2(3) 5 ± √――― 25 + 24 __
x=
x=
2 x = 4 or
-1 ± √― 81 x = __ 2 -1 ± 9 x=_ 2 -1 - 9 -1 + 9 x = _ or x = _
-(-5)± √(-5) - 4(3)(-2) 2
x=
6 5 ± √― 49 _ 6 5±7 x=_ 6 5+7 5-7 _ or x = _ x=
6 x = 2 or
correctness by substituting the values into the original equations and verifying that both solutions make the equation true. When checking solutions, remind students that the square of a negative number is a positive number.
2
x = -5
2
The solutions are 4 and -5.
6
_
x = -1 3 The solutions are 2 and - 1 . 3
_
11. 4x 2 = 4 - x
12. 9x 2 + 3x - 2 = 0
4x 2 + x - 4 = 0
――――― -1± √1 - 4(4)(-4) x = ___ 2(4) ――― √ -1± 1 + 64 __
x=
2
x=
8
8
The solutions are
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-1- √― 65 _ 8 -1+ √― -1- √― 65 65 _ _
or x =
8
and
8
.
390
2
― __ _ _ _
18 -3 ± √81 x= 18 -3 ± 9 x= 18 -3 + 9 -3 - 9 x= or x = 18 18 2 1 or x = - x= 3 3 2 1 The solutions are and - . 3 3
© Houghton Mifflin Harcourt Publishing Company
-1± √― 65 x=_ 8 -1+ √― 65 _
x=
x=
――――― -3 ± √3 - 4(9)(-2) ___ 2(9) ――― -3 ± √9 + 72 __
_ _ _ _
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13. 14x + 3 = -8x 2
INTEGRATE MATHEMATICAL PRACTICES Focus on Modeling MP.4 To help students follow the derivation of the
14x + 3 = -8x
x=
a = 8, b = 14, c = 3 x= x=
――――― -14 ± √14 - 4(8)(3) ___ 2(8) ―――― -14 ± √196 - 96 __ 2
x=
Students may not write solutions involving radicals in simplest form. Show them examples, such as
16
4
―
――――― -3 ± √3 - 4(1)(1) __ 2(1) ―― √ -3 ± 9 -4 __ 2
-3 ± √― 5 x=_ 2 -3 + √― -3 - √― 5 5 x = _ or x = _ 2 2 -3 + √― -3 - √― 5 5 The solutions are _ and _.
-14 ± √―― 100 x = __ 16 -14 ± 10 x=_ 16 -14 + 10 -14 - 10 _ x= or x = _ 16 16 3 1 or x = - _ x = - _ 4 2 3 1 and - _ The solutions are - _ .
AVOID COMMON ERRORS
2
2
2
2
For each problem, use the discriminant to determine the number of real solutions for the equation. Then, find the solutions and check to see if they make sense in the context of the problem.
3 ± √63 , which can be simplified to 1 ± √7 . ________ 3
a = 1, b = 3, c = 1
8x 2 + 14x + 3 = 0
quadratic formula, you may want to have them assign specific values to a, b, and c and show what each step of the derivation looks like in terms of those values in order to check the reasonableness of the steps.
―
14. x 2 + 3x 2 + 1 = 0 2
© Houghton Mifflin Harcourt Publishing Company · Image Credits: ©Dan Galic/ Alamy
15. Sports A soccer player kicks the ball to a height of 1 meter inside the goal. The equation for the height h of the ball at time t is h = −4.9t 2 - 5t + 2. Find the time the ball reached the goal.
The ball is 1 meter off the ground when h = 1. Substitute for h in the equation. h = −4.9t 2 - 5t + 2 1 = −4.9t 2 - 5t + 2 0 = −4.9t 2 - 5t + 1 Find the discriminant.
(-5)2 - 4(−4.9)(1) = 44.6 Since b 2 - 4ac > 0, the equation has two real solutions. Use the quadratic formula to find the solutions of the quadratic equation. 5 ± √―― 44.6 __ -9.8 5 ± 6.68 t≈_
t=
-9.8 t ≈ -1.19 or t ≈ 0.17 Disregard the negative solution because there is no negative time in this context. The soccer ball reached the goal after about t ≈ 0.17 seconds.
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16. The length and width of a rectangular patio are, (x + 8) feet and (x + 6) feet, respectively. If the area of the patio is 160 square feet, what are the dimensions of the patio?
QUESTIONING STRATEGIES If the discriminant of a quadratic equation that models a real-world situation is greater than 0, will the problem have two solutions? Only when both solutions make sense in the context; for example, if there is a positive and a negative solution to a problem that asks for a length, the negative solution does not make sense, because length cannot be negative.
A = (x + 6) (x + 8)
A = (x + 6)(x + 8)
160 = (x + 6)(x + 8) 160 = x 2 + 14x + 48
―― __ __
0 = x 2 + 14x - 112 -14 ± √644 x= 2 -14 ± 25.38 x≈ 2 x ≈ 5.69 or x ≈ -19.69 Disregard the negative solution because it yields a negative length. Length: 5.69 + 8 = 13.69 feet Width: 5.69 + 6 = 11.69 feet 17. Chemistry A scientist is growing bacteria in a lab for study. One particular type of bacteria grows at a rate of y = 2t 2 + 3t + 500. A different bacteria grows at a rate of y = 3t 2 + t + 300. In both of these equations, y is the number of bacteria after t minutes. When is there an equal number of both types of bacteria?
3t 2 + t + 300 = 2t 2 +3t + 500 3t 2 + t + 300 - 2t 2 - 3t - 500 = 0 t -2 ± √―― 804 __ t= 2 -2 ± 28.35 t ≈ __
2
- 2t - 200 = 0
© Houghton Mifflin Harcourt Publishing Company
2 t ≈ -15.18 or t ≈ 13.18 Disregard the negative solution because there is no negative time in this context. There is an equal number of both types of bacteria at t ≈ 13.175 minutes.
Use this information for Exercises 18 and 19. A gymnast, who can stretch her arms up to reach 6 feet, jumps straight up on a trampoline. The height of her feet above the trampoline can be modeled by the equation h = -16x 2 + 12x, where x is the time in seconds after her jump. 18. Do the gymnast’s hands reach a height of 10 feet above the trampoline? Use the discriminant to explain. (Hint: Since h = height of feet, you must use the difference between the heights of the hands and feet.)
Evaluating the discriminant for h = 10 - 6 = 4 gives d = -112 < 0, so there are no real solutions for this height. Thus, her feet never reach 4 feet, so her hands never reach 10 feet. Module 9
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19. Which of the following are possible heights she achieved? Select all that apply. 9 a. h = _ Use the discriminant to determine the maximum height h m the gymnast 4 reaches. When the discriminant equals 0, there is one real solution, which b. h = 4 will be when the maximum height occurs. Write the equation in standard c. h = 3 form and set the discriminant equal to zero. Solve for -c = h m. d. h = 0.5 2 h m = -16x + 12x 1 e. h = _ 4 0 = -16x 2 + 12x - h m
INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 Let students know that any quadratic equation can be solved by using the quadratic formula and that it will always give exact solutions— but that this may not always be the quickest method. Encourage students to recall earlier lessons and try the other methods, as needed.
b 2 - 4ac = 12 2 - 4(-16)(-h m)
0 = 12 - 4(-16)(-h m) 2
-144 = -64(h m) 9 = hm 4 The letters A, D, and E are possible heights of the gymnast since they are 9 all less or equal to h m = . 4 20. Explain the Error Dan said that if a quadratic equation does not have any real solutions, then it does not represent a function. Explain Dan’s error. Having no real solutions means that the value of the function can never be zero. That is, the function has no zeros (or x-intercepts). The quadratic expression is a function as it is a polynomial.
_
_
MULTIPLE REPRESENTATIONS Remind students that the solutions to a quadratic equation are also the zeros of its related function.
H.O.T. Focus on Higher Order Thinking
JOURNAL
21. Communicate Mathematical Ideas Explain why a positive discriminant results in two real solutions.
Have students write a quadratic equation and use the discriminant to find the number of real solutions. Have them use the Quadratic Formula to find the real solution(s), if any exist. If there are no real solutions, have them explain what this means.
Because every positive number has two square roots, a positive discriminant results in two solutions to a quadratic equation.
© Houghton Mifflin Harcourt Publishing Company
22. Multi-Step A model rocket is launched from the top of a hill 10 meters above ground level. The rocket’s initial speed is 10 meters per second. Its height h can be modeled by the equation h = -4.9t 2 + 10t + 10, where t is the time in seconds. a. When does the rocket achieve a height of 100 meters? h = -4.9t 2 + 10t + 10
100 = -4.9t 2 + 10t + 10 0 = -4.9t 2 + 10t - 90
(10) 2 -4(-4.9)(-90) = -1664 < 0
Since b 2 - 4ac < 0, the equation has no real solutions. The rocket never reaches a height of 100 meters.
―― __ __
b. How long does it take the rocket to reach ground level? -10 ± √296 t= -9.8 -10 ± 17.20 t≈ -9.8 t ≈ -0.74 or t ≈ 2.78
Taking only the positive solution, the rocket will reach ground level after t ≈ 2.78 seconds. Module 9
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Lesson Performance Task
AVOID COMMON ERRORS
Over the course of a baseball game, several batters have hit balls directly at the center fielder. The center fielder stands 200 feet from home plate. He can catch the ball when it is 7 feet off of the ground, and he can sprint 60 feet in 4.1 seconds. Given this information, can the center fielder catch the hits made by the players in the following table?
Height Ball is Hit at in Feet
Player 1
2
3
Some students may assume that a parabola modeling the height of a ball as a function of time describes the path of the ball through space. Encourage students to graph the parabola, and point out the units of the axes. The x-axis measures time and does not represent the horizontal distance the ball has travelled.
Initial Ball Speed in Feet per Second
5 feet
54
4 feet 6 inches
72
4 feet 9 inches
86
5 feet 3 inches
74
5 feet
81
5 feet 3 inches
83
4 feet 6 inches
97
5 feet 3 inches
68
4 feet 9 inches
62
QUESTIONING STRATEGIES How many real values for t are possible when using the formula for height? two
The height of the ball and the distance it travels are given by the general models below, where h is the height, d is the distance traveled, v 0 is the initial speed of the ball, s is the height the ball is hit from, and t is the time in seconds. Height
Do you think both values for t will make sense for the problem in the Lesson Performance Task? No; the path the ball takes is a parabola; one value for t will be right after the ball is hit, before it reaches its highest point, and the other will be in the outfield, after the ball reaches its highest point.
Distance
h = -16t 2 + v 0t + s
7 = -16t + 54t + 5 -54 ± √2788 t= -32 -54 + √2788 t= -32 ≈ 0.0375
―― __ ―― __
d = v 0t
2
t=
≈ 3.3375
-54 - √―― 2788 __
© Houghton Mifflin Harcourt Publishing Company · Image Credits: ©Action Sports Photography/Shutterstock
or
-32
At 0.0375 seconds, the ball has just been hit with the bat. d = 54t = 54(3.34) ≈ 180 Can the center fielder can move 200 feet to 180 feet in 3 seconds? 60 20 > The player can field the ball. 4.1 3 Initial Ball Speed in Player Feet per Second Flight Time Fielded Y/N 1 54 3.34 yes 72 4.47 no 86 5.35 no 2 74 4.60 no 81 5.04 no 83 5.17 no 3 97 6.04 no 68 4.22 no 62 3.84 yes
_ _
How many real values for d are possible when using the formula for distance? 1
Player 3’s first hit is an out of the park home run. Module 9
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Lesson 3
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Have students investigate the longest home run ever recorded in regular season major-league baseball, hit by Mickey Mantle on September 10, 1960. Ask them to find its maximum height and greatest horizontal distance from home plate, and then ask them what other information they would need to model the height of the ball using a quadratic equation. Maximum height: about 145 ft; greatest horizontal distance from home plate: about 643 ft; sample answer: the height at which the ball was hit, and the time it took the ball to hit the ground.
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Scoring Rubric 2 points: Student correctly solves the problem and explains his/her reasoning. 1 point: Student shows good understanding of the problem but does not fully solve or explain his/her reasoning. 0 points: Student does not demonstrate understanding of the problem.
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LESSON
9.4
Name
Choosing a Method for Solving Quadratic Equations
Class
9.4
Choosing a Method for Solving Quadratic Equations
Essential Question: How can you choose a method for solving a given quadratic equation? Resource Locker
Comparing Solution Methods for Quadratic Equations
Explore
Common Core Math Standards The student is expected to: COMMON CORE
7x 2 - 3x - 5 = 0
A-REI.B.4b
Try to solve the equation by factoring.
Solve quadratic equations ... as appropriate to the initial form of the equation …
Mathematical Practices COMMON CORE
Find the factors of 7 and -5 to complete the table:
MP.5 Using Tools
Factors of 7
Factors of –5
Outer Product + Inner Product
1, 7
1, -5
2
1, 7
5, -1
34
1, 7
-1, 5
-2
1, 7
-5, 1
-34
Language Objective Explain to a partner how you decide whether to use square roots, completing the square, or the quadratic formula to solve a quadratic equation.
No .
Now, try to solve the equation by completing the squares. © Houghton Mifflin Harcourt Publishing Company
The choice of method usually depends on the form of the equation, the coefficient values, or personal preference. For example, you can solve ax 2 - c = 0 or a(x 2 + b) = c by taking square roots. You can solve any quadratic equation ax 2 + bx + c = 0 by completing the square or by using the quadratic equation, but one or the other may be preferable depending on the values of a, b, and c.
None of the sums of the inner and outer products of the factor pairs 7 and -5 equal -3. Does this mean the equation cannot be solve
ENGAGE Essential Question: How can you choose a method for solving a given quadratic equation?
Date
The leading coefficient is not a perfect square. Multiply both sides by a value that makes the coefficient a perfect square.
7 (7x 2 - 3x - 5) = (0) 7 49 x 2 - 21 x - 35 =
0
Add or subtract to move the constant term to the other side of the equation.
49 x 2 - 21 x = 35
b and reduce to simplest form. Find _ 4a 2
2
21 441 9 b 2 = _ = _ =_ _ 4a 196 4 4 49
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IN2_MNL
Lesson 9.4
Watch for the hardcover student edition page numbers for this lesson.
ing.
ete the table: -5 to compl
1, 7
395
HARDCOVER PAGES 395410
Solu
on by factor
7 and factors of
Factors of
Harcour t
IN2_MNLESE389830_U4M09L4 395
Comparing Equations
Explore
Resource Locker
hods tion Met
CORE
n Mifflin
View the Engage section online. Discuss the photo, asking how a quadratic equation could be used to help a landscaper design a stone patio. Then preview the Lesson Performance Task.
a Method Choosing Equations Quadratic
Name
© Houghto
PREVIEW: LESSON PERFORMANCE TASK
for Solving
)
4
Lesson 4 395 09/04/14
6:53 PM
07/04/14 6:39 PM
F
G
9 9 49 x 2 - 21 x + _ = 35 + _ 4 4
b to both sides of the equation, Add _ 4a 2
(
)
EXPLORE
9 149 ____ 49 x 2 - 21 x + _ = 4 4
Comparing Solution Methods for Quadratic Equations
Factor the perfect-square trinomial on the left side of the equation. 2
149 3 7 x-_ =_ 2 4
3 7 x-_=± 2
―――
H
Take the square root of both sides.
I
Subtract the constant from both sides, and then divide by a. Find both solutions for x
3
3
2
2
―――
1494
INTEGRATE TECHNOLOGY
149 _ 4
Students have the option of completing the Explore activity either in the book or online.
3
7 x=_+_=± _+_
―――
149
QUESTIONING STRATEGIES
2
Which solution method works best when there is no x-term? Taking square roots is quickest when there is no x-term. Isolate, x 2, then take the square root of both sides of the equation.
3
± _+_
2 4 7 x _ = __ 7
7
――― 149
3
4
2
_+_
±
x = __
――― 149
3
4
2
―
_+_
x = __ or
3
4
2
x = - __
7
7
J
149
_+_ © Houghton Mifflin Harcourt Publishing Company
7
Solve both equations to three decimal places using your calculator. x=
x = -0.658
or
1.086
Now use the quadratic formula to solve the same equation.
K
Identify the values of a, b, and c. a = 7 , b = -3 , c = -5
L
Substitute values into the quadratic formula.
―――――――
(
)
- -3 ± 3 - 4(7) -5 x = ___ 2
(7)
2
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Learning Progressions In this lesson, students use different solution methods to solve quadratic equations, including taking the square root, completing the square, using the quadratic formula, and factoring. It is important that students understand that the different solution methods result in the same solution. Students should choose the method that best fits the situation at hand when solving quadratic equations. This is an important skill for college readiness.
Choosing a Method for Solving Quadratic Equations
396
EXPLAIN 1
Simplify the discriminant and the denominator.
√
_
3 ± 149 x=_
14
Solving Quadratic Equations Using Different Methods
Use your calculator to finish simplifying the expression for x. x=
QUESTIONING STRATEGIES
or
1.086
x = -0.658
Reflect
1.
How is the first step in solving a quadratic equation by completing the square different from solving one by using the quadratic formula? The first step in completing the square is to place the variable terms on one side of the equation and the constant on the other. The first step in completing the square is to write the equation in standard form, ax 2 + bx + c = 0.
Discussion Another method you learned for solving quadratics is taking square roots. Why would that not work in this case? Taking square roots only works when there are is no x term in the standard form (as in
4x 2 - 9 = 0), or when the quadratic is already expressed as a perfect square of a binomial
(
)
plus constant terms as in (x + 4) - 5 = 10 .
Explain 1
2
Solving Quadratic Equations Using Different Methods
You have seen several ways to solve a quadratic equation, but there are reasons why you might choose one method over another. Factoring is usually the fastest and easiest method. Try factoring first if it seems likely that the equation is factorable. Both completing the square and using the quadratic formula are more general. Quadratic equations that are solvable can be solved using either method.
© Houghton Mifflin Harcourt Publishing Company
Example 1
Speculate which method is the most appropriate for each equation and explain your answer. Then solve the equation using factoring (if possible), completing the square, and the quadratic formula.
x 2 + 7x + 6 = 0 Factor the quadratic. Set up a factor table adding factors of c.
Substitute in factors. Use the Zero Product Property
Factors of c
Sum of Factors
1, 6
7
2, 3
5
(x + 1)(x + 6) = 0 x+1=0
Solve both equation for x.
x = -1
or or
x+6=0 x = -6
Complete the square.
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Peer-to-Peer Activity Have students work in pairs. Instruct each student to write three different equations, each one as an example of an equation that is best solved using one of the solution methods learned in this module: using square roots, completing the square, and using the quadratic formula. Have the students switch problems and solve each equation. Have them discuss why they did or did not use the methods preferred by their partners.
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x 2 + 7x = -6
Move the constant term to the right side. b to both sides. Add _ 4a Simplify. 2
Factor the perfect-square trimonial on the left.
_
7 =± 5 Take the square root of both sides. x + _ 2 2 Write two equations.
AVOID COMMON ERRORS
49 = -6 + _ 49 x 2 + 7x + _ 4 4 49 = _ 25 x 2 + 7x + _ 42 4 7 25 x+ =_ 4 2
(
Solve both equations.
Students often forget how to apply the quadratic formula to an equation that does not have an x 2-term, x-term, or constant. Remind students that the missing term has a coefficient of 0, so they can substitute 0 for the corresponding variable: a, b, or c.
_)
7 =_ 5 or x + _ 7 = -_ 5 x+_ 2 2 2 2 x = -1 or x = -6
Apply the quadratic formula. a = 1, b = 7, c = 6 -7 ± 7 2 - 4(1)(6) x = __ 2(1) -7 ± 25 x= _ 2 -7 + 5 -7 - 5 x = _ or x=_ 2 2 x = -1 or x = -6
―――――
Identify the values of a, b, and c. Substitute values into the quadratic formula.
―
Simplify the discriminant and denominator. Evaluate the square root and write as two equations. Simplify.
Because the list of possible factors that needed to be checked was short, it makes sense to try factoring x 2 + 7x + 6 first, even if you don’t know if you will be able to factor it. Once factored, the remaining steps are fewer and simpler than either completing the squares or using the quadratic formula.
B
2x 2 + 8x + 3 = 0 Factor the quadratic.
Factors of c
Sum of Inner and Outer Products
1, 2
1, 3
5
1, 2
3, 1
7
Can the quadratic be factored?
© Houghton Mifflin Harcourt Publishing Company
Factors of c
No .
Complete the square. 2x 2 + 8x = -3
Move the constant term to the right side.
Multiply both sides by 2 to make a perfect square. 4 x 2 + 16x = -6 b to both sides. Add _ 4a 2
4x 2 + 16x + 16 = 10
(2
Factor the left side.
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x+ 4
) = 10 2
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Multiple Representations Students should realize that no matter what method they use to solve a quadratic equation, the solutions are always the same. They should, therefore, choose the easiest method or the one they feel most comfortable with. Students should keep in mind, however, that not all quadratic equations can be solved by taking the square root.
Choosing a Method for Solving Quadratic Equations
398
―
Take the square root of both sides.
2x + 4 = ± √10
Write two equations.
2x + 4 = √10
―
―
or 2x + 4 = -10
―
10 x = -2 + _ or 2
Solve both equations.
―
10 x = -2 - _ 2
Apply the quadratic formula. Identify the values of a, b, and c.
a= 2 ,b= 8 ,c= 3
Substitute values into the quadratic formula.
-8 ± 8 - 4 2 3 x = ___
―――――――― 2
( 2)
( )( )
――― 2
40 -8 ± x = __ 4
Simplify the discriminant and denominator.
―――
Evaluate the square root and write as two equations.
10 -8 ± 2 x = __ 4
―
10 x = -2 ± _
Simplify.
2
Reflect
2.
What are the advantages and disadvantages of solving a quadratic equation by taking square roots? Taking square roots is the easiest approach and least likely to result in an error due to
the minimal number of computations, but it is also the most restrictive, only working for values of b = 0 in the standard form. 3.
What are the advantages and disadvantages of solving a quadratic equation by factoring? Sample answer: If the equation is not difficult to factor, factoring will usually be easier
© Houghton Mifflin Harcourt Publishing Company
than completing the squares or applying the quadratic formula. It usually requires fewer calculations, allowing for fewer opportunities to make a mistake. However, many quadratic equations that can be solved for two real solutions are not solvable by factoring. 4.
What are the advantages and disadvantages of solving a quadratic equation by completing the square? Sample answer: The method can be used on any solvable quadratic equation, but is b usually more time consuming than either factoring or the quadratic formula. If ___ is a 2a
small integer, it will involve fewer computations and simpler numbers than the quadratic formula. It may be easier to remember than the quadratic formula, because the process can be checked along the way.
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Connect Vocabulary The disregarded solution of a problem is called an extraneous solution. Help students understand what this means. Analyze the words disregard (not give regard to) and extraneous (its root word is extra).
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5.
What are the advantages and disadvantages of solving a quadratic equation by using the quadratic formula? Sample answer: The quadratic formula works for all solvable equations and is usually
EXPLAIN 2
less work than completing the square. It will be more work and more likely to result in an error from an incorrect computation than factoring or taking square roots. It is also more
Choosing Solution Methods for Quadratic Equation Models
likely to result in errors than any other technique due to misremembering the formula or copying it incorrectly (the negative signs often lead to sign errors).
AVOID COMMON ERRORS
Your Turn
Solve the quadratic equations by any method you chose. Identify the method and explain why you chose it. Sample explanations given. 6.
9x - 100 = 0 Take the square roots because b = 0. 2
7.
9x 2 = 100
3x = ± 10 10 x=± 3
_
When using the quadratic formula, some students may forget to first write the equation in standard form. Remind students that the form in which they should write the equation depends on the method they plan to use to solve it.
x 2 + 4x - 7 = 0 Complete the square, because it is not factorable, but the coefficients are small and will not lead to a lot of fractional terms.
x 2 + 4x = 7
x + 4x + 4 = 11 2
(x + 2) 2 = 11
―
x + 2 = ± √11
―
x = -2 ± √11
Explain 2
Choosing Solution Methods for Quadratic Equation Models
Recall that the formula for height, in feet, of a projectile under the influence of gravity is given by h = -16t 2 + vt + s, where t is the time in seconds, v is the upward initial velocity (at t = 0), and s is the starting height. © Houghton Mifflin Harcourt Publishing Company
Example 2
Marco is throwing a tennis ball at a kite that is stuck 42 feet up in a tree, trying to knock it loose. He can throw the ball at a velocity of 45 feet per second upward at a height of 4 feet. Will his throw reach the kite? How hard does Marco need to throw the ball to reach the kite?
Analyze Information The initial velocity is:
45
The starting height is:
4
The height of the kite is:
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Choosing a Method for Solving Quadratic Equations
400
Formulate a Plan
QUESTIONING STRATEGIES
Use the projectile motion formula to write an equation for the height of the ball t seconds after Marco throws it.
How can you use completing the square to solve a quadratic equation? Write the 2 b equation in the form x 2 + bx = c, and add to 2 both sides. Then, factor the perfect trinomial, and find the square root of both sides of the equation.
h = -16 t 2 +
( _)
45 t + 4
To determine if the ball can reach the height of the kite, set up the equation to find the time it takes the ball to reach the height of the kite. -16t 2 + 45t + 4 = 42
Convert the equation to standard form. -16t 2 + 45t + -38 = 0
How can you use the quadratic formula to solve a quadratic equation? What is one advantage of using it? Sample answer: Write the equation in standard form. Then, substitute a, b, ――― -b ± √b 2 - 4ac ____________ and c into the quadratic formula, , and 2a simplify. One advantage of using the quadratic formula is that you can quickly evaluate the discriminant, b 2 - 4ac, to determine the number of real solutions.
This problem will be easiest to solve by using the quadratic formula. To check if the ball reaches the kite, begin by calculating the discriminant . To determine the velocity that Marco must throw the ball to reach the kite, we should find the velocity where the discriminant is equal to 0 , which is the exact moment at which the ball changes direction and falls back to earth.
Solve a = -16 , b = 45 , c = -38
Identify values of a, b, and c. Evaluate the discriminant first.
2
b 2 - 4ac =
(
45 - -16
=
-407
)( -38 )
A negative discriminant means that there are 0 solutions to the equation. Rudy’s throw will/will not reach as high as the kite. The velocity with which Rudy needs to throw the ball to reach the kite is the coefficient b of the x-term of the quadratic equation.
© Houghton Mifflin Harcourt Publishing Company
b= v Substitute v into the discriminant and solve for a discriminant equal to 0 to find the velocity at which Rudy needs to throw the ball. Identify values of a, b, and c.
a = -16 , b = v , c = -38 2
Evaluate the discriminant first.
v
(
- 4 -16
)( -38 ) = 0
v 2 - -2432 = 0
Simplify.
This quadratic equation should be solved by taking square roots because it has no x-term. v2 =
Move the constant term to the right.
2432
v ≈ ± 49.3 The negative velocity represents a downward throw and will not result in the ball hitting the kite. The tennis ball must have a velocity of at about 49.3 feet per second to reach the kite. Take square roots of both sides. Use your calculator.
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Justify and Evaluate Plot the graph of Marco’s throw on your graphing calculator to see that the conclusion you reached (no solution) makes sense because the graphs do not intersect. Sketch the graph.
Then plot the height of the ball when the discriminant is equal to zero. The graphs intersect in one point.
Your Turn
8.
0 = -16t 2 + 24t + 40
0 = -8(-2t 2 - 3t - 5)
0 = -8(t + 1)(2t - 5) t+1=0
2t - 5 = 0 5 t = or 2.5 2 The wheel will not hit the ground before it falls off, so the answer must be the positive, and the time is 2.5 seconds. I chose to solve by factoring because after 8 was factored out, the number were easy to factor. t = -1
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or
or
_
402
© Houghton Mifflin Harcourt Publishing Company
The wheel of a remote controlled airplane falls off while the airplane is climbing at 40 feet in the air. The wheel starts with an initial upward velocity of 24 feet per second. How long does it take to fall to the ground? Set up the equation to determine the time and pick one method to solve it. Explain why you chose that method. h = -16t 2 + vt + s
Lesson 4
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Choosing a Method for Solving Quadratic Equations
402
9.
ELABORATE
Marco’s brother, Jessie, is helping Marco knock a kite from the tree. He can throw the ball 50 feet per second upwards, from a height of 5 feet. Is he throwing the ball hard enough to reach the kite, and if so, how long does it take the ball to reach the kite?
-16t 2 + 50t + 5 = 42
QUESTIONING STRATEGIES
-16t 2 + 50t - 37 = 0
Which method of solving a quadratic equation do you like best? Why? Answers will vary.
b - 4ac = 50 2 - 4(-16)(-37) = 132 2
t=
SUMMARIZE THE LESSON
t=
Complete the graphic organizer shown below with students to discuss and summarize lesson concepts. Write one equation that would be best solved using each solution method and explain why.
-50 ± √―― 132 __ 2(-16)
-50 + √―― 132 __ -32
or
t ≈ 1.20s
t=
-50 - √―― 132 __ -32
t ≈ 1.92s
The ball reaches the same height as the kite at about 1.20 seconds, and then again on the way back down at 1.92 seconds.
Elaborate
Solution Methods for Quadratic Equations
10. Which method do you think is best if you are going to have to use a calculator? If a calculator is needed, the best method in this situation is probably to either use the
quadratic formula or to solve graphically.
4x2 = 12 There is no x-term.
Completing the Square
9x2 + 9x = 15 it’s in the form ax2 + bx = c and a is a perfect square.
Using Quadratic Formula
2x2 - 6x + 4 = 0 it’s in the form ax2 + bx + c = 0
© Houghton Mifflin Harcourt Publishing Company • Image credits ©Jamie Wilson/Shutterstock
Taking Square Roots
11. Some factorable quadratic expressions are still quite difficult to solve by factoring rather than using another method. What makes an equation difficult to factor? Sample answer: When the a and c terms have many factors, the number of combinations of
factor pairs to try can be very large. If either a or c is a negative number, this also increases the number of combinations of factor pairs to try. 12. You are taking a test on quadratic equations and you can’t decide which method would be the fastest way to solve a particular problem. How could looking at a graph of the equation on a calculator help you decide which method to use? By graphing the related function on a calculator, I can estimate where the zeros are. If the
zeros are integers, then the equation would probably be easy to factor. If there are no zeros, I know that there is no solution. 13. Essential Question Check-In How should you determine a method for solving a quadratic equation? The choice of method depends on the equation. If taking square roots or factoring can be
used, those should be the first choices. Convert to standard form first (unless the equation starts with squares on both sides of the equation) and then determine which method to use based on the coefficients.
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Evaluate: Homework and Practice 1.
EVALUATE • Online Homework • Hints and Help • Extra Practice
Look at this quadratic equation and explain what you think will be the best approach to solving it. Do not solve the equation. 3.38x 2 + 2.72x - 9.31 = 0
This formula is expressed with two digit decimal coefficients. Factoring and completing the squares could be attempted by multiplying the equation by 100 to get to integer coefficients, but the amount of work in either method would be unreasonable. The quadratic equation, on the other hand, can be used easily with a calculator to evaluate the coefficients without any further manipulation.
ASSIGNMENT GUIDE
Solve the quadratic equation by any means. Identify the method and explain why you chose it. Irrational answers may be left in radical form or approximated with a calculator (round to two decimal places). 2.
x 2 - 7x + 12 = 0
3.
36x 2 - 64 = 0
Taking square roots, because b = 0.
Factoring, because there are not many factors to check.
36x 2 = 64
x 2 - 7x + 12 = 0
6x = ±8 4 x=± 3
_
(x - 3)(x - 4) = 0 x-3=0
or
x=3
x-4=0
or
x=4
Concepts and Skills
Practice
Explore Comparing Solution Methods for Quadratic Equations
Exercise 1
Example 1 Solving Quadratic Equations Using Different Methods
Exercises 2–17, 22–25
Example 2 Choosing Solution Methods for Quadratic Equation Models
Exercises 18–21
QUESTIONING STRATEGIES 4x 2 - 4x - 3 = 2
5.
Completing the square, because it is not factorable, but the coefficients are small and will not lead to a lot of fractional terms.
Factoring, because there are not many factors to check, and in this case, it works. 8x 2 + 9x + 1 = 0
(8x + 1)(x + 1) = 0
4x 2 - 4x = 5
4x 2 - 4x + 1 = 6
(2x - 1) 2 = 6
8x 2 + 9x + 2 = 1
8x + 1 = 0
_
x = -1 8
―
x+1=0
or or
x = -1
2x - 1 = ± √6
― _ _―
2x = 1 ± √6 √6 1 x = ± 2 2
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Exercise
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Depth of Knowledge (D.O.K.)
© Houghton Mifflin Harcourt Publishing Company
4.
Which of the methods for solving a quadratic equation give an answer every time? Using the quadratic formula and completing the square give an answer every time. Using square roots is not always possible.
COMMON CORE
Mathematical Practices
1–17
2 Skills/Concepts
MP.2 Reasoning
18–21
2 Skills/Concepts
MP.1 Problem Solving
22
2 Skills/Concepts
MP.2 Reasoning
23
3 Strategic Thinking
MP.3 Logic
24
3 Strategic Thinking
MP.3 Logic
25
3 Strategic Thinking
MP.3 Logic
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Choosing a Method for Solving Quadratic Equations
404
VISUAL CUES
6.
Have students write the general quadratic equation in standard form and then write the quadratic formula. Have them indicate how the coefficients and constants move from the equation to the formula. They can use colors, highlighting, arrows, or whatever appeals to them as long as the process is clear. A sample is shown below.
5x 2 + 0x - 13 = 0 Taking square roots, because b = 0.
7.
5x 2 = 13 √5x = ± √13
― √― 13 x = ±_ √― 5 √― 65 _
―
x=±
7x 2 - 5x - 5 = 0 The quadratic formula, because the equation cannot be factored and completing the square will be complicated.
a = 7, b = -5, c = -5 x= x=
5
――――― -(-5) ± √7 - 4(7)(-5) ___ 2(7) ―― √ 5 ± 165 _ 2
14 x = 1.27
x = -0.56
or
ax2 + bx + c = 0 8.
x=
-b ±
b2 - 4ac 2a
3x 2 - 6x = 0 Factoring, because it is easy to factor a quadratic expression with no constant term.
3x(x - 2) = 0
3x = 0
or
x=0
or
9.
2x 2 + 4x - 3 = 0 Completing the square, because it is not factorable, but the coefficients are small and will not lead to a lot of fractional terms.
2x 2 + 4x = 3
x-2=0
4x 2 + 8x = 6
x=2
4x + 8x + 4 = 10 2
(2x + 2) 2 = 10
―
© Houghton Mifflin Harcourt Publishing Company
2x + 2 = ± √10 2x = -2 ± √10 √10 x = -1 ± 2
2 2 11. (2x - 1) = x 10. (x - 5) = 16 2 Taking square roots, because the equation 4x - 4x + 1 = x is already expressed as a squared binomial 4x 2 - 5x + 1 = 0 and a constant. Factoring, because there are not too many (x - 5) 2 = 16 factors to check, and in this case, it works. x - 5 = ±4 (x - 1)(4x - 1) = 0
x-5=4
x=9
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Lesson 9.4
― ― _
or
or
x - 5 = -4
x-1=0
x=1
x=1
405
or or
4x - 1 = 0 1 x= 4
_
Lesson 4
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12. 2(x + 2) - 5 = 3
2 13. (2x - 3) = 4x
2
(2x - 3) 2 = 4x
Taking square roots, because the equation can be easily converted to a squared binomial and a constant.
4x - 12x + 9 = 4x 2
4x 2 - 16x + 9 = 0
2(x + 2) = 8 2
Completing the square, because it is not factorable, but the coefficients are small and will not lead to a lot of fractional terms.
(x + 2) 2 = 4
x + 2 = ±2 x+2=2 x=0
or
x + 2 = -2
4x 2 - 16x = -9
x = -4
or
4x - 16x + 16 = -9 + 16 2
(2x - 4)2 = 7
― √― 7 x=2±_
2x - 4 = ± √7
2
14. 6x 2 - 5x + 12 = 0
15. 3x 2 + 6x + 2 = 0
The quadratic formula, because the equation cannot be factored and completing the square will be complicated.
Completing the square, because it is not factorable, but the coefficients are small and will not lead to a lot of fractional terms.
―――――― x = ___ 2(5) 5 ± √―― -263 x = __ a = 6, b = -5, c = 12
3x 2 + 6x = -2
2 -(-5) ± √(-5) - 4(6)(12)
9x 2 + 18x = -6
9x + 18x + 9 = -6 + 9 2
(3x + 3) 2 = 3
10 The discriminant is negative, so there is no real solution.
5 =0 1 x 2 + 3x + _ 16. _ 2 2
x = -1 ±
6x 2 + 13x - 19 = 0 The quadratic formula, because factoring and completing the square may be timecomsuming.
or or
x+5=0
a = 6, b = 13, c = -19
x = -5
――――――― -(13) ± √(13) - 4(6)(-19) ___ 2(6) ―― -13 ± √625 __ 2
x= x=
12
-13 + 25 x =_
or
x=1
or
12
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406
-13 - 25 _ 12 19 x = -_
© Houghton Mifflin Harcourt Publishing Company
x = -1
3
6x 2 + 13x + 7 = 26
x 2 + 6x + 5 = 0 x+1=0
√― 3 _
17. (6x + 7)(x + 1) = 26
Factoring, because after multiplying by 2 to eliminate the fractional coefficients, it is apparent that will be easy to factor.
(x + 1)(x + 5) = 0
―
3x + 3 = ± √3
x=
6
Lesson 4
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Choosing a Method for Solving Quadratic Equations
406
Use the projectile motion formula and solve the quadratic equation by any means. Identify the method and explain why you chose it. Irrational answers and fractions should be converted to decimal form and rounded to two places. 18. Gary drops a pair of gloves off of a balcony that is 64 feet high down to his friend on the ground. How long does it take the pair of gloves to hit the ground?
v=0
s = 64 h = 0
-16t 2 + 64 = 0
Because there is no initial velocity, there is no t term and the solution can be solved by finding square roots. -16t 2 = -64 16t 2 = 64
4t = ±8 4t = 8 t=2
4t = -8
or
t = -2
or
The balloon hits the ground after it is dropped, not before, so the answer must be the positive time: 2 seconds after it is dropped. 19. A soccer player jumps up and heads the ball while it is 7 feet above the ground. It bounces up at a velocity of 20 feet per second. How long will it take the ball to hit the ground?
v = 20 s = 7 h = 0
-16t 2 + 20t + 7= 0
© Houghton Mifflin Harcourt Publishing Company • Image credits ©Ijansempoi/Shutterstock
The quadratic formula, because the equation cannot be factored and completing the square will be complicated. a = -16, b = 20, c = 7
t=
-32 t = -0.29
-20 - √―― 848 _
or
t=
or
t = 1.54
-32
The ball hits the ground after it is headed, not before, so the answer must be the positive time: 1.54 seconds after it is headed.
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Lesson 9.4
2(-16)
-20 ± √―― 848 t = __ -32 -20 + √―― 848 __
Module 9
407
―――――― -(20) ± √(-5) - 4(16)(7) ___ 2
t=
407
Lesson 4
4/2/14 1:31 AM
20. A stomp rocket is a toy that is launched into the air from the ground by a sudden burst of pressure exerted by stomping on a pedal. If the rocket is launched at 24 feet per second, how long will it be in the air?
INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 Point out to students that completing the
v = 24 s=0
h=0
-16t 2 + 24t = 0
square and using the quadratic formula to solve a quadratic equation should produce the same solutions. Remind them that using multiple solution methods is a good way to check their answers.
Factoring, because the quadratic expression does not have a constant term. -8t(2t - 3) = 0 -8t = 0
or
t=0
2t - 3 = 0
t = 1.5
or
The rocket goes up at first, so the answer cannot be 0. It is in the air for 1.5 seconds. 21. A dog leaps off of the patio from 2 feet off of the ground with an upward velocity of 15 feet per second. How long will the dog be in the air? v = 15 s=2
h=0
-16t 2 + 15t + 2 = 0 The quadratic formula, because the equation cannot be factored and completing the square will be complicated. a = -16, b = 15, c = 2
――――――― -(15) ± √(15) - 4(-16)(2) ___ 2(-16) ―― √ -15 ± 353 __ 2
t= t=
-32
―― __
or
t=
t = 1.06
-15 - √―― 353 _ -32
© Houghton Mifflin Harcourt Publishing Company
-15 + √353 t = -32 t = -0.12 or
The dog will be in the air for a positive amount of time, so the answer is 1.06 seconds. 22. Multipart Classification Indicate whether the following statements about finding solutions to quadratic equations with integer coefficients are true or false. a. Any quadratic equation with a real solution can be solved by using the quadratic formula.
True
False
b. Any quadratic equation with a real solution can be solved by completing the square.
True
False
c. Any quadratic equation with a real solution can be solved by factoring.
True
False
d. Any quadratic equation with a real solution can be solved by taking the square root of both sides of the equation.
True
False
True
False
True
False
e. If the equation can be factored, it has rational solutions. f. If the equation has only one real solution, it cannot be factored. Module 9
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JOURNAL
H.O.T. Focus on Higher Order Thinking
23. Justify Reasoning Any quadratic equation with a real solution can be solved with the quadratic formula. Describe the kinds of equations where that would not be the best choice, and explain your reasoning.
Have students write about their preferred methods of solving quadratic equations. Have them describe the method’s advantages and disadvantages.
Equations that can be solved by taking square roots or by factoring are usually solved that way because there is less computation involved in both of those methods. The answer will usually be found in less time and with fewer errors. If an equation cannot be factored, but can be solved by completing the square without large or fractional terms, it will probably be easier to solve by completing the square rather than using the quadratic formula.
24. Critique Reasoning Marisol decides to solve the quadratic equation by factoring 21x 2 + 47x - 24 = 0. Do you think she chose the best method? How would you solve this equation?
Answers will vary: Some students will say to factor because it can be factored, others will notice that it requires checking a lot of factor pairs and point out that it may be better to go straight to the quadratic equation. 25. Communicate Mathematical Ideas Explain the difference between the statements “The quadratic formula can be used to solve any quadratic equation with a real solution” and “Every quadratic equation has a real solution.”
© Houghton Mifflin Harcourt Publishing Company
The second statement is false. Some quadratic equations cannot be solved for a real value of x. If there are real solutions, then the quadratic formula can be used to find them. Either way, the first statement is true.
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Lesson 9.4
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Lesson Performance Task AVOID COMMON ERRORS
A landscaper is designing a patio for a customer who has several different ideas about what to make.
Students may make computational errors under the radical, given the various negative signs. Some students may find it helpful to circle everything after b 2, and evaluate it separately before adding.
Use the given information to set up a quadratic equation modeling the situation and solve it using the quadratic formula. Then determine if another method for solving quadratic equations would have been easier to use and explain why.
a. One of the customer’s ideas is to buy bluestone tiles from a home improvement store using several gift cards he has received as presents over the past few years. If the total value of the gift cards is $6500 and the bluestone costs $9 per square foot, what are the dimensions of the largest patio that can be made that is 12 feet longer than it is wide?
INTEGRATE MATHEMATICAL PRACTICES Focus on Technology MP.5 Have students check their answers for
b. Another of the customer’s ideas is simply a quadratic equation scrawled on a napkin.
reasonableness by graphing the equations. The x-intercepts should appear to be close to their calculated answers.
x 2 - 54x + 720 = 0 c. The third idea is also a somewhat random quadratic polynomial. x 2 - 40x + 257 = 0
a. The amount of bluestone is
_ 6500 _ = 722.2 feet. 9
The area of the patio is x 2 + 12x = 722 or x 2 + 12x - 722 = 0. x=
―――――― -12 ± √12 - 4(1)(-722) ___
≈ 21.53
2
x ≈ -6 - 27.53
or
© Houghton Mifflin Harcourt Publishing Company • Image credits ©Ozgur Coskun/Shutterstock
x ≈ -6 + 27.53
2
≈ -33.53
Therefore, the patio will be approximately 21.5 feet by 33.53 feet.
――――――― -(-54) ± √(-54) - 4(1)(720) ___ b. x = 2
x = 27 + 3 = 30
2 or
x = 27 - 3 = 24
Therefore, the patio will be 30 feet by 24 feet. The factored polynomial is given as (x - 24)(x - 30) = 0.
――――――― -(-40) ± √(-40) - 4(1)( 397) ____ 2(1) 40 ± √――――― 1600 - 1588 __ = 2 40 ± √― 12 _ = x = 20 + √― 3 2 ― 40 ± 2 √3 _ ≈ 20 + 1.73 = 2 ― ≈ 21.73 = 20 ± √3 2
c. x =
―
x = 20 - √3 or
≈ 20 - 1.73 ≈ 18.27
Therefore, the patio will be 21.7 feet by 18.3 feet. Module 9
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Lesson 4
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Have students use what they have learned about linear inequalities to graph the solutions to Part B in the Lesson Performance Task had the equation been the inequality x 2 - 54x + 720 ≤ 0. Students will find that because this is an inequality in one variable, a number line can be used to graph its solution, 24 ≤ x ≤ 30.
4/2/14 1:31 AM
Scoring Rubric 2 points: Student correctly solves the problem and explains his/her reasoning. 1 point: Student shows good understanding of the problem but does not fully solve or explain his/her reasoning. 0 points: Student does not demonstrate understanding of the problem.
Choosing a Method for Solving Quadratic Equations
410
LESSON
9.5
Name
Solving Nonlinear Systems
9.5
Solving Nonlinear Systems
Explore
The student is expected to: A-REI.C.7
Solve a simple system consisting of a linear equation and a quadratic equation in two variables algebraically and graphically.
Determining the Possible Number of Solutions of a System of Linear and Quadratic Equations
The graph of the quadratic function ƒ(x) = x 2 - 2x - 2 is shown. On the same coordinate plane, graph the following linear functions:
MP.1 Problem Solving
g (x) = -x - 2, h (x) = 2x - 6, j (x) = 0.5x - 5
Language Objective 2 -4
-2
PREVIEW: LESSON PERFORMANCE TASK View the Engage section online. Discuss students’ experiences of having another driver enter traffic in front of them while they are driving or riding in a car. Then preview the Lesson Performance Task.
h(x) = 2x -6 x 2
-2
ENGAGE
4 j(x) = 0.5x -5
-4
© Houghton Mifflin Harcourt Publishing Company
You can graph both equations and find the points of intersection, if there are any. Create a graph by hand and read approximate solutions from it, or use a graphing calculator to find precise intersection points. You can also find the solution algebraically by setting the two functions equal and solving the resulting quadratic equation.
y
4
g(x) = -x -2
Explain to a partner how to solve a system consisting of a quadratic equation and a linear equation by graphing.
Essential Question: How can you solve a system of equations when one equation is linear and the other is quadratic?
Look at the graph of the system consisting of the quadratic function, ƒ (x), and the linear function, g (x). Based on the intersections of these two graphs, how many solutions exist in a 2 system consisting of these two functions?
Look at the graph of the system consisting of the quadratic function, ƒ (x), and the linear function, h (x). Based on the intersections of these two graphs, how many solutions exist in a 1 system consisting of these two functions?
Look at the graph of the system consisting of the quadratic function, ƒ (x), and the linear function, j (x). Based on the intersections of these two graphs, how many solutions exist in a 0 system consisting of these two functions?
Reflect
1.
A system consisting of a quadratic equation and a linear equation can have solutions.
Module 9
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A system of one linear and one quadratic equation may have zero, one, or two solutions.
Mathematical Practices COMMON CORE
Date
Essential Question: How can you solve a system of equations when one equation is linear and the other is quadratic?
Common Core Math Standards COMMON CORE
Class
411
09/04/14
7:25 PM
07/04/14 6:37 PM
Solving a System of Linear and Quadratic Equations Graphically
Explain 1
EXPLORE
A system of equations consisting of a linear and quadratic equation can be solved graphically by finding the points where the graphs intersect.
Determining the Possible Number of Solutions of a System of Linear and Quadratic Equations
Solve the system of equations graphically.
Example 1
⎧ y = (x + 1) 2 - 4 ⎨ ⎩ y = 2x - 2 Graph the quadratic function. The vertex is the point (–1, –4). The x-intercepts are the points where y = 0.
4
(x + 1) 2 - 4 = 0
INTEGRATE TECHNOLOGY
2
Introduce students to solving systems of equations by using a graphing calculator to graph the systems and find the intersection points. To verify each point of intersection, students can substitute its x-value into both equations and check that the y-value is the same for both equations.
x
(x + 1) 2 = 4
-4
x + 1 = ±2 x=1
y
-2
0
2
-2
4
x = -3
or
Graph the linear function on the same coordinate plane. The solutions of the system are the points where the graphs intersect. The solutions are (–1, –4) and (1, 0). ⎧ y = 2(x - 2) - 2 ⎨ ⎩ y = -x - 1 Graph the quadratic function. The vertex is the point ⎛ ⎞ ⎜2, -2 ⎟. ⎝ ⎠ The x-intercepts are the points where y = 0. 2
6
2
2
x
2(x - 2) = 2 2
-4
0 -2
2
4
-4
= ±1 or
-2
x= 1
Graph the linear function on the same coordinate plane. There are
0
intersection points. This system has
0
If the graph of a quadratic function intersects a horizontal line at exactly one point, what must be true about that point? It must be the vertex of the parabola.
© Houghton Mifflin Harcourt Publishing Company
(x - 2) 2 = 1 x= 3
How does graphing a system of equations help you find the number of solutions of the system? The number of points at which the graphs intersect is the number of solutions.
4
2(x - 2) - 2 = 0
x-2
QUESTIONING STRATEGIES
y
solution(s).
EXPLAIN 1 Solving a System of Linear and Quadratic Equations Graphically
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Integrate Mathematical Practices
This lesson provides an opportunity to address Mathematical Practice MP.1, which calls for students to “make sense of problems and persevere in solving them.” Students consider the meaning of a problem while planning how to find a solution. Students use graphs to determine the number of solutions for a system of equations. They solve systems by graphing manually or with a calculator, and they also solve systems algebraically. Finally, students interpret the solutions in the context of real-world situations.
4/2/14 1:33 AM
INTEGRATE MATHEMATICAL PRACTICES Focus on Patterns MP.8 Discuss with students how symmetry can be useful in graphing a quadratic function in vertex form. Students should understand that chosen points can be reflected across the axis of symmetry.
Solving Nonlinear Systems 412
Your Turn
QUESTIONING STRATEGIES Why do the points where the graphs of two functions intersect represent the solutions to the system of equations? The coordinates of the intersection points are the pairs of values that satisfy both equations.
2.
QUESTIONING STRATEGIES When finding solutions algebraically, how can you check your answers? Substitute the x-values of the solutions into both original equations and verify that the resulting y-values are the same for both equations.
413
Lesson 9.5
4
x
3.
-6
-4
0
-2
2
y
⎧ y = (x + 1) 2 - 9 ⎨ ⎩ y = 6x - 12
x -4
Solution: (2, 0)
-2
0 -2
2
4
-4 -6 -8
INTEGRATE MATHEMATICAL PRACTICES Focus on Modeling MP.4 Students who are comfortable with solving
Explain 2
Solving a System of Linear and Quadratic Equations Algebraically
Systems of equations can also be solved algebraically by using the substitution method to eliminate a variable. If the system is one linear and one quadratic equation, the equation resulting after substitution will also be quadratic and can be solved by selecting an appropriate method. Example 2 © Houghton Mifflin Harcourt Publishing Company
When solving a system, students sometimes stop after finding the value(s) of one of the variables. Remind them that solutions of a system are ordered pairs. They need to substitute the value(s) they find into one of the original equations and solve for the other variable.
Solution: (-4, 0) or (-2, 8)
6
-8
Solving a System of Linear and Quadratic Equations Algebraically
AVOID COMMON ERRORS
⎧ y = -2(x + 2) + 8 ⎨ ⎩ y = 4x + 16
2
EXPLAIN 2
systems of linear and quadratic equations by graphing may not see the value in using algebra to find solutions. Discuss reasons for solving algebraically: solutions that are not on grid lines can be hard to read, and the intersection may be located outside of the portion of the coordinate grid shown.
y 8
2
Solve the system of equations algebraically.
⎧ y = (x + 1) 2 - 4 ⎨ ⎩ y = 2x - 2 Set the two the expressions for y equal to each other, and solve for x.
(x + 1) 2 - 4 = 2x - 2 x 2 + 2x - 3 = 2x - 2 x2 - 1 = 0 x2 = 1 x = ±1 Substitute 1 and -1 for x to find the corresponding y-values. y = 2x - 2
y = 2x - 2
y = 2(1) - 2 = 0
y = 2(-1) - 2 = -4
The solutions are (1, 0) and (-1, -4). Module 9
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Peer-to-Peer Activity Have students work in pairs. Ask each student to write a quadratic function, and have partners graph each other’s functions. Next, have students draw three parallel lines on the graphs they created: one that does not intersect the graph of the quadratic function, one that intersects it at one point, and one that intersects it at two points. Have students write an equation for each line and identify the coordinates of each point of intersection.
4/2/14 1:33 AM
B
⎧ y = (x + 4)(x + 1) ⎨ ⎩ y = -x - 5 Set the two the expressions for y equal to each other, and solve for x.
(x + 4)(x + 1) = -x - 5 x + 5 x + 4 = -x - 5 2
x2 + 6 x + 9 = 0
(x + 3 )(x + 3) = 0
x = -3
Substitute -3 for x to find the corresponding y-value. y = -x - 5
( )
y = - -3 - 5 = -2 The solution is (-3, -2). Reflect
4.
Discussion After finding the x-values of the intersection points, why use the linear equation to find the y-values rather than the quadratic? What if the quadratic equation is used instead? The linear equation will usually be easier to evaluate, having fewer operations. Both equations give the same y-values.
Your Turn
5.
⎧ y = 2x 2 + 9x + 5 ⎨ ⎩ y = 3x - 3
2x 2 + 9x + 5 = 3x - 3 x 2 + 3x + 4 = 0 x = x =
© Houghton Mifflin Harcourt Publishing Company
2x 2 + 6x + 8 = 0
―――――
2 -(3) ± √(3) - 4(1)(4) 2(1)
――
-(3) ± √- 7 2
The discriminant is negative, so there are no real solutions.
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Critical Thinking
4/2/14 1:33 AM
Initiate a discussion about how many points of intersection are possible when a system consists of two quadratic equations. Have students sketch graphs representing all of the possibilities. Students should find that the graphs of two quadratic functions can intersect at 0, 1, or 2 points. Discuss what similarities exist among systems that have the same number of solutions.
Solving Nonlinear Systems 414
Explain 3
EXPLAIN 3
Systems of equations can be solved by graphing both equations on a graphing calculator and using the Intersect feature.
Solving a Real-World Problem with a System of Linear and Quadratic Equations
Example 3
AVOID COMMON ERRORS When using a graphing calculator to graph a linear-quadratic system, students may find that the graph shows no intersection points and conclude that the system has no solutions. It is possible that the two graphs actually do intersect but that the domain or range of the graphs needs to be extended to include the intersection point. Caution students that they need to use a suitable viewing window to look for intersections before they come to a conclusion.
students that solutions involving negative numbers can often be discarded, but not always. Give students examples in which negative solutions are valid, such as distance above or below sea level or time before or after an event. Remind students that it is important to look at each context to check whether or not a solution is valid.
Create and solve a system of equations to solve the problem.
A rock climber is pulling his pack up the side of a cliff that is 175.5 feet tall at a rate of 2 feet per second. The height of the pack in feet after t seconds is given by h = 2t. The climber drops a coil of rope from directly above the pack. The height of the coil in feet after t seconds is given by h = -16t 2 + 175.5. At what time does the coil of rope hit the pack? Create the system of equations to solve. ⎧ h = -16t 2 + 175.5 ⎨ ⎩ h = 2t Graph the functions together and find any points of intersection.
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Andresr/ Shutterstock
INTEGRATE MATHEMATICAL PRACTICES Focus on Modeling MP.4 When solving real-world problems, remind
Solving a Real-World Problem with a System of Linear and Quadratic Equations
The intersection is at (-3.375, -6.75).
The intersection is at (3.25, 6.5).
The x-value represents time, so this solution is not reasonable.
This solution indicates that the coil hits the pack after 3.25 seconds.
A window washer is ascending the side of a building that is 520 feet tall at a rate of 3 feet per second. The elevation of the window washer after t seconds is given by h = 3t. The supplies are lowered to the window washer from the top of the building at the same time that he begins to ascend the building. The height of the supplies in feet after t seconds is given by h = -2t 2 + 520. At what time do the supplies reach the window washer? Create the system of equations to solve. ⎧
⎨ ⎩
h = -2 t 2 + 520 h =
3 t
QUESTIONING STRATEGIES How can you determine in which direction a parabola will open? If the coefficient of the x 2 term of the equation is positive, the parabola will open upward. If the coefficient of the x 2 term is negative, the parabola will open downward.
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Lesson 5
LANGUAGE SUPPORT IN2_MNLESE389830_U4M09L5.indd 415
Connect Vocabulary When solving real-world problems, students may encounter words describing occupations or pastimes. Point out to English learners that in English, nouns that name an occupation often end with -er. Have them think of occupation words that could be used in a word problem, such as teacher, bus driver, and football player.
4/2/14 1:33 AM
Graph the functions together and find any points of intersection.
⎛ ⎞ The intersection is at about ⎜ -16.9 , -50.7 ⎟. ⎝ ⎠ The x-value represents time , so this
solution is
not reasonable
⎛ ⎞ The intersection is at about ⎜ 15.4 , 46.2 ⎟. ⎝ ⎠ This solution indicates that the supplies
reach the window washer about .
15 seconds later
Reflect
6.
How did you know which intersection to use in the example problems? In both cases, the other intersection has a negative value of t before the moving objects
in each problem begin to move. The equations describing the object heights only apply after t = 0. Your Turn
7.
A billboard painter is using a pulley system to hoist a can of paint up to a scaffold at a rate of half a meter per second. The height of the can of paint as a function of time is given by h(t) = 0.5t. Five seconds after he starts raising the can of paint, his partner accidentally kicks a paint brush off of the scaffolding, which 2 falls to the ground. The height of the falling paint brush can be represented by h(t) = -4.9 (t - 5) + 30. When does the brush pass the paint can?
The paint brush passes by the can about 7.3 seconds after the painter starts hoisting it up or about 2.3 seconds after the paintbrush starts to fall.
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© Houghton Mifflin Harcourt Publishing Company
⎧ h(t) = -4.9(t - 5)2 + 30 ⎨ ⎩ h(t) = 0.5t
Lesson 5
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Elaborate
ELABORATE
8.
INTEGRATE MATHEMATICAL PRACTICES Focus on Patterns MP.8 Review the standard form and vertex form of
Discussion When solving a system of equations consisting of a quadratic equation and a linear equation by graphing, why is it difficult to be sure there is one solution as opposed to 0 or 2? For the system to have one solution, the line must just touch the parabola without crossing it. A slight translation closer and they will cross twice, while a slight translation away means they will miss entirely.
9.
a quadratic function. Understanding how to quickly find the vertex and axis of symmetry from the vertex form of the equation can make it easier for students to solve nonlinear systems by graphing.
How can you use the discriminant to determine how many solutions a linear-quadratic system has? After using substitution, you will have one quadratic equation to solve. The number of
solutions of the system is equal to the number of solutions of that resulting quadratic equation. 10. Essential Question Check-in How can the graphs of two functions be used to solve a system of a quadratic and a linear equation? The x and y coordinates of the points of intersection of the two curves are solutions to the
system of equations.
SUMMARIZE THE LESSON
Evaluate: Homework and Practice 1.
The graph of the function ƒ (x) = -_14 (x - h) + 4 is shown. Graph the functions g(x) = x + 1, h(x) = x + 2, and j(x) = x + 3 with the graph of ƒ (x), and determine how many solutions each system has. 2
1
ƒ (x) and h (x): ƒ (x) and j (x):
0
2 0 -2
2
4
6
x
Solve each system of equations graphically. 2.
⎧ y = (x + 3) 2 - 4 ⎨ ⎩ y = 2x + 2
4
y
3.
2
(-3, -4), (-1, 0)
x -6
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Lesson 9.5
• Online Homework • Hints and Help • Extra Practice
y
4
-2
-4
-2
⎧ y = x2 - 1 ⎨ ⎩y = x - 2
4
-2
y
2
No Solutions
x -4
0
-2
0 -2
2
4
-4
-4
417
6
2
ƒ (x) and g (x): © Houghton Mifflin Harcourt Publishing Company
How can you solve a system of a quadratic and a linear equation algebraically? Rewrite each equation so that y is given in terms of x, then set the two expressions for y equal to each other. Simplify the resulting quadratic equation, then solve it for x by factoring, taking a square root, completing the square, or using the quadratic formula. Substitute each x-value into one of the original equations to find the corresponding y-value.
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4.
⎧ y = (x - 4) 2 - 2 ⎨ ⎩ y = -2
4
y
5.
2
(4, -2)
0
2
-2
4
4
⎧ y = -(x - 2)2 + 9 ⎨ ⎩ y = 3x + 3
-2
0 -2
⎧ y = 3(x + 1) - 1 ⎨ ⎩y = x - 4 2
7.
8 6
4
4 2
x -4
-2
2 2
0
2
-2
4
-4
x 0
ASSIGNMENT GUIDE
y
No Solution
4
-2
2
-4
y
(-1, 0) (2, 9)
EVALUATE x
-4
6
-4
6.
y
2
(2, 0) (1,3)
x -2
⎧ y = -x 2 + 4 ⎨ ⎩ y = -3x + 6
4
Solve the system of equations algebraically. 8.
⎧ y = x2 + 1 ⎨ ⎩y = 5
9.
⎧ y = x 2 - 3x + 2 ⎨ ⎩ y = 4x - 8
x 2 - 3x + 2 = 4x - 8
x2 + 1 = 5
x 2 - 7x + 10 = 0
x2 = 4
(x - 2)(x - 5) = 0
x = ±2
x=2
y=5
(2, 0) (5, 12)
⎧ y = (x - 3)2 10. ⎨ ⎩y = 4
-x 2 + 4x = x + 2 x - 3x + 2 = 0
x - 3 = ±2 or
(1, 4) (5, 4)
y = 4(5) - 8 = 12
⎧ y = -x 2 + 4x 11. ⎨ ⎩y = x + 2
(x - 3)2 = 4 x=5
x=5
2
(x - 2)(x - 1) = 0
x=1
x=2
or
y=2+2=4
(2, 4) (1, 3) Module 9
Exercise
Depth of Knowledge (D.O.K.)
y=1+2=3
Mathematical Practices
1–7
2 Skills/Concepts
MP.7 Using Structure
8–15
2 Skills/Concepts
MP.6 Precision
16–19
2 Skills/Concepts
MP.4 Modeling
20
2 Skills/Concepts
MP.6 Precision
21
2 Skills/Concepts
MP.3 Logic
22
2 Skills/Concepts
MP.7 Using Structure
Explore Determining the Possible Number of Solutions of a System of Linear and Quadratic Equations
Exercises 1–22, 26
Example 1 Solving a System of Linear and Quadratic Equations Graphically
Exercises 2–7
Example 2 Solving a System of Linear and Quadratic Equations Algebraically
Exercises 8–15, 20, 25
Example 3 Solving a Real-World Problem with a System of Linear and Quadratic Equations
Exercises 16–19, 21, 23–24
method to use for solving a system of equations. For example, they may prefer to solve algebraically when the equations involve small, easily factored numbers, and to graph with a calculator when the problem involves decimals or very large numbers. The choice of whether to solve a quadratic equation by factoring or by using the quadratic formula may also depend on the coefficients in the equations, but approaches will vary from one student to another.
Lesson 5
COMMON CORE
Practice
INTEGRATE MATHEMATICAL PRACTICES Focus on Communication MP.5 Have students share how they decide which
x=1
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(2, 5) (-2, 5)
or
y = 4(2) - 8 = 0
Concepts and Skills
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Solving Nonlinear Systems 418
⎧ y = 2x 2 - 5x + 6 12. ⎨ ⎩ y = 5x - 6
MODELING
⎧ y = x 2+ 7 13. ⎨ ⎩ y = -9x + 29
2x 2 - 5x + 6 = 5x - 6
Review the different forms of a linear equation. Students should understand that linear equations presented in slope-intercept form and in point-slope form will provide them with information that can be helpful in graphing the function.
x 2 + 7 = -9x + 29
x 2 - 5x + 6 = 0
x 2 + 9x - 22 = 0
(x - 2)(x - 3) = 0 x=2
or
y = 5 (2 ) - 6 = 4
(2, 4) (3, 9)
AVOID COMMON ERRORS
(x - 2)(x + 11) = 0
x=3
x=2
y = 5 (3 ) - 6 = 9
y = -9(2) + 29 = 11
(2, 11) (-11, 128)
x = -11
y = -9(-11) + 29 = 128
⎧ y = (x + 2)(x + 4) 15. ⎨ ⎩ y = 3x + 2
⎧ y = 4x 2 + 45x + 83 14. ⎨ ⎩ y = 5x - 17
When students are solving a system of equations algebraically, remind them that the solutions will be ordered pairs. Students may feel they have found the solution after finding the value of one variable, but they must substitute the value(s) they find into one of the original equations in order to find the other value of the ordered pair.
or
(x + 2)(x + 4) = 3x + 2
4x 2 + 45x + 83 = 5x - 17 x + 10x + 25 = 0
x 2 + 3x + 6 = 0
2
(x + 5)(x + 5) = 0
x=
-3 ± √―― -15 __
2 There are no real solutions.
x = -5
y = 5(-5) - 17 = -42
(-5, -42)
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Germanskydiver/Shutterstock
Create and solve a linear quadratic system to solve the problem. 16. The height in feet of a skydiver t seconds after deploying her parachute is given by h(t) = -300t + 1000. A ball is thrown up toward the skydiver, and after t seconds, the height of the ball in feet is given by h(t) = -16t 2 + 100t. When does the ball reach the skydiver?
⎧h(t) = -16t 2 + 100t
⎨
⎩ h(t) = -300t + 1000
Two intersections: (2.82, 154.74) and (22.18, -5654.74) The ball reaches the skydiver at 2.82 seconds.
17. A wildebeest fails to notice a lion that is charging from behind at 65 feet per second until the lion is 40 feet away. The lion’s position as a function of time is given by p(t) = 65t - 40. The wildebeest has to begin accelerating from a standstill until it is captured or reaches a top speed fast enough to stay ahead of the lion. The wildebeest’s position as a function of time is given by d(t) = 35t 2. Does the wildebeest escape?
⎧p(t) = 65t - 40
⎨
⎩ d(t) = 35t
2
The lion will catch the wildebeest if their positions are ever the same, which is indicated by an intersection. The graphs to do not intersect, so the wildebeest escapes. Module 9
Exercise
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Lesson 9.5
Lesson 5
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Depth of Knowledge (D.O.K.)
COMMON CORE
Mathematical Practices
23–24
3 Strategic Thinking
MP.4 Modeling
25
3 Strategic Thinking
MP.6 Precision
26
3 Strategic Thinking
MP.3 Logic
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18. An elevator in a hotel moves at 20 feet per second. Leaving from the ground floor, its height in feet after t seconds is given by the formula h(t) = -20t. A bolt comes loose in the elevator shaft above, and its height in feet after falling for t seconds is given by h(t) = -16t 2 + 200. At what time and at what height does the bolt hit the elevator?
KINESTHETIC EXPERIENCE After they create graphs for a system of equations, have students trace the paths of both graphs using their index fingers or colored pencils. They should note the shape of each graph and how the parabola turns toward and away from the line at different points. This process may help students recognize situations in which the graphs are likely to intersect if they are extended beyond the grid.
⎧h(t) = -16t 2 + 200
⎨
⎩ h(t) = 20t
2 intersections: (2.97, 59.31) and (-4.22, -84.31)
The bolt hits the elevator 2.97 seconds later, at a height of 59.31 feet.
19. A bungee jumper leaps from a bridge 100 meters over a gorge. Before the 40-meter-long bungee begins to slow him down, his height is characterized by h(t) = -4.9t 2 + 100. Two seconds after he jumps, a car on the bridge blows out a tire. The sound of the tire blow-out moves down from the top of the bridge at the speed of sound and has a height given by h(t) = -340(t - 2) + 100. How high will the bungee jumper be when he hears the sound of the blowout?
⎧h(t) = -4.9t 2 + 100
CRITICAL THINKING
⎨
⎩ h(t) = -340t + 780
When students are working with real-world situations, remind them that they have to read the problem carefully and analyze the given information to determine whether the x-values or the y-values represent the solution to the problem.
The graphs intersect at about (2.1, 79.2), so the jumper hears the tire blow-out at a height of about 79.2 meters. (There is another intersection at a negative height, so it is unreasonable). 20. Explain the Error A student is asked to solve the system of equations y = x 2 + 2x - 7 and y - 2 = x + 1. For the first step, the student sets the right hand sides equal to each other to get the equation x 2 + 2x - 7 = x + 1. Why does this not give the correct solution?
Before equating the two right sides, both equations must be solved for y. 21. Explain the Error After solving the system of equations in Evaluate 18 (the elevator and the bolt), a student concludes that there are two different times that the bolt hits the elevator. What is the error in the student’s reasoning?
22. Multi-part Classification
The functions listed are graphed here. 2 3 (x - 2) 2 + 3 f 1(x) = 2(x + 3) + 1 and f 2(x) = -_ 4 1x + 1 g 1(x) = x + 3 and g 2(x) = 3 and g 3(x) = -_ 2 Use the graph to classify each system as having 0, 1, or 2 solutions.
a. ⎧ y = ƒ 1(x)
b. ⎧ y = ƒ 1(x)
c. ⎧ y = ƒ 1(x)
0
2
2
⎨ ⎩ y = g 1(x)
d. ⎧ y = ƒ 2(x)
e.
0
1
⎨ ⎩ y = g 1(x)
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⎨ ⎩ y = g 2(x) ⎧ y = ƒ 2(x) ⎨ ⎩ y = g 2(x)
f. 2
4
y
x -2
0
2
⎨ ⎩ y = g 3(x)
© Houghton Mifflin Harcourt Publishing Company
One solution has a negative value of t, which would mean the bolt hit the elevator before it began to fall.
⎧ y = ƒ 2(x) ⎨ ⎩ y = g 3(x)
420
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Solving Nonlinear Systems 420
JOURNAL
H.O.T. Focus on Higher Order of Thinking
23. Explain the Error After solving the system of equations in Evaluate 16 (the skydiver and the ball), a student concludes there are two valid solutions because they both have positive times. The ball must pass by the skydiver twice. What is the error in the student’s reasoning? The second intersection point at 22.18 seconds corresponds to a height -5654.74 feet, or under the ground. The second intersection point is outside the range of the model even though it has a positive time, and it cannot be a valid solution.
In their journals, have students sketch graphs of linear-quadratic systems with no solutions, with one solution, and with two solutions. Then have them describe how to solve the systems algebraically as well as by graphing.
24. Multi-Part Problem The path of a baseball hit for a home run can be modeled by x 2 + x + 3, where x and y are in feet and home plate is at the origin. y = -_ 484 The ball lands in the stands, which are modeled by 4y - x = -352 for x ≥ 400. Use a graphing calculator to graph the system. a. What do the variables x and y represent? x is the horizontal distance from home plate. y is the height above the ground. b. About how far is the baseball from home plate when it lands?
The only intersection point with non-negative coordinates is (458.96, 26.74). So the baseball is about 459 feet from the plate when it lands.
c.
About how high up in the stands does the baseball land? 26.7 feet.
25. Draw Conclusions A certain system of a linear and a quadratic equation has two solutions, (2, 7) and (5, 10). The quadratic equation is y = x 2 - 6x + 15. What is the linear equation? Justify your answer.
The two solutions are points on the line and can be used to solve for the line. y = mx + b 10 - 7 =1 m= 10 = 1(5) + b 5-2 b=5 The line is y = x + 5. Two points define a line, so there is no need to use the quadratic equation.
© Houghton Mifflin Harcourt Publishing Company
_
26. Justify Reasoning It is possible for a system of two linear equations to have infinitely many solutions. Explain why this is not possible for a system with one linear and one quadratic equation.
Two linear equations can have infinite solutions if the linear functions associated represent the same line, which will overlap itself at every point on the line. A parabola and a line are different shapes and can intersect only at certain points. Since the parabola can only change direction once, it cannot cross the line more than twice.
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CONNECT VOCABULARY
Lesson Performance Task
Some students may not be familiar with the terms pit row and pace car. Ask volunteers familiar with motor racing to help define these terms. The pit row is the area where teams set up garages during races. Each team may have one pit per car. The pace car is a car that leads the racing cars at a safe speed either at the start of the race or during a caution period, such as when debris is on the track. The pace car enters the track ahead of the leading car in the race, and race cars are not allowed to pass the pace car during a caution period.
A race car leaves pit row at a speed of 40 feet per second and accelerates at a constant rate of 44 feet per second. Its distance from the pit exit is given by the function d r (t) = 22t 2 + 40t. The race car leaves ahead of an approaching pace car traveling at a constant speed of 120 feet per second. In each case, find out if the pace car will catch up to the race car, and if so, how far down the track it will catch up. If there is more than one solution, explain how you know which one to select.
a. The pace car passes by the exit to pit row 1 second after the race car exits. The race car is traveling at a constant speed, so its position is a linear function of
time, d p (t) = mt + b. The slope of the line is the speed of the pace car, m = 120. To find b, use the time and position when the pace car passes the pit row exit. d p (1) = 0
120(1) + b = 0 120 + b = 0 b = -120
The position of the pace car is given by d p (t) = 120t - 120.
INTEGRATE MATHEMATICAL PRACTICES Focus on Modeling MP.4 Have students compare the y-intercepts for
To find out if the pace car catches up to the race car, plot both functions on a graphing calculator, with x for time and y for distance. The line does not intersect the parabola, so the pace car never catches up to the race car.
the two cars and explain what they represent. Students should recognize that they represent the positions of the two cars at time t = 0. The starting position of the race car is 0. The pace car does not reach the pit exit until after the race car leaves, so its starting position is represented by a negative number.
b. The pace car passes the exit half a second after the race car.
To solve part b, find the function for the pace car again: m = 120.
() 1 +b=0 120(_ 2) 1 =0 dp _ 2
© Houghton Mifflin Harcourt Publishing Company
60 + b = 0 b = -60
The position of the pace car is given by d p (t) = 120t - 60. Plot both functions on the graphing calculator. Two intersections are found, one at (1.06, 66.9) and the other at (2.58, 249.5). There are two solutions because the pace car first passes the race car at 1.06 seconds, and then the race car passes in front again at 2.58 seconds. The pace car initially catches up to the race car at the
INTEGRATE MATHEMATICAL PRACTICES Focus on Communication MP.3 Show students a graph of the two equations
first intersection, or 66.9 feet down the track.
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for part A that shows x ranging from 0 to 5 and y ranging from 0 to 500. Have students explain how they know that the graphs will never intersect, even though they see only a small section of the graph. Students should recognize that the distance between the two graphs increases as you extend the curves in either direction.
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Have students find and graph an equation for the position of a pace car that catches up to the race car exactly once, but does not pass it. Students need to find a line with slope 120 that is tangent to the parabola. They can sketch a line that appears to be tangent, then adjust the y-intercept until they find a line with only one intersection point. Alternatively, they can set the equation d p(t) = 120t + b equal to d r(t) = 22t 2 + 40t, put the resulting quadratic equation in standard form, and find the value of b that will make the discriminant equal to zero. When the equation is d p(t) = 120t - 72.7, the pace car will catch up to the race car at about t = 1.82 seconds, 145.7 feet from the pit exit.
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Scoring Rubric 2 points: Student correctly solves the problem and explains his/her reasoning. 1 point: Student shows good understanding of the problem but does not fully solve or explain his/her reasoning. 0 points: Student does not demonstrate understanding of the problem.
Solving Nonlinear Systems 422
MODULE
9
MODULE
STUDY GUIDE REVIEW
9
Using Square Roots to Solve Quadratic Equations
Study Guide Review
Essential Question: How can you use quadratic equations to solve real-world problems?
ASSESSMENT AND INTERVENTION
KEY EXAMPLE
(Lesson 9.1)
Solve (x - 8) = 49 by taking the square root. 2
x - 8 = ±7
Equations in the form a(x + b) = c can be solved by taking square roots. Take the square root of both sides.
x = 7 + 8 and x = -7 + 8
Solve both cases.
2
(x - 8) 2 = 49
Key Vocabulary
completing the square (completar el cuadrado) discriminant (discriminante) quadratic formula (fórmula cuadrática) square root (raíz cuadrada)
x = ±7 + 8
Assign or customize module reviews.
x = 15 and x = 1
MODULE PERFORMANCE TASK
KEY EXAMPLE
(-6) _ =1 4(9) 2 9x - 6x + 1 = 20 + 1 2
COMMON CORE
KEY EXAMPLE
SUPPORTING STUDENT REASONING
• How fast does the firework need to go to reach a height above 70 m? Students should choose a velocity and then check to see whether it works. Some will not. • How do I find the maximum height of a firework? The maximum height of the firework is at the vertex of the parabola.
(Lesson 9.3)
Solve 8x - 8x + 2 = 0 using the quadratic equation. 2
© Houghton Mifflin Harcourt Publishing Company
• What equation should I use for the parabola? Students can use h = -4.9t 2 + vt + h 0. The values for v and h 0 are provided in the problem.
b2 Find _. 4a Complete the square.
(3x - 1) 2 = 21 _ _ √ 21 + 1 -√ 21 + 1 x = _ and x = _ 3 3
Mathematical Practices: MP.1, MP.2, MP.4, MP.6 A-CED.A.2, F-BF.A.1, F-IF.B.4
Students should begin this problem by focusing on what information they will need. Here is some of the information they may ask for.
(Lesson 9.2)
Solve 9x 2 - 6x = 20 by completing the square.
a = 8, b = -8, c = 2
__
x=
-b ± √ b 2 - 4ac __ 2a
Identify a, b, and c. Use the quadratic formula.
___
8 ± √ (-8) - (4)(8)(2) x = ___ 2(8) 2
_
8 ± √0 x=_ 16 1 x=_ 2
Since b 2 - 4ac = 0, the equation has one real solution.
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Study Guide Review
SCAFFOLDING SUPPORT
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• For students needing more guidance, provide the formula h = -4.9t 2 + vt + h 0 with the problem context. • If students have trouble finding the maximum heights of different parabolas, b is the x-coordinate of the vertex. remind them that -__ 2a • Encourage students to find the maximum height of the firework at a particular velocity before working on the firing delay. (If the fire officials won’t approve a particular velocity, then they can’t use it in the show.)
423
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• For the part of the problem dealing with firing delays, you may wish to remind them that the t-coordinate of the vertex represents the time from the firing until the time of the explosion.
4/12/14 9:16 AM
EXERCISES Solve each equation. (Lessons 9.1, 9.2, 9.3, 9.4) 1.
x + 12x = - 17 2
SAMPLE SOLUTION 2.
4x 2 + 8x = 10 _
√ 14 - 2 x = _ and x = 2
Assume the initial velocity 37 m/s. The equation describing this velocity is y = -4.9x 2 + 3x + 1.9. The vertex of the graph has an x-value of:
21 1 and x = ___ x = __
_ _ x = √19 - 6 and x = - √ 19 - 6
3.
(4x - 11) = 100 2
4
4. _ - √14 - 2 _
4
b = -________ 37 -___ 2a 2(-4.9) ≈ 3.776
3x 2 + 17x + 10 = 0
2 x = -5 and x = -_ 3
2
and the height at the vertex is about: y = -4.9x 2 + 37x + 1.9
5.
A diver jumps off a high diving board that is 33 feet above the surface of the pool with an initial upward velocity of 6 feet per second. The height of the diver above the surface of the pool can be represented by the equation - 16t 2 + 6t + 33 = 0. How long will the diver be in the air, to the nearest hundredth of a second? Identify the method you used to solve the quadratic equation, and explain why you chose it. (Lesson 9.4)
= -4.9(3.776) + 373.776 + 1.9 2
≈ -69.865 + 139.712 + 1.9 ≈ 71.75
1.64 seconds; Possible Answer: I used the quadratic formula because the equation cannot be factored and completing the square would be time-consuming with these numbers.
So, this initial velocity is acceptable to the fire officials. Similarly, at an initial velocity of 40 m/s, the equation of the parabola is y = -4.9x 2 + 40x + 1.9, the x-coordinate of the 40 vertex is at -_ ≈ 4.082, and the maximum 2(-4.9) height is:
MODULE PERFORMANCE TASK
Fireworks Display You are planning a fireworks show for a local Fourth of July celebration. Fire officials require that all fireworks explode at a height greater than 70 meters so that debris has a chance to cool off as it falls.
y = -4.9(4.082) + 40(4.082) + 1.9 2
© Houghton Mifflin Harcourt Publishing Company
The firing platform for the fireworks is 1.9 meters off the ground. You have the option of firing your fireworks at an initial velocity of anywhere between 35 and 42 meters/second. If every firework is timed to explode when it reaches its maximum height, find two different initial velocities that are acceptable to the local fire officials. Then, figure out how long to delay the firing of the slower firework so that it will explode at the same time as the faster firework. Use your own paper to complete the task. Be sure to write down all your data and assumptions. Then use graphs, tables, or algebra to explain how you reached your conclusion.
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≈ 83.53 This is also high enough for the fire officials. The slower firework explodes about 3.776 seconds after firing, and the faster firework explodes about 4.082 seconds after firing. For both fireworks to explode at the same time, the slower firework should be delayed by 4.082 - 3.776, or 0.306 second.
Study Guide Review
DISCUSSION OPPORTUNITIES
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4/12/14 9:16 AM
• Students may wish to combine their findings into a plan for a fireworks show, in which fireworks explode at different heights and different times.
Assessment Rubric 2 points: Student correctly solves the problem and explains his/her reasoning. 1 point: Student shows good understanding of the problem but does not fully solve or explain. 0 points: Student does not demonstrate understanding of the problem.
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Ready to Go On?
Ready to Go On?
ASSESS MASTERY
9.1–9.5 Using Square Roots to Solve Quadratic Equations
Use the assessment on this page to determine if students have mastered the concepts and standards covered in this module.
• Online Homework • Hints and Help • Extra Practice
Find the discriminant of each quadratic equation, and determine the number of real solutions of each equation. (Lesson 9.4) 3x 2 + 2x + 6 = 0
1.
ASSESSMENT AND INTERVENTION
2.
-68; no real solutions
4x 2 + 6x = 8
164; two real solutions
Solve each equation using the given method. (Lessons 9.1, 9.2, 9.3, 9.4) 8x 2 - 72 = 0; square root
3.
4.
Differentiated Instruction Resources • Reading Strategies • Success for English Learners • Challenge Worksheets Assessment Resources
―
3x 2 + 7x + 8 = 0; quadratic formula
no real solution
⎧y = x 2 + 2 Find the solution or solutions of the system of equations ⎨ . (Lesson 9.5) ⎩y = x + 4 -1, 3 and 2, 6 ) ( ( )
7. © Houghton Mifflin Harcourt Publishing Company
• Reteach Worksheets
6.
x = -6 and x = -1
ADDITIONAL RESOURCES Response to Intervention Resources
2x 2 + 14x + 12 = 0; factoring
5.
―
√10 - 2 - √10 - 2 and x = __________ x = ___________ 5 5
-3 and 3
Access Ready to Go On? assessment online, and receive instant scoring, feedback, and customized intervention or enrichment.
25x 2 + 20x = 6; completing the square
ESSENTIAL QUESTION What are the methods of solving a quadratic equation without factoring? When can you use each method?
8.
Possible Answer: You can use square roots, completing the square, or the quadratic 2 formula. You can use square roots only when the equation is of the form a(x + b) = c, but you can use the other two methods with any quadratic equation.
• Leveled Module Quizzes
Module 9
COMMON CORE IN2_MNLESE389830_U4M09MC 425
425
Module 9
Study Guide Review
425
Common Core Standards
4/12/14 9:16 AM
Content Standards Mathematical Practices
Lesson
Items
9.4
1
A-REI.B.4
MP.2
9.4
2
A-REI.B.4
MP.2
9.1
3
A-REI.B.4
MP.1
9.2
4
A-REI.B.4
MP.1
9.4
5
A-REI.B.4
MP.1
9.3
6
A-REI.B.4
MP.1
9.5
7
A-REI.C.7
MP.2
MODULE MODULE 9 MIXED REVIEW
MIXED REVIEW
Assessment Readiness
Assessment Readiness
1. Is the given expression a perfect-square trinomial? Select Yes or No for each expression. A. x 2 + 24x + 144 Yes No B. 4x 2 + 36x + 9 C. 9x 2 - 6x + 1
9
Yes Yes
ASSESSMENT AND INTERVENTION
No No
2. Consider the following statements. Choose True or False for each. True False A. 4x 2 - 64 = 0 has 2 real solutions. B. x 2 - 5x - 9 = 0 has only 1 real solution. True False C. 3x 2 + 4x + 2 = 0 has no real solutions.
True
False
3. Solve -2x 2 - 9x = -4. What are the solutions? Explain how you solved the problem.
Assign ready-made or customized practice tests to prepare students for high-stakes tests.
――
――
9 + √113 - √113 x = 9_______ and x = _______ ; I used the quadratic formula and set a equal to -4 -4
-2, b equal to -9, and c equal to 4.
4. A landscaper is making a garden bed in the shape of a rectangle. The length of the garden bed is 2.5 feet longer than twice the width of the bed. The area of the garden bed is 62.5 square feet. Find the perimeter of the bed. Show your work.
ADDITIONAL RESOURCES Assessment Resources
x(2x + 2.5) = 62.5
• Leveled Module Quizzes: Modified, B
2x 2 + 2.5x - 62.5 = 0
――――――― -2.5 ± √2.5 - 4(2)(-62.5)
x = ___________________ 2
2(2)
―――
AVOID COMMON ERRORS
± 506.25 ___________ x = -2.5 4 √
± 22.5 ________ x = -2.5 4
The width cannot be negative. The perimeter of the bed is 2(5 + 12.5) = 35 feet.
Module 9
COMMON CORE
Item 2 Some students may use the quadratic formula for each equation, which is more timeconsuming than necessary. Remind students that they could use the determinant or graph the equations on their calculators and look for the x-intercepts.
© Houghton Mifflin Harcourt Publishing Company
x = 5 and x = -6.25
Study Guide Review
426
Common Core Standards
IN2_MNLESE389830_U4M09MC 426
4/12/14 9:16 AM
Content Standards Mathematical Practices
Lesson
Items
8.3
1*
A-SSE.A.2
MP.1
9.3
2
A-REI.B.4
MP.1
9.4
3
A-REI.B.4
MP.2
5.1, 9.4
4*
A-CED.A.1, A-APR.A.1, A-REI.B.4
MP.4
* Item integrates mixed review concepts from previous modules or a previous course.
Study Guide Review 426
MODULE
10
Linear, Exponential, and Quadratic Models
Linear, Exponential, and Quadratic Models
Essential Question: How can you use linear,
ESSENTIAL QUESTION:
exponential, and quadratic models to solve real-world problems?
Answer: You can make a table of values or you can graph a model for the real-world problem to see how the values change as the x-values of the function increase.
10 MODULE
LESSON 10.1
Fitting a Linear Model to Data LESSON 10.2
Graphing Exponential Functions LESSON 10.3
Modeling Exponential Growth and Decay This version is for PROFESSIONAL DEVELOPMENT Algebra 1 and Geometry only VIDEO
LESSON 10.4
Modeling with Quadratic Functions
Professional Development Video Author Juli Dixon models successful teaching practices in an actual high-school classroom.
Professional Development my.hrw.com
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Kip Evans/Alamy
LESSON 10.5
Comparing Linear, Exponential, and Quadratic Models
REAL WORLD VIDEO The Kemp’s Ridley sea turtle is an endangered species of turtle that nests along the Texas coast. Functions can be used to model the survivorship curve of the Kemp’s Ridley sea turtle.
MODULE PERFORMANCE TASK PREVIEW
What Model Fits a Survivorship Curve? Survivorship curves are graphs that show the number or proportion of individuals in a particular population that survive over time. Survivorship curves are used in diverse fields such as actuarial science, demography, biology, and epidemiology. How can you determine what mathematical model best fits a certain type of survivorship curve? Let’s find out!
Module 10
DIGITAL TEACHER EDITION IN2_MNLESE389830_U4M10MO 427
Access a full suite of teaching resources when and where you need them: • Access content online or offline • Customize lessons to share with your class • Communicate with your students in real-time • View student grades and data instantly to target your instruction where it is needed most
427
Module 10
427
PERSONAL MATH TRAINER Assessment and Intervention Assign automatically graded homework, quizzes, tests, and intervention activities. Prepare your students with updated, Common Core-aligned practice tests.
4/12/14 3:21 PM
Are YOU Ready?
Are You Ready?
Complete these exercises to review skills you will need for this chapter.
Scatter Plots
ASSESS READINESS
Tell whether the correlation is positive or negative, or if there is no correlation.
Example 1
y
Use the assessment on this page to determine if students need strategic or intensive intervention for the module’s prerequisite skills.
• Online Homework • Hints and Help • Extra Practice
Scatter plots can help you see relationships between two variables.
• In a positive correlation, as the value of one variable increases, the value of the other variable increases. • In a negative correlation, as the value of one variable decreases, the value of the other variable increases. • Sometimes there is no correlation, meaning there is no relationship between the variables.
x
ASSESSMENT AND INTERVENTION
The scatter plot has a negative correlation.
Tell whether the correlation is positive or negative, or if there is no correlation. y
1.
y
2.
3 2
x
1
x
no correlation
Personal Math Trainer will automatically create a standards-based, personalized intervention assignment for your students, targeting each student’s individual needs!
positive correlation
Constant Rate of Change Example 2
Tell if the rate of change is constant.
+1 +1
Time (hr)
1
2
3
4
Distance (mi)
45
90
135
180
change in miles rate of change = _______________ change in hours 45 = ___ 1
Tell if the rate of change is constant. Age (mo)
3
6
9
12
Weight (lb)
12
16
18
20
4.
Hours
2
4
6
8
Pay ($)
16
32
48
64
Module 10
Tier 1 Lesson Intervention Worksheets Reteach 10.1 Reteach 10.2 Reteach 10.3 Reteach 10.4 Reteach 10.5
ADDITIONAL RESOURCES See the table below for a full list of intervention resources available for this module. Response to Intervention Resources also includes: • Tier 2 Skill Pre-Tests for each Module • Tier 2 Skill Post-Tests for each skill
yes
no
IN2_MNLESE389830_U4M10MO 428
© Houghton Mifflin Harcourt Publishing Company
+1
+45 +45 +45 The rate of change is constant.
3.
TIER 1, TIER 2, TIER 3 SKILLS
428
Response to Intervention Tier 2 Strategic Intervention Skills Intervention Worksheets
Tier 3 Intensive Intervention Worksheets available online
9 Constant Rate... 13 Linear Associations 14 Linear Functions 11 Graphing Linear Nonproportional... 12 Graphing Linear... 4 Scatter Plots
Building Block Skills 5, 22, 23, 27, 40, 42, 46, 63, 65, 68, 70
Differentiated Instruction
4/12/14 3:07 PM
Challenge worksheets Extend the Math Lesson Activities in TE
Module 10
428
LESSON
10.1
Name
Fitting a Linear Model to Data
Class
Date
10.1 Fitting a Linear Model to Data Essential Question: How can you use the linear regression function on a graphing calculator to find the line of best fit for a two-variable data set? Resource Locker
Common Core Math Standards The student is expected to: COMMON CORE
Explore 1
S-ID.B.6b
For any set of data, different lines of fit can be created. Some of these lines will fit the data better than others. One way to determine how well the line fits the data is by using residuals. A residual is the signed vertical distance between a data point and a line of fit.
Informally assess the fit of a function by plotting and analyzing residuals. Also S-ID.B.6c, S-ID.C.8, F-LE.B.5
Mathematical Practices COMMON CORE
Plotting and Analyzing Residuals
After calculating residuals, a residual plot can be drawn. A residual plot is a graph of points whose x-coordinates are the variables of the independent variable and whose y-coordinates are the corresponding residuals.
MP.5 Using Tools
Looking at the distribution of residuals can help you determine how well a line of fit describes the data. The plots below illustrate how the residuals may be distributed for three different data sets and lines of fit.
Language Objective Demonstrate to a partner how to find and plot the residuals for a line of fit. Explain what the residuals tell you about the quality of fit.
Essential Question: How can you use the linear regression function on a graphing calculator to find the line of best fit for a two-variable data set? Using the STAT, STAT PLOT, Y=, and GRAPH keys, you can enter data into the calculator and use it to perform linear regression on a function.
PREVIEW: LESSON PERFORMANCE TASK View the Engage section online. Discuss why some of the hottest recorded temperatures on Earth were in the deserts nearest the equator and how the angle of the Sun at different latitudes causes temperature differences. Then preview the Lesson Performance Task.
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©bikeriderlondon/Shutterstock
ENGAGE
Distribution of residuals about the x-axis is random and tight. The line fits the data very well.
Distribution of residuals about the x-axis is random but loose. The line fits the data, but the modelFPO is weak.
The table lists the median age of females living in the United States, based on the results of the United States Census over the past few decades. Follow the steps listed to complete the task.
A
Use the table to create a table of paired values for x and y. Let x represent the time in years after 1970 and y represent the median age of females.
Year
Median Age of Females
1970
29.2
1980
31.3
1990
34.0
2000
36.5
2010
38.2
x
0
10
20
30
y
29.2
31.3
34.0
36.5
Module 10
40 38.2
ges must EDIT--Chan DO NOT Key=NL-A;CA-A Correction
Lesson 1
429
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Date Class
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Quest Essential
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Credits:
IN2_MNLESE389830_U4M10L1 429
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36.5 38.2
x
0
y
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0L1 429 30_U4M1
ESE3898
of the United
Watch for the hardcover student edition page numbers for this lesson.
34.0
1990
Module 10
IN2_MNL
the results
HARDCOVER PAGES 429444
a table and y to create after 1970 Use the table the time in years ent s. Let x repres n age of female the media of Females Median Age Year 29.2 1970 31.3
2010
Lesson 10.1
of Distribution about the residuals random. x-axis is not not fit The line does well. the data
on States, based the United s living in ete the task. n age of female listed to compl lists the media s. Follow the steps The table x and y. few decade values for over the past of paired represent
2000
429
Distribution of residuals about the x-axis is not random. The line does not fit the data well.
10 31.3
20 34.0
30 36.5
40 38.2
Lesson 1
429 4/9/14
7:09 PM
07/04/14 7:26 PM
B
EXPLORE 1
Residual Subtract Predicted from Actual to Find the Residual.
x
Actual y
Predicted y based on y = 0.25x + 29
0
29.2
29.0
0.2
10
31.3
31.5
-0.2
20
34.0
34.0
0
30
36.5
36.5
0
40
38.2
39.0
-0.8
Plotting and Analyzing Residuals INTEGRATE TECHNOLOGY Graphing calculators can be used to compute and plot residuals. After students are comfortable doing this work on their own, they can use the graphing calculator to check their work.
Plot the residuals.
Residuals
C
Use residuals to calculate the quality of fit for the line y = 0.25x + 29, where y is median age and x is years since 1970.
0.8 0.4 0 -0.4 -0.8
QUESTIONING STRATEGIES When looking at a plot of the residuals, how do you determine whether your line is a good model for the data? Explain. If the residual values are on or near the x-axis and evenly distributed on both sides of the axis, it means that the residuals are small and therefore the line is close to being ideal for the data. If the residual values are large (not near the x-axis) and/or not evenly distributed on both sides of the axis, it means that the line is not close to being ideal for the data.
10 30 50 x-values
D
Evaluate the quality of fit to the data for the line y = 0.25x + 29. Since all of the residuals are small and the points on the residual plot are tightly
distributed around the x-axis, the line y = 0.25x + 29 is a good model for this situation. Reflect
Discussion When comparing two lines of fit for a single data set, how does the residual size show which line is the best model? The better model has residuals closer to 0.
2.
Discussion What would the residual plot look like if a line of fit is not a good model for a data set? The residual plot would show points randomly scattered both above and below the x-axis, and the distances from the x-axis would be large.
Module 10
430
© Houghton Mifflin Harcourt Publishing Company
1.
Lesson 1
PROFESSIONAL DEVELOPMENT IN2_MNLESE389830_U4M10L1.indd 430
Math Background
4/2/14 1:45 AM
Why does linear regression use the sum of the squares of the residuals rather than just the sum? First, adding just the residuals can give misleading results. Large positives and negatives can cancel each other out, allowing a very loose fit of data points to have a very small sum. Summing the absolute values of the residuals might solve this problem but presents other complications. Using squares of residuals solves the positive-negative cancellation problem and also ensures that very large residuals affect the sum more than small residuals.
Fitting a Linear Model to Data
430
Explore 2
EXPLORE 2
Analyzing Squared Residuals
When different people fit lines to the same data set, they are likely to choose slightly different lines. Another way to compare the quality of a line of fit is by squaring residuals. In this model, the closer the sum of the squared residuals is to 0, the better the line fits the data.
Analyzing Squared Residuals
In the previous section, a line of data was fit for the median age of females over time. After performing this task, two students came up with slightly different results. Student A came up with the equation y = 0.25x + 29.0 while Student B came up with the equation y = 0.25x + 28.8, where x is the time in years since 1970 and y is the median age of females in both cases.
AVOID COMMON ERRORS
A
In squaring decimals, some students may forget to move the decimal point in their answers. Remind students that the square of a decimal should have twice as many decimal places as the original number.
Complete each table below.
y = 0.25x + 29.0
QUESTIONING STRATEGIES What must be true for the square of a residual to be less than the residual? The absolute value of the residual must be less than 1.
x
y (Actual)
y (Predicted)
Residual
0
29.2
29.0
0.2
Square of Residual 0.04
10
31.3
31.5
-0.2
0.04
20
34.0
34.0
0
30
36.5
36.5
0
40
38.2
39.0
-0.8
x
y (Actual)
y (Predicted)
Residual
0
29.2
28.8
0.4
0.16
10
31.3
31.3
0
0
20
34.0
33.8
0.2
0.04
30
36.5
36.3
0.2
0.04
40
38.2
38.8
-0.6
0.36
0 0 0.64
© Houghton Mifflin Harcourt Publishing Company
y = 0.25x + 28.8
B
Find the sum of squared residuals for each line of fit.
C
Square of Residual
Which line has the smaller sum of squared residuals?
y = 0.25x + 28.8
y = 0.25x + 29.0 : 0.72 y = 0.25x + 28.8 : 0.60 Reflect
3.
How does squaring a residual affect the residual’s value? If a residual is between 0 and 1, squaring the residual makes the residual smaller. If a
residual is negative, squaring the residual makes the residual positive. 4.
Are the sums of residuals or the sum of the squares of residuals a better measure of quality of fit? The sum of the squares of residuals is a better measure of quality of fit.
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Peer-to-Peer Activity Have students work in pairs. Give each pair a set of bivariate data. Each student will choose a line of fit, find its equation, and calculate the sum of the squares of the residuals. Then the two students should compare their results to determine which line is a better fit for the data. Students can then use a graphing calculator to compare their equations to the equation for the line of best fit.
431
Lesson 10.1
4/2/14 1:45 AM
Assessing the Fit of Linear Functions from Residuals
Explain 1
EXPLAIN 1
The quality of a line of fit can be evaluated by finding the sum of the squared residuals. The closer the sum of the squared residuals is to 0, the better the line fits the data. Example 1
Assessing the Fit of Linear Functions from Residuals
The data in the tables are given along with two possible lines of fit. Calculate the residuals for both lines of fit and then find the sum of the squared residuals. Identify the lesser sum and the line with better fit.
x
2
4
6
8
y = x + 2.2
y
7
8
4
8
y = x + 2.4
INTEGRATE MATHEMATICAL PRACTICES Focus on Reasoning MP.2 Explain to students the theoretical
a. Find the residuals of each line.
x
y (Actual)
2
7
y Predicted by y = x + 2.4
Residual for y = x + 2.4
4.4
y Predicted by y = x + 2.2
2.6
Residual for y = x + 2.2
4.2
4
8
6.4
2.4
6.2
2.6
6
4
8.4
-4.4
8.2
-4.2
8
8
10.4
-2.4
10.2
underpinning for wanting the sum of squared residuals to be as small as possible. Point out that the residuals are a measure of how far the data points are from the line. Small values for the residuals mean that the data points are closer to the line, which in turn means that the line has a tight fit to the data.
2.8
-2.2
b. Square the residuals and find their sum. 2 2 2 2 y = x + 2.4 (2.6) + (2.4) + (-4.4) + (-2.4) = 6.76 + 5.76 + 19.36 + 5.76 = 37.64
y = x + 2.2 (2.8) + (2.6) + (-4.2) + (-2.2) = 7.84 + 6.76 + 17.64 + 4.84 = 37.08 2
2
2
2
The sum of the squared residuals for y = x + 2.2 is smaller, so it provides a better fit for the data.
x
1
2
3
4
y = 2x + 3
y
5
4
6
10
y = 2x + 2.5 © Houghton Mifflin Harcourt Publishing Company
a. Find the residuals of each line.
y Predicted by y=2x+3
Residual for y=2x+3
y Predicted by y = 2 x + 2.5
5
0
4.5
0.5
7
-3
6.5
-2.5
x
y (Actual)
1
5
2
4
3
6
9
4
10
11
-3
8.5
-1
b. Square the residuals and find their sum.
10.5
Residual for y = 2 x + 2.5
-2.5
-0.5
( 0 ) + ( -3 ) + ( -3 ) + ( -1 ) = 0 + 9 + 9 + 1 = 19 y = 2x + 2.5 : ( 0.5 ) + ( -2.5 ) + ( -2.5 ) + ( -0.5 ) = 0.25 + 6.25 + 6.25 + 0.25 = y = 2x + 3 :
2
2
2
2
2
2
2
2
13
The sum of the squared residuals for y = 2 x + 2.5 is smaller, so it provides a better fit for the data. Module 10
432
Lesson 1
DIFFERENTIATE INSTRUCTION IN2_MNLESE389830_U4M10L1.indd 432
Technology
4/2/14 1:45 AM
Encourage students to try performing regression analysis using an online application. A quick browser search will turn up several online regression calculators. Make sure that students pay attention to the format in which ordered pairs must be entered for the application they choose. Applications typically will plot the data points, show the line of best fit, and provide an equation for the line.
Fitting a Linear Model to Data
432
Reflect
QUESTIONING STRATEGIES When comparing different lines of fit, how do you know which line has the best fit? Explain. Find the residuals for each line, square them, and find their sums. The line with the smallest sum for the squares of its residuals is the best fit.
5.
How do negative signs on residuals affect the sum of squared residuals? Negative residuals do not affect the sum of squared residuals.
6.
What quadrant is not used in finding the sum of squared residuals that is used in finding the sum of residuals? Quadrant IV
Your Turn
How can you find the value of a residual by looking at the coordinates of the point on a residual plot? The y-coordinate of the point will be the value of the residual.
7.
The data in the table are given along with two possible lines of fit. Calculate the residuals for both lines of fit and then find the sum of the squared residuals. Identify the lesser sum and the line with better fit.
x
1
2
3
4
y=x+4
y
4
7
8
6
y = x + 4.2
x
y (Actual)
y Predicted by y=x+4
Residual for y=x+4
y Predicted by y = x + 4.2
Residual for y = x + 4.2
1
4
5
–1
5.2
–1.2
2
7
6
1
6.2
0.8
3
8
7
1
7.2
0.8
4
6
8
–2
8.2
–2.2
y = x + 4 : (-1) + (1) + (1) + (-2) = 7 2
2
2
2
y = x + 4.2 : (-1.2) + (0.8) + (0.8) + (-2.2) = 7.56 2
2
2
2
© Houghton Mifflin Harcourt Publishing Company
The sum of the squared residuals for y = x + 4 is smaller, so it provides a better fit.
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Lesson 1
LANGUAGE SUPPORT IN2_MNLESE389830_U4M10L1.indd 433
Connect Vocabulary In this lesson, students work with residuals to find the line of best fit for a data set. Relate the word residual to the word residue, which students may have used in a science context. Students should be able to make a connection between a small amount that is left over (residue) and the amount that a data value differs from the predicted value (residual). Suggest they think of wanting residuals to be as small as possible as wanting to avoid having a lot of “leftovers.”
433
Lesson 10.1
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Performing Linear Regression
Explain 2
EXPLAIN 2
The least-squares line for a data set is the line of fit for which the sum of the squared residuals is as small as possible. Therefore the least-squares line is a line of best fit. A line of best fit is the line that comes closest to all of the points in the data set, using a given process. Linear regression is a method for finding the least-squares line. Example 2
Performing Linear Regression
Given shows latitudes and average temperatures in degrees Celsius for several cities, use your calculator to find an equation for the line of best fit. Then interpret the correlation coefficient and use the line of best fit to estimate the average temperature of another city using the given latitude.
Latitude
Average Temperature (°C)
Barrow, Alaska
71.2°N
-12.7
Yakutsk, Russia
62.1°N
-10.1
London, England
51.3°N
10.4
Chicago, Illinois
41.9°N
10.3
San Francisco, California
37.5°N
13.8
Yuma, Arizona
32.7°N
22.8
Tindouf, Algeria
27.7°N
22.8
Dakar, Senegal
14.0°N
24.5
Mangalore, India
12.5°N
27.1
City
INTEGRATE MATHEMATICAL PRACTICES Focus on Technology MP.5 Make sure that students understand the purpose of each step when using a calculator to find a line of best fit. They first input the data, then create a scatter plot, then use the linear regression feature to find the equation for the line of best fit. Next they input the equation and graph it. Students should be able to explain what they are doing at each step in the process.
Estimate the average temperature in Vancouver, Canada at 49.1°N. Create a scatter plot of the data.
Enter the data into data lists on your calculator. Enter the latitudes in column L1 and the average temperatures in column L2.
© Houghton Mifflin Harcourt Publishing Company
Use the Linear Regression feature to find the equation for the line of best fit using the lists of data you entered. Be sure have the calculator also display values for the correlation coefficient r and r 2. The correlation coefficient is about -0.95, which is very strong. This indicates a strong correlation, so we can rely on the line of fit for estimating average temperatures for other locations within the same range of latitudes.
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The equation for the line of best fit is y ≈ -0.693x + 39.11.
QUESTIONING STRATEGIES
Graph the line of best fit with the data points in the scatter plot.
The calculator’s linear regression feature gives you an equation in the form y = ax + b. What form of linear equation is that? What do a and b represent? Slope-intercept form; a is the slope of the line of best fit, and b is the y-intercept.
Use the TRACE function to find the approximate average temperature in degrees Celsius for a latitude of 49.1°N. The average temperature in Vancouver should be around 5°C.
How does the line of fit displayed on the calculator differ from lines of fit found in previous exercises? Explain. The lines of fit found in previous exercises were estimates; the line of fit found using a calculator is the best possible fit for the data points given.
Latitude
Average Temperature (°F)
Fairbanks, Alaska
City
64.5°N
30
Moscow, Russia
55.5°N
39
Ghent, Belgium
51.0°N
46
Kiev, Ukraine
50.3°N
49
Prague, Czech Republic
50.0°N
50
Winnipeg, Manitobia
49.5°N
52
Luxembourg
49.4°N
53
Vienna, Austria
48.1°N
56
Bern, Switzerland
46.6°N
59
Estimate the average temperature in degrees Fahrenheit in Bath, England, at 51.4°N. Enter the data into data lists on your calculator.
© Houghton Mifflin Harcourt Publishing Company
Use the Linear Regression feature to find the equation for the line of best fit using the lists of data you entered. Be sure to have the calculator also display values for the correlation coefficient r and r 2. The correlation coefficient is about 0.95 , which indicates a very strong correlation. The correlation coefficient indicates that the line of best fit [is/ is not] reliable for estimating temperatures of other locations within the same range of latitudes. The equation for the line of best fit is y ≈ y≈-
1.60
x + 131.05
The average temperature in degrees Fahrenheit in Bath, England, should be around
49
°F.
Graph the line of best fit with the data points in the scatter plot. Then use the TRACE function to find the approximate average temperature in degrees Fahrenheit for a latitude of 51.4°N.
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x + 131.05 .
Use the equation to estimate the average temperature in Bath, England at 46.6°N.
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Reflect
8.
9.
ELABORATE
Interpret the slope in terms of the context for Example 2A. The average temperature in degrees Celsius decreases at a rate of about 0.69 degrees per
degree in latitude.
CONNECT VOCABULARY
Interpret the y-intercept with regards to the context of the first part of this example. At 0 degrees in latitude or the equator, the average temperature is 39 degrees Celsius.
Discuss with students the differences and similarities between a line of fit and a line of best fit. Students should realize that while there are many lines of fit for a data set, there is only one line of best fit.
Your Turn
10. Use the given data and your calculator to find an equation for the line of best fit. Then interpret the correlation coefficient and use the line of best fit to estimate the average temperature of another city using the given latitude.
SUMMARIZE THE LESSON
y ≈ -0.93x + 77.7
City
Latitude
Average Temperature (°F)
Anchorage, United States
61.1°N
18
≈ 0.93(40.1) + 77.7
Dublin, Ireland
53.2°N
29
≈ 40.41
Zurich, Switzerland
47.2°N
34
The temperature in Trenton,
Florence, Italy
43.5°N
37
New Jersey, should be around
Trenton, New Jersey
40.1°N
40
40 degrees Fahrenheit.
Algiers, Algeria
36.5°N
46
El Paso, Texas
31.5°N
49
Dubai, UAE
25.2°N
56
Manila, Philippines
14.4°N
61
How do you use a graphing calculator to determine a line of best fit for a data set? First, enter the x- and y-values into the calculator. Then, plot the points. Next, find the equation for the line of best fit. Finally, use the calculator to draw the line of best fit.
The correlation coefficient is about 0.99, which indicates a very strong correlation. Therefore, the line of best fit is reliable for estimating temperatures within the same range of latitudes.
11. What type of line does linear regression analysis make? A linear regression makes a line of best fit.
12. Why are squared residuals better than residuals? Squared residuals are better than normal residuals because they provide a more exact
measure of the accuracy of a chosen line of best fit.
© Houghton Mifflin Harcourt Publishing Company
Elaborate
13. Essential Question Check-In What four keys are needed on a graphing calculator to perform a linear regression? STAT, PLOT, Y=, and GRAPH
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Evaluate: Homework and Practice
EVALUATE
The data in the tables below are shown along with two possible lines of fit. Calculate the residuals for both lines of fit and then find the sum of the squared residuals. Identify the lesser sum and the line with better fit. 1.
ASSIGNMENT GUIDE
x
2
4
6
8
y=x+ 5
y
1
3
5
7
y = x + 4.9
• Online Homework • Hints and Help • Extra Practice
2 2 2 2 y = x + 5: (-6) + (-6) + (-6) + (-6) = 36 + 36 + 36 + 64 = 144
Concepts and Skills
Practice
Explore Plotting and Analyzing Residuals
Exercise 21
Explore 2 Analyzing Squared Residuals
Exercises 23–24
Example 1 Assessing the Fit of Linear Functions from Residuals
Exercises 1–14, 19–20, 22
Example 2 Performing Linear Regression
Exercises 15–18
2 2 2 2 y = x + 4.9: (-5.9) + (-5.9) + (-5.9) + (-5.9) = 139.24
The sum of the squared residuals for y = x + 4.9 is smaller, so it provides a better fit for the data.
2.
x
1
2
3
4
y = 2x + 1
y
1
7
3
5
y = 2x + 1.1
y = 2x + 1: 4 + 4 + 16 + 16 = 40
y = 2x + 1.1: 4.41 + 4.41 + 16.81 + 16.81 = 42.44
The sum of the squared residuals for y = 2x + 1 is smaller, so it provides a better fit for the data. 3.
x
2
4
6
8
y = 3x + 4
y
2
8
4
6
y = 3x + 4.1
2 2 2 2 y = 3x + 4: (-8) + (-8) + (-18) + (-22) = 936
2 2 2 2 y = 3x + 4.1: (-8.1) + (-8.1) + (-18.1) + (-22.1) = 947.24
© Houghton Mifflin Harcourt Publishing Company
The sum of the squared residuals for y = 3x + 4 is smaller, so it provides a better fit. 4.
x
1
2
3
4
y=x+1
y
2
1
4
3
y = x + 0.9
2 2 2 2 y = x + 1: (0) + (-2) + (0) + (-2) = 8
2 2 2 2 y = x + 0.9: (0.1) + (-1.9) + (0.1) + (-1.9) = 7.24
The sum of the squared residuals for y = x + 0.9 is smaller, so it provides a better fit. 5.
x
2
4
6
8
y = 3x + 1.2
y
1
5
4
3
y = 3x + 1
2 2 2 2 y = 3x + 1.2: (-6.2) + (-8.2) + (-15.2) + (-22.2) = 829.56 2 2 2 2 y = 3x + 1: (-6) + (-8) + (-15) + (-22) = 809
The sum of the squared residuals for y = 3x + 1 is smaller, so it provides a better fit.
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Depth of Knowledge (D.O.K.)
COMMON CORE
Mathematical Practices
1–14
1 Recall of Information
MP.2 Reasoning
15–18
1 Recall of Information
MP.5 Using Tools
19–20
1 Recall of Information
MP.4 Modeling
21
1 Recall of Information
MP.2 Reasoning
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6.
x
1
2
3
4
y=x+ 5
y
4
1
3
2
y = x + 5.3
INTEGRATE MATHEMATICAL PRACTICES Focus on Critical Thinking MP.3 Discuss with students the reasons why the
2 2 2 2 y = x + 5: (-2) + (-6) + (-5) + (-7) = 114
2 2 2 2 y = x + 5.3: (-2.3) + (-6.3) + (-5.3) + (-7.3) = 126.36
The sum of the squared residuals for y = x + 5 is smaller, so it provides a better fit. 7.
x
2
4
6
8
y = 2x + 1
y
3
6
4
5
y = 2x + 1.4
squares of the residuals are used to perform a linear regression. Students should understand that using the squares guarantees that the values will be nonnegative and their sum will not be 0 unless the points are perfectly linear.
y = 2x + 1: (-2) 2 + (-3) 2 + (-9) 2 + (-12) 2 = 238
y = 2x + 1.4: (-2.4) 2 + (-3.4) 2 + (-9.4) 2 + (-12.4) 2 = 259.44
The sum of the squared residuals for y = 2x + 1 is smaller, so it provides a better fit. 8.
x
1
2
3
4
y=x+ 2
y
5
3
6
4
y = x + 2.2
y = x + 2: (2) 2 + (-1) 2 + (1) 2 + (-2) 2 = 10
y = x + 2.2: (1.8) 2 + (-1.2) 2 + (1.8) 2 + (-2.2) 2 = 12.76
The sum of the squared residuals for y = x + 2 is smaller, so it provides a better fit.
9.
x
2
4
6
8
y=x+ 3
y
1
5
7
3
y = x + 2.6
© Houghton Mifflin Harcourt Publishing Company
y = x + 3: (-4) 2 + (-2) 2 + (-2) 2 + (-8) 2 = 88
y = x + 2.6: (-3.6) 2 + (-1.6) 2 + (-1.6) 2 + (-7.6) 2 = 75.84
The sum of the squared residuals for y = x + 2.6 is smaller, so it provides a better fit. 10.
x
1
2
3
4
y = x + 1.5
y
2
5
4
3
y = x + 1.7
y = x + 1.5: (-.5) 2 + (1.5) 2 + (-.5) 2 + (-2.5) 2 = 9
y = x + 1.7: (-.7) 2 + (1.3) 2 + (-.7) 2 + (-2.7) 2 = 9.96
The sum of the squared residuals for y = x + 1.5 is smaller, so it provides a better fit.
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Depth of Knowledge (D.O.K.)
COMMON CORE
Mathematical Practices
22
2 Skills/Concepts
MP.2 Reasoning
23–24
2 Skills/Concepts
MP.2 Reasoning
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11.
AVOID COMMON ERRORS When squaring residuals to find a line of fit, students may forget to move the decimal point in their answers. Remind students to check that their answers have the correct number of decimal places before finding the sum of the squares of the residuals.
x
1
2
3
4
y = 2x + 3.1
y
2
9
7
12
y = 2x + 3.5
y = 2x + 3.1: (-3.1) 2 + (1.9) 2 + (-2.1) 2 + (0.9) 2 = 18.44
y = 2x + 3.5: (-3.5) 2 + (1.5) 2 + (-2.5) 2 + (0.5) 2 = 21
The sum of the squared residuals for y = 2x + 3.1 is smaller, so it provides a better fit. 12.
x
1
3
5
7
y = 1.6x + 4
y
2
6
8
13
y = 1.8x + 4
y = 1.6x + 4: (-3.6) 2 + (-2.8) 2 + (-4.0) 2 + (-2.2) 2 = 41.64 y = 1.8x + 4: (-3.8) 2 + (-3.4) 2 + (-5.0) 2 + (-3.6) 2 = 63.96
The sum of the squared residuals for y = 1.6x + 4 is smaller, so it provides a better fit.
13.
x
1
2
3
4
y=x+ 5
y
7
5
11
8
y = 1.3x + 5
y = x + 5: (1) 2 + (-3) 2 + (1) 2 + (-4) 2 = 27
y = 1.3x + 5: (0.7) 2 + (-3.9) 2 + (-0.5) 2 + (-6.1) 2 = 53.16
The sum of the squared residuals for y = x + 5 is smaller, so it provides a better fit. 14.
x
1
2
3
4
y = 2x + 3
y
4
11
5
15
y = 2.4x + 3
© Houghton Mifflin Harcourt Publishing Company
y = 2x + 3: (-1) 2 + (4) 2 + (-4) 2 + (4) 2 = 49
y = 2.4x + 3: (-1.4) 2 + (3.2) 2 + (-5.2) 2 + (2.4) 2 = 45
The sum of the squared residuals for y = 2.4x + 3 is smaller, so it provides a better fit.
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Use the given data and your calculator to find an equation for the line of best fit. Then interpret the correlation coefficient and use the line of best fit to estimate the average temperature of another city using the given latitude. 15.
City
Latitude
Average Temperature (°F)
Calgary, Alberta
51.0°N
24
Munich, Germany
48.1°N
26
Marseille, France
43.2°N
29
St. Louis, Missouri
38.4°N
34
Seoul, South Korea
37.3°N
36
Tokyo, Japan
35.4°N
38
New Delhi, India
28.4°N
43
Honululu, Hawaii
21.2°N
52
Bangkok, Thailand
14.2°N
58
8.6°N
63
Panama City, Panama
GRAPHIC ORGANIZERS Students can create a flow chart or step-by-step diagram to serve as a visual guide to the process of finding a line of best fit. One chart could describe finding a line of fit by hand, and another could explain how to find a line of best fit using a calculator.
80°
Calgary
Marseille
40°
0°
Munich
Seoul
St Louis
Tokyo
Panama City
Honolulu
Bangkok
New Delhi
40°
160°
120°
80°
40°
0°
40°
80°
120°
© Houghton Mifflin Harcourt Publishing Company
80°
160°
y = –0.95x + 71.13
y = –0.95(8.6) + 71.13
y ≈ 63
The correlation coefficient is about 0.99, which indicates a very strong correlation. Therefore, the line of best fit is reliable for estimating temperatures within the same range of latitudes. The temperature in Panama City should be around 63 degrees Fahrenheit.
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16.
Latitude
Average Temperature (°F)
Oslo, Norway
59.6°N
21
Warsaw, Poland
52.1°N
28
Milan, Italy
45.2°N
34
Vatican City, Vatican City
41.5°N
41
Beijing, China
39.5°N
42
Tel Aviv, Israel
32.0°N
48
Kuwait City, Kuwait
29.2°N
48
Key West, Florida
24.3°N
55
Bogota, Columbia
4.4°N
64
Mogadishu, Somalia
2.0°N
66
City
y = -0.78x + 70.69
y = -0.78(29.2) + 70.69 y = 48
The correlation coefficient is about 0.99, which indicates a very strong correlation. Therefore, the line of best fit is reliable for estimating temperatures within the same range of latitudes. The temperature in Kuwait City should be around 48 degrees Fahrenheit.
© Houghton Mifflin Harcourt Publishing Company
17.
Latitude
Average Temperature (°F)
Tornio, Finland
65.5°N
28
Riga, Latvia
56.6°N
36
Minsk, Belarus
53.5°N
39
Quebec City, Quebec
46.5°N
45
Turin, Italy
45.0°N
47
Pittsburgh, Pennsylvania
40.3°N
49
Lisbon, Portugal
38.4°N
52
Jerusalem, Israel
31.5°N
58
New Orleans, Louisiana
29.6°N
60
Port-au-Prince, Haiti
18.3°N
69
City
y = -0.86x + 85.19
y = -0.86(31.5) + 85.19 y ≈ 58
The correlation coefficient is about 0.99, which indicates a very strong correlation. Therefore, the line of best fit is reliable for estimating temperatures within the same range of latitudes. The temperature in Jerusalem should be around 58 degrees Fahrenheit. Module 10
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Latitude (°N)
Average Temperature (°F)
Juneau, Alaska
58.2
15
Amsterdam, Netherlands
52.2
24
Salzburg, Austria
47.5
36
Belgrade, Serbia
44.5
38
Philadelphia, Pennsylvania
39.6
41
Tehran, Iran
35.4
44
Nassau, Bahamas
25.0
52
Mecca, Saudi Arabia
21.3
56
Dakar, Senegal
14.4
?
6.5
65
18.
City
Georgetown, Guyana
INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 Students often have the wrong sign for their residuals because they subtract the actual value from the predicted value instead of the other way around. While finding the squares of the residuals will mask this error because the square of a real number is always nonnegative, students should be careful to calculate the residuals correctly if they want to know whether the model’s predicted value is an underestimate (positive residual) or an overestimate (negative residual) of the actual value.
y = -0.91x + 74.74
y = -0.91(14.4) + 74.74 y = 62
The correlation coefficient is about 0.97, which indicates a very strong correlation. Therefore, the line of best fit is reliable for estimating temperatures within the same range of latitudes. The temperature in Dakar should be around 62 degrees Fahrenheit.
19.
Year
Median age of men
1970
25.3
1980
26.8
1990
29.1
2000
31.4
2010
35.6
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© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Martin Allinger/Shutterstock
Demographics Each table lists the median age of people living in the United States, based on the results of the United States Census over the past few decades. Use residuals to calculate the quality of fit for the line y = 0.5x + 20, where y is median age and x is years since 1970.
Since the residuals are large, the line y = 0.5x + 20 is not a good fit.
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20.
JOURNAL Have students summarize the steps for evaluating a line of best fit by finding the sum of the squares of the residuals.
Year
Median Age of Texans
1970
27.1
1980
29.3
1990
31.1
2000
33.8
2010
37.6
Since the residuals are large, the line y = 0.5x + 20 is not a good fit.
21. State the residuals based on the actual y and predicted y values. a. Actual: 23, Predicted: 21
2
b. Actual: 25.6, Predicted: 23.3 2.3 c. Actual: 24.8, Predicted: 27.4 -2.6 d. Actual: 34.9, Predicted: 31.3 3.6 H.O.T. Focus on Higher Order Thinking
22. Critical Thinking The residual plot of an equation has x-values that are close to the x-axis from x = 0 to x = 10, but has values that are far from the axis from x = 10 to x = 30. Is this a strong or weak relationship?
The strength of the relationship is weak because of the weak correlation between x = 10 and x = 30. A short distance of weak correlation cannot be offset by an otherwise strong correlation.
© Houghton Mifflin Harcourt Publishing Company
23. Communicate Mathematical Ideas In a squared residual plot, the residuals form a horizontal line at y = 6. What does this mean? This means that either all of the residuals are either the same as each other or the opposite of each other. 24. Interpret the Answer Explain one situation other than those in this section where squared residuals are useful.
Sample Answer: Squared residuals are useful when the velocity equation of a car is unknown but needs to be approximated as well as possible. Using squared residuals will determine how well the approximated line fits the data.
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Lesson Performance Task
CONNECT VOCABULARY Some students may not understand the meaning of Earth’s hemispheres, the equator, or latitude. Explain that the equator is an imaginary horizontal line that encircles the Earth, dividing it into two equal halves called the Northern and Southern Hemispheres. A series of lines parallel to the equator, called latitude lines, are used to indicate how far north or south of the equator any point on Earth is located. Latitude values are based on a scale that puts the equator at 0° latitude, the north pole at latitude 90° North, and the south pole at latitude 90° South. On a globe of Earth, point out the equator, the lines of latitude and longitude, and the locations of the cities listed in the table.
The table shows the latitudes and average temperatures for the 10 largest cities in the Southern Hemisphere.
City
Latitude (°S)
Average Temperature (°F)
Sao Paulo, Brazil
23.9
69
Buenos Aires, Argentina
34.8
64
Rio de Janeiro, Brazil
22.8
76
Jakarta, Indonesia
6.3
81
Lodja, DRC
3.5
73
Lima, Peru
12.0
68
Santiago de Chile, Chile
33.2
58
Sydney, Australia
33.4
64
Melbourne, Australia
37.7
58
Johannesburg, South Africa
26.1
61
a. Use a graphing calculator to find a line of best fit for this data set. What is the equation for the best-fit line? Interpret the meaning of the slope of this line. b. The city of Piggs Peak, Swaziland, is at latitude 2.0 °S. Use the equation of your best-fit line to predict the average temperature in Piggs Peak. The actual average temperature for Piggs Peak is 65.3 °F. How might you account for this difference in predicted and actual values?
a. The equation of the best-fit line is y ≈ -0.495x + 78.772. The slope of the line is about -0.5, which indicates that the average temperature decreases about 0.5 °F for every degree of increase in south latitude. b. Students should find that the predicted average temperature for Piggs Peak is about 65.9 °F. This divergence of the actual value from the predicted value is due at least in part to the fact that the correlation coefficient is 0.62, which is not a very strong correlation.
© Houghton Mifflin Harcourt Publishing Company
c. Assume that you graphed the latitude and average temperature for 10 cities in the Southern Hemisphere. Predict how the line of best fit for that data set might compare with the best-fit line for the Northern Hemisphere cities.
c. Students should predict that the line for 10 different cities in the Southern Hemisphere would be similar to this best-fit line. Interested students could research the latitude and average temperature for 10 different cities at a range of latitudes in the Southern Hemisphere and carry out the linear regression.
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Have students research the latitude and average temperature for the city in which they live. Then have students use the equation of the line of best fit they found in Part A of the Lesson Performance Task to predict the average temperature in their city and compare it to the known average temperature. Discuss possible reasons why the predicted value is or is not close to the actual average temperature. Students might consider whether altitude, proximity to a large body of water, or other factors influence the temperature in their city.
4/2/14 1:45 AM
Scoring Rubric 2 points: Student correctly solves the problem and explains his/her reasoning. 1 point: Student shows good understanding of the problem but does not fully solve or explain his/her reasoning. 0 points: Student does not demonstrate understanding of the problem.
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LESSON
10.2
Name
Graphing Exponential Functions
Class
Date
10.2 Graphing Exponential Functions Essential Question: How do you graph an exponential function of the form f(x) = ab x?
Common Core Math Standards The student is expected to: COMMON CORE
Exponential functions follow the general shape y = ab x.
Graph exponential… functions, showing intercepts and end behavior… Also F-IF.C.8b
Mathematical Practices COMMON CORE
Exploring Graphs of Exponential Functions
Explore
F-IF.C.7e
Resource Locker
Graph the exponential functions on a graphing calculator, and match the graph to the correct function rule. 1.
MP.4 Modeling
2.
Language Objective
3. 4.
Explain the domain, range, and end behavior of the graphs of exponential functions of the form ƒ(x) = ab x with a < 0 and 0 < b < 1.
y = 3(2)
x
y = 0.5(2) y = 3(0.5)
a x
c
x
b
y = -3(2)
x
a.
b.
c.
d.
d
Essential Question: How do you graph an exponential function of the form f(x) = abx? Use the value of a to find the y-intercept. Choose several values of x other than 0 to plot a few more ordered pairs and connect them with a smooth curve. Use the values of a and b to determine the end behavior of the function.
© Houghton Mifflin Harcourt Publishing Company
ENGAGE
In all the functions 1–4 above, the base b > 0. Use the graphs to make a conjecture: State the domain and range of y = ab x if a > 0.
In all of the functions 1–4 above, the domain values are all real numbers, or -∞ < x < ∞. If a > 0 and b > 0, then the range values are all positive, or y > 0.
In all the functions 1–4 above, the base b > 0. Use the graphs to make a conjecture: State the domain and range of y = ab x if a < 0.
In all of the functions 1–4 above, the domain values are all real numbers, or -∞ < x < ∞. If a < 0 and b > 0, then the range values are all negative, or y < 0.
PREVIEW: LESSON PERFORMANCE TASK View the Engage section online. Discuss the photo and the fact that the weight of a pumpkin might grow exponentially. Then preview the Lesson Performance Task.
What is the y-intercept of ƒ(x) = 0.5(2) ? x
0.5 Module 10
Lesson 2
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© Houghto
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0.5 Module 10
ESE3898
0L2 445 30_U4M1
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445
Lesson 10.2
445 4/9/14
7:10 PM
07/04/14 7:21 PM
Note the similarities between the y-intercept and a. What is their relationship?
EXPLORE
The value of a in y = ab x is the y-intercept.
Exploring Graphs of Exponential Functions
Reflect
1.
Discussion What is the domain for any exponential function y = ab x? all real numbers, or -∞ < x < ∞.
2.
Discussion Describe the values of b for all functions y = ab x The base b is positive, or b > 0, in functions of the form y = ab x
Explain 1
INTEGRATE TECHNOLOGY Have students complete the Explore activity in either the book or online lesson.
Graphing Increasing Positive Exponential Functions
The symbol ∞ represents infinity. We can describe the end behavior of a function by describing what happens to the function values as x approaches positive infinity (x → ∞) and as x approaches negative infinity (x → -∞). Example 1
QUESTIONING STRATEGIES
Graph each exponential function. After graphing, identify a and b, the y-intercept, and the end behavior of the graph. Use inequalities to discuss the behavior of the graph.
What is the relationship between the value of a and the y-intercept of the graph of ƒ(x) = ab x ? The y-intercept is equal to a.
ƒ(x) = 2 x Choose several values of x and generate ordered pairs.
x -1
f(x) = 2 x
EXPLAIN 1
0.5
0
1
1
2
2
4
Graphing Increasing Positive Exponential Functions
Graph the ordered pairs and connect them with a smooth curve.
y
f(x) = 2x (2, 4)
4
b=2
2 (-1, 0.5)
y-intercept: (0, 1) End Behavior: As x-values approach positive infinity (x → ∞), y-values approach positive infinity (y → ∞). As x-values approach negative infinity (x → -∞), y-values approach zero (y → 0).
-4
-2
0
(1, 2) (0, 1) 2
x 4
-2 -4
Using symbols only, we say: As x → ∞, y → ∞, and as x → -∞, y → 0.
Module 10
446
© Houghton Mifflin Harcourt Publishing Company
a=1
AVOID COMMON ERRORS When generating ordered pairs for exponential functions of the form ƒ(x) = ab x, students may multiply a by b and then raise that product to x. Remind students to use the correct order of operations.
Lesson 2
PROFESSIONAL DEVELOPMENT IN2_MNLESE389830_U4M10L2.indd 446
Integrate Mathematical Practices
4/2/14 1:47 AM
This lesson provides an opportunity to address Mathematical Practice MP.4, which calls for students to use “modeling.” Students use tables and graphs to represent exponential functions. They can use the graphs to determine the end behavior of the functions, and make generalizations about the effect of parameters a and b on the end behavior of an exponential function of the form f(x) = ab x.
Graphing Exponential Functions 446
B
QUESTIONING STRATEGIES
ƒ(x) = 3(4)
x
Choose several values of x and generate ordered pairs.
If an exponential function is of the form ƒ(x) = ab x, both a and b are positive real numbers, and the domain is the set of all real numbers, what is the range of the function? The range is set of all positive real numbers.
x -1
f(x) = 3(4)
x
0.75 3 12 48
0 1 2
Graph the ordered pairs and connect them with a smooth curve.
y 45
b= 4
30
f(x) = 3(4)x
15 (-1, 0.75)
x
y-intercept:
(0,3)
End Behavior: As x → ∞, y → ∞ and as x → -∞, y → 0 .
-4
-2
(1, 12) (0, 3) 0 2 4
-15
Reflect
3.
(2, 48)
a= 3
If a > 0 and b > 1, what is the end behavior of the graph? If a > 0 and b > 1, as x approaches infinity y approaches infinity, and as x approaches
negative infinity y approaches 0. 4.
Describe the y-intercept of the exponential function ƒ(x) = ab x in terms of a and b. A graph of the exponential f(x) = ab x will have a y-intercept at (0, a).
Your Turn
5.
ƒ(x) = 2(2)
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x -1 0 1 2
x
f(x) = 2(2)
x
9
1 2 4 8
6 (-1, 1) -4
a=2
b=2
y-intercept: (0, 2)
-2
3 0
y (2, 8) f(x) = 2(2)x (1, 4) (0, 2) 2
x 4
-3 -6
End Behavior: As x →∞, y →∞ and as x → -∞, y → 0.
Module 10
447
Lesson 2
COLLABORATIVE LEARNING IN2_MNLESE389830_U4M10L2.indd 447
Peer-to-Peer Activity In pairs, have students develop and record their own rules for determining whether a graph represents one of the four main types of exponential functions: increasing positive, decreasing negative, decreasing positive, and increasing negative. For example, the rules might be a > 0 and b > 1, a < 0 and 0 < b < 1, a > 0 and 0 < b < 1, and a < 0 and b > 1. Then, give each pair one of each type of function and its graph. Have the students use their rules to determine which type of exponential function each graph represents. As a class, discuss the rules that worked.
447
Lesson 10.2
4/2/14 1:47 AM
Graphing Decreasing Negative Exponential Functions
Explain 2
EXPLAIN 2
You can use end behavior to discuss the behavior of a graph. Example 2
Graphing Decreasing Negative Exponential Functions
Graph each exponential function. After graphing, identify a and b, the y-intercept, and the end behavior of the graph. Use end behavior to discuss the behavior of the graph.
ƒ(x) = -2(3)
x
QUESTIONING STRATEGIES
Choose several values of x and generate ordered pairs.
x -1
f(x) = -2(3) -0.7 -2
0
-6
1
-18
2
Graph the ordered pairs and connect them with a smooth curve.
y
a = -2
6
b=3
x -4
y-intercept: (0, -2)
ƒ(x) = -3(4)
-1 1 2
x
Graph the ordered pairs and connect them with a smooth curve. a = -3 b= 4 y-intercept:
(2, -18)
40
y
20
( 0 , -3 )
End Behavior: As x → ∞, y → -∞ and as x → -∞, y → 0 .
f(x) = -3(4)x (0, -3) -4 -2 0 2 4 (1, -12) -20 -40 -60
Module 10
448
x
© Houghton Mifflin Harcourt Publishing Company
0
f(x) = -3(4) -0.75 -3 -12 -48
Students may forget that by the order of operations x rules, ƒ(x) = -2 x means that ƒ(x) = -1(2 ), and therefore the negative sign is not raised to the power. Tell students that the base of an exponential function cannot be negative.
f(x) = -2(3)x
-18
Choose several values of x and generate ordered pairs.
AVOID COMMON ERRORS
2 4 (1, -6)
-12
x
x
-2
(0, -2) -6
End Behavior: As x → ∞, y → -∞ and as x → -∞, y → 0.
How can the value of a tell you in which quadrants the graph of an exponential function in the form ƒ(x) = ab x is located? If a is greater than 0, the graph is located in quadrants I and II. If a is less than 0, the graph is located in quadrants III and IV.
x
(2, -48)
Lesson 2
DIFFERENTIATE INSTRUCTION IN2_MNLESE389830_U4M10L2.indd 448
Cognitive Strategies
4/2/14 1:47 AM
Guide students to see that for exponential functions of the form f (x) = ab x , b > 1 represents growth (the graph is increasing), and b < 1 represents decline (the graph is decreasing). Remind students that if b = 1, the function is linear and not exponential; the function represents no change: no growth or decline.
Graphing Exponential Functions
448
Reflect
EXPLAIN 3
6.
Graphing Decreasing Positive Exponential Functions
If a < 0 and b > 1, what is the end behavior of the graph? If a < 0 and b > 1, as x approaches infinity y approaches negative infinity, and as
x approaches negative infinity y approaches 0. Your Turn
7.
QUESTIONING STRATEGIES
ƒ(x) = -3(3)
x
f(x) = -3(3)
x
If 0 < b < 1 and a is a positive constant, how can you alter b to make the graph of y = ab x decrease more gradually? Increase the value of b.
-1 0 1 2
x
20
-1 -3 -9 -27
y
10 -4
a = -3
(0, -3) 2 4 (1, -9) -10
-2
-20
b=3
f(x) = -3(3)x (2, -27)
-30
y-intercept: (0, -3)
x
0
End Behavior: As x → ∞, y → -∞ and as x → -∞, y → 0.
Explain 3 Example 3
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Graphing Decreasing Positive Exponential Functions
Graph each exponential function. After graphing, identify a and b, the y-intercept, and the end behavior of the graph. Use inequalities to discuss the behavior of the graph.
ƒ(x) = (0.5)
x
Choose several values of x and generate ordered pairs.
x -1
f(x) = (0.5)
x
2
0
1
1
0.5
2
0.25
Graph the ordered pairs and connect them with a smooth curve.
y f(x) = 0.5x
(-1, 2) 2 (1, 0.5) (2, 0.25) (0, 1)
a=1 -4
b = 0.5 y-intercept: (0, 1) End Behavior: As x → ∞, y → 0 and as x → -∞, y → ∞. Module 10
449
4
-2
0
2
x
4
-2 -4
Lesson 2
LANGUAGE SUPPORT IN2_MNLESE389830_U4M10L2 449
Connect Vocabulary Make sure that students understand what is meant by the end behavior of a function. The end behavior of a function describes what happens to the function values as x gets larger and larger (for example, as x becomes 100, 1000, and 1,000,000) and what happens to the function values as the values of x get smaller and smaller (for example, as x becomes –100, –1000, and –1,000,000). So, end behavior describes what happens to f(x) when x is farther and farther to the right and what happens to f(x) when x is farther and farther to the left.
449
Lesson 10.2
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B
ƒ(x) = 2(0.4)
x
COGNITIVE STRATEGIES
Choose several values of x and generate ordered pairs.
x -1
f(x) = 2(0.4) 5 2
0
()
1 y = 3 __ 2
0.8 0.32
1 2
Graph the ordered pairs and connect them with a smooth curve. a= 2 b = .4 y-intercept:
f(x) = 2(0.4)x
-4
-2
x
0
2
x 4
-2 -4
Reflect
8.
and y = 3(2) .
4 2 (0, 2) (1, 0.8)
End Behavior: As x → ∞, y → 0 and as x → -∞, y → ∞ .
x
y
6 (-1, 5)
(0,2)
( )
x
1 is the Show students that the graph of y = a __ b reflection of y = ab x over the y-axis by graphing
x
If a > 0 and 0 < b < 1, what is the end behavior of the graph? If a > 0 and 0 < b < 1, as x approaches infinity y approaches zero, and as x approaches
negative infinity y approaches infinity. Your Turn
9.
Graph the exponential function. After graphing, identify a and b, the y-intercept, and the end behavior of the graph. Use inequalities to discuss the behavior of the graph. ƒ(x) = 3(0.5) x
x
x
(-1, 6) 6 f(x) = 3(0.5)x 4 (0, 3) 2
6 3 1.5 0.75
-4
a=3
b = 0.5
-2
0
y
(1, 1.5) (2, 0.75) 2
x
4
-2 -4
y-intercept: (0, 3)
End Behavior: As x →∞, y → 0 and as x → -∞, y → ∞.
Module 10
IN2_MNLESE389830_U4M10L2.indd 450
450
© Houghton Mifflin Harcourt Publishing Company
-1 0 1 2
f(x) = 3(0.5)
Lesson 2
4/2/14 1:47 AM
Graphing Exponential Functions
450
Graphing Increasing Negative Exponential Functions
Explain 4
EXPLAIN 4
Graph each exponential function. After graphing, identify a and b, the y-intercept, and the end behavior of the graph. Use inequalities to discuss the behavior of the graph.
Graphing Increasing Negative Exponential Functions
Example 4
ƒ(x) = -0.5 x Choose several values of x and generate ordered pairs.
QUESTIONING STRATEGIES
f(x) = -0.5 x
x
-2
-1
Why does it make sense for the graph of f (x) = ab x to stretch when a > 1, and for the graph to shrink when 0 < a < 1, compared to the graph of f (x) = b x ? Multiplying by a factor a greater than 1 increases the outputs, compared with the function f (x) = b x, and multiplying by a factor a between 0 and 1 decreases the outputs, compared with the function f (x) = b x.
-1
0
-0.5
1
-0.25
2
Graph the ordered pairs and connect them with a smooth curve.
2
a=1
1
b = 0.5
y-intercept: (0, 1)
-2
End Behavior: As x → ∞, y → 0 and as x → -∞, y → -∞.
ƒ(x) = -3(0.4)
x
-1
f(x) = -3(0.4)
x
-3
© Houghton Mifflin Harcourt Publishing Company
451
Lesson 10.2
x
-1.2 -0.48
1 2
Graph the ordered pairs and connect them with a smooth curve.
6
a = -3
y
3
(
0 , -3
)
x -2
End Behavior: As x → ∞, y → 0 and as x → -∞, y → -∞ .
IN2_MNLESE389830_U4M10L2.indd 451
1
-3
0
Module 10
0
-1
-7.5
-1
y-intercept:
(2, -0.25) x
(1, -0.5) (0, -1) f(x) = -0.5x (-1, -2) -2
Choose several values of x and generate ordered pairs.
b = 0.4
y
451
-1
0
1
2 (1, -1.2) -3 (0, -3) f(x) = -3(0.4)x -6 (-1, -7.5) -9
Lesson 2
4/2/14 1:47 AM
Reflect
INTEGRATE TECHNOLOGY
10. If a < 0 and 0 < b < 1, what is the end behavior of the graph? If a < 0 and 0 < b < 1, as x approaches infinity y approaches zero, and as x approaches
Using graphing calculators can allow students to compare the graphs of exponential functions quickly, but they must enter the functions
negative infinity y approaches negative infinity. Your Turn
()
11. Graph the exponential function. After graphing, identify a and b, the y-intercept, and the end behavior of the graph. Use inequalities to discuss the behavior of the graph. x ƒ(x) = -2(0.5)
x -1 0 1 2
f(x) = -2(0.5)
x
4
-4 -2 -1 -0.5
y
2
ELABORATE
x -2
-1
0
2 (1, -1) -2 (0, -2) f(x) = -2(0.5)x
a = -2
(-1, -4)
b = 0.5
y-intercept: (0, -2)
1
QUESTIONING STRATEGIES
-4
Which of the following functions has the greatest y-intercept and which has the least 1 (2.5)x, y = 2 (2.5)x, y-intercept: y = 2.5 x, y = __ 2 x x or y = 4(2.5) ? Explain. y = 4 (2.5) has the 1 (2.5)x has the least greatest y-intercept (4), and y = __ 2 1 . y-intercept __ 2
-6
End Behavior: As x → ∞, y → 0 and as x → -∞, y → -∞.
Elaborate
()
12. Why is ƒ(x) = 3(-0.5) not an exponential function? x f(x) = 3(-0.5) is not an exponential function because its values alternate between x
negative and positive, creating a line that jumps between increasingly smaller positive
that takes on increasingly smaller positive or negative values as x approaches infinity.
13. Essential Question Check-In When an exponential function of the form ƒ(x) = ab x is graphed, what does a represent? a represents the y-intercept of the function.
SUMMARIZE THE LESSON
© Houghton Mifflin Harcourt Publishing Company
and negative values as x approaches infinity. An exponential function has a smooth curve
Copy and complete the graphic organizer with your students. In each box, give an example of an appropriate exponential function and sketch its graph.
Exponential Functions: y = ab x a > 0, b > 1 y = 3(2)x
IN2_MNLESE389830_U4M10L2.indd 452
452
a < 0, b > 1 y = -1(3)x
y
12
Module 10
()
x
2 is correctly. Show students that y = -3 ∗ __ 3 x 2 by graphing not the same as y = -3 ∗ __ 3 both functions on a calculator.
y
-3
x
Lesson 2
-3
4/2/14 1:47 AM
0
2
12
0
a < 0, 0 < b < 1 1 x 3
y = -2
y
x
y -3
x -3
3
x
3
-12
3
a > 0, 0 < b < 1 x y=2 1
0
0
3
-12
Graphing Exponential Functions
452
Evaluate: Homework and Practice
EVALUATE
State a, b, and the y-intercept then graph the function on a graphing calculator. 1.
ƒ(x) = 2(3)
x
2.
ƒ(x) = -6(2)
x
Graph:
Graph:
a=2
a = -6
y-intercept: (0, 2)
y-intercept: (0, -6)
• Online Homework • Hints and Help • Extra Practice
ASSIGNMENT GUIDE Concepts and Skills
Practice
Explore Exploring Graphs of Exponential Functions
Exercises 1–2, 19
Example 1 Graphing Increasing Positive Exponential Functions
Exercises 5, 9, 13, 15, 18, 20–22
Example 2 Graphing Decreasing Negative Exponential Functions
Exercises 8, 12, 16, 23
Example 3 Graphing Decreasing Positive Exponential Functions
Exercises 4, 7, 10, 17, 25
b=2
b=3
3.
ƒ(x) = -5(0.5)
x
Graph:
a = -5
a=3
On a graphing calculator, have students graph
()
x
()
x
x x 1 , and y = -3 __ 1 . y = 3(2) , y = -3(2) , y = 3 __ 2 2 Have them keep track of the changes by drawing sketches of each graph on the same coordinate grid.
© Houghton Mifflin Harcourt Publishing Company
INTEGRATE TECHNOLOGY
5.
ƒ(x) = 6(3)
x
y-intercept: (0, 3) 6.
ƒ(x) = -4(0.2)
x
Graph:
Graph:
a=6
a = -4
y-intercept: (0, 6)
y-intercept: (0, -4)
b=3
Module 10
Exercise
IN2_MNLESE389830_U4M10L2.indd 453
b = 0.2
Lesson 2
453
Depth of Knowledge (D.O.K.)
COMMON CORE
Mathematical Practices
1–8
1 Recall of Information
MP.2 Reasoning
9–18
2 Skills/Concepts
MP.2 Reasoning
2 Recall of Information
MP.5 Using Tools
20–21
1 Skills/Concepts
MP.4 Modeling
22–25
3 Strategic Thinking
MP.3 Logic
19
Lesson 10.2
x
b = 0.8
y-intercept: (0, -5)
Exercises 3, 6, 11, 14, 24
ƒ(x) = 3(0.8)
Graph:
b=5
Example 4 Graphing Increasing Negative Exponential Functions
453
4.
4/2/14 1:47 AM
7.
ƒ(x) = 7(0.9)
x
8.
ƒ(x) = -3(2)
QUESTIONING STRATEGIES
x
Graph:
Graph:
a=7
a = -3
y-intercept: (0, 7)
y-intercept: (0,-3)
b = 0.9
How do you know the exponent in ab x applies to b and not ab? By the order of operations, bases are raised to exponents before multiplying.
b=2
State a, b, and the y-intercept then graph the function and describe the end behavior of the graphs. 9.
ƒ(x) = 3(3)
10. ƒ(x) = 5(0.6)
x
Ordered pairs:
x
Ordered pairs: f(x) = 3(3)
x -1 0 1 2
x
1 3 9 27
-1 0 1 2
Graph the ordered pairs and connect them with a smooth curve.
30
y f(x) = 3(3)x
-1
9 6 (0, 5) 3
(1, 9) x
-2
2
-1
-10
0
y f(x) = 5(0.6)x (1, 3) 1
© Houghton Mifflin Harcourt Publishing Company
-2
8.3 5 3 1.8
(-1, 8.3)
(2, 27)
(0, 3) 0 1
x
Graph the ordered pairs and connect them with a smooth curve.
20 10 (-1, 1)
f(x) = 5(0.6)
x
(2, 1.8) 2 x
-3 -6
a=3
a=5
y-intercept: (0, 3)
y-intercept: (0, 5)
End Behavior: As x → ∞, y → ∞ and as x → -∞, y → 0.
End Behavior: As x → ∞, y → 0 and as x → -∞, y → ∞.
b=3
Module 10
IN2_MNLESE389830_U4M10L2.indd 454
b = 0.6
454
Lesson 2
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Graphing Exponential Functions
454
11. ƒ(x) = -6(0.7)
AVOID COMMON ERRORS Students may have trouble raising a number to a negative power. Remind them that a number raised to a negative power is the reciprocal of the number raised to the opposite power. For example,
12. ƒ(x) = -4(3)
x
f(x) = -6(0.7)
x -1 0 1 2
()
x
2
6
f(x) = -4(3)
x -1 0 1 2
-8.6 -6 -4.2 -2.9
1 = ___ 1 . 4-2 = __ 4 16
x
x
-1.3 -4 -12 -36
y
y
8
3
4 x
-2
-1
0
x -2 -1 (-1, -1.3)
1 2 (2, -2.9)
-3 f(x) = -6(0.7)x (1, -4.2) -6 (0, -6) (-1, -8.6) -9
a = -4; b = 3
f(x) = 5(2)
x -1 0 1 2
© Houghton Mifflin Harcourt Publishing Company
End Behavior: As x → ∞, y → -∞ and as x → -∞, y → 0.
14. ƒ(x) = -2(0.8)
x
x
21
y
-1 0 1 2
1
x -2
x
-7
0
1 2 (2, -1.3)
a = -2; b = 0.8
y-intercept: (0, -2)
End Behavior: As x → ∞, y → 0 and as x → -∞, y → -∞.
End Behavior: As x → ∞, y → ∞ and as x → -∞, y → 0.
Lesson 10.2
-1
-1 f(x) = -2(0.8)x (1, -1.6) -2 (0, -2) (-1, -2.5) -3
2
y-intercept: (0, 5)
455
y
1
a = 5; b = 2
IN2_MNLESE389830_U4M10L2.indd 455
x
-2.5 -2 -1.6 -1.3
2
-14
Module 10
f(x) = -2(0.8)
(2, 20)
14 f(x) = 5(2)x (1, 10) 7 (-1, 2.5) (0, 5) 0
x
x
2.5 5 10 20
-1
(1, -12)
y-intercept: (0, -4)
End Behavior: As x → ∞, y → 0 and as x → -∞, y → -∞.
-2
1 2 -4 (0, -4) f(x) = -4(3)x -8
-12
a = -6; b = 0.7 y-intercept: (0, -6)
13. ƒ(x) = 5(2)
0
455
Lesson 2
4/2/14 1:47 AM
15. ƒ(x) = 9(3)
16. ƒ(x) = -5(2)
x
f(x) = 9(3)
x -1 0 1 2
x
90
f(x) = -5(2)
x -1 0 1 2
3 9 27 81 y
COGNITIVE STRATEGIES
x
-2.5 -5 -10 -20
(2, 81)
14 7
f(x) = 9(3)x (-1, 3)
30 (0, 9)
(1, 27)
-2 (-1, -2.5)
x
-30 -60
-21
0
1
2
a = 9; b = 3
f(x) = 7(0.4)
f(x) = 6(2)
x -1 0 1 2
17.5 7 2.8 1.1
3 6 12 24
y
y
18
30
12 f(x) = 7(0.4)x (0, 7) 6 (1, 2.8)
20
0
1
(-1, 3)
2
-2
-6
a = 7; b = 0.4
IN2_MNLESE389830_U4M10L2.indd 456
(2, 24) (1, 12) x
(0, 6)
-1
0
1
2
-10
a = 6; b = 2
y-intercept: (0, 7) End Behavior: As x → ∞, y → 0 and as x → -∞, y → ∞.
f(x) = 6(2)x
10
(2, 1.1) x
x
© Houghton Mifflin Harcourt Publishing Company
-1 0 1 2
Module 10
(2, -20)
x
x
x
-1
asymptote. A horizontal asymptote is a line that the graph gets closer and closer to, but never reaches. The line y = 0 is the horizontal asymptote of the graphs in this lesson.
(1, -10)
End Behavior: As x → ∞, y → -∞ and as x → -∞, y → 0. 18. ƒ(x) = 6(2)
x
-2
2
a = -5; b = 2
End Behavior: As x → ∞, y → ∞ and as x → -∞, y → 0.
(-1, 17.5)
x 1 (0, -5)
y-intercept: (0, -5)
y-intercept: (0, 9)
17. ƒ(x) = 7(0.4)
0
-7 f(x) = -5(2)x -14
-1
INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 Introduce students to the term horizontal
y
60
-2
Show students that the graph of y = -ab x is the reflection of y = ab x over the x-axis.
x
y-intercept: (0, 6)
456
End Behavior: As x → ∞, y → ∞ and as x → -∞, y → 0.
Lesson 2
4/2/14 1:47 AM
Graphing Exponential Functions
456
19. Identify the domain and range of each function. Make sure to provide these answers using inequalities. ⎫ ⎫ ⎧ ⎧ x Domain: ⎨x⎢ -∞ < x 0⎬ a. ƒ(x) = 3(2) ⎭ ⎭ ⎩ ⎩ x ⎫ ⎫ ⎧ ⎧ Domain: ⎨x⎢ -∞ < x 0⎬ b. ƒ(x) = 7(0.4) ⎭ ⎭ ⎩ ⎩ x ⎫ ⎫ ⎧ ⎧ c. ƒ(x) = -2(0.6) Domain: ⎨x⎢ -∞ < x 1 and a > 0, increases without bound as x increases without bound, and approaches 0 as x decreases without bound. The value of an exponential decay function f(x) = ab x, where 0 < b < 1 and a > 0, approaches 0 as x increases without bound, and increases without bound as x decreases without bound. Such end behavior can also be described using the notation f(x) → ∞ as x → ∞ and f(x) → 0 as x → ∞, where the symbol ∞ represents infinity.
4/2/14 1:50 AM
QUESTIONING STRATEGIES Before you graph it, how can you tell that the x function f(x) = 500(0.8) will decrease as x increases? A positive number is being multiplied x times by a number between 0 and 1, so it will get smaller as x increases.
Modeling Exponential Growth and Decay
460
Recall that a function of the form y = ab x represents exponential growth when a > 0 and b > 1. If b is replaced by t 1 + r and x is replaced by t, then the function is the exponential growth model y = a(1 + r) , where a is the initial amount, the base (1 + r) is the growth factor, r is the growth rate, and t is the time interval. The value of the model increases with time.
What does it mean for a function to approach 0 as x increases without bound? Is 0 a value of the function? The value of the function gets closer and closer to 0 as x increases, but it never reaches 0.
Example 1
EXPLAIN 1
A painting is sold for $1800, and its value increases by 11% each year after it is sold. Find the value of the painting in 30 years.
y = a(1 + r)
t
= 1800(1 + 0.11) = 1800(1.11)
QUESTIONING STRATEGIES
t
t
Find the value in 30 years. y = 1800(1.11) = 1800(1.11)
t
30
≈ 41,206.13 After 30 years, the painting will be worth approximately $41,206.13. Create a table of values to graph the function. © Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Tetra Images/Corbis
Why doesn’t the graph of an exponential growth model intersect the y-axis at the origin? When x = 0, y is the initial value of the increasing quantity.
Write an exponential growth function for each situation. Graph each function and state its domain, range and an asymptote. What does the y-intercept represent in the context of the problem?
Write the exponential growth function for this situation.
Modeling Exponential Growth
How do you use a percent increase amount when writing an exponential growth equation? Give an example. Convert the percent increase to a decimal and add it to 1 to find the number that is raised to a power. Possible answer: For a 4% increase, the decimal 1.04 will be raised to a power.
Modeling Exponential Growth
Explain 1
QUESTIONING STRATEGIES
t
y
0
1800.00
(t, y)
y
(0, 1800)
44,000
8
4148.20
(8, 4148.20)
16
9559.60
(16, 9559.60)
24
22,030.00
(24, 22,030)
32
50,770.00
(32, 50,770.00)
Determine the domain, range and an asymptote of the function.
(32, 50,770.00)
33,000 22,000
(24, 22,030)
11,000 (0, 1800) (16, 9559.60) t (8, 4148.20) 0 8 16 24 32
The domain is the set of real numbers t such that t ≥ 0. The range is the set of real numbers y such that y ≥ 1800. An asymptote for the function is y = 0. The y-intercept is the value of y when t = 0, which is the value of the painting when it was sold.
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Lesson 3
COLLABORATIVE LEARNING IN2_MNLESE389830_U4M10L3.indd 461
Small Group Activity Divide students into groups of three or four. Have students research several banks, either locally or online, for interest rates on savings accounts, money market accounts, and CDs. Have them work as a group to develop exponential growth models for each type of account at each bank on an initial deposit of $500. Ask students to make a chart comparing the rates and potential earnings after 5 years for the different types of accounts at each bank.
461
Lesson 10.3
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B
A baseball trading card is sold for $2, and its value increases by 8% each year after it is sold. Find the value of the baseball trading card in 10 years. Write the exponential growth function for this situation. y = a(1 + r)
t
(1 + 0.08 ) 2 ( 1.08 )
= 2 =
t
t
Find the value in 10 years. y = a(1 + r)
t
( 1.08 ) 2 ( 1.08 )
= 2 =
t
10
≈ 4.32
After 10 years, the baseball trading card will be worth approximately $ 4.32 million dollars . Create a table of values to graph the function.
t
(t, y)
y
6
(0, 2)
0
2
3
2.5194
6
3.1737
9
3.998
12
5.0363
4.5
(3, 2.5194) (6, 3.1737) (9, 3.998) (12, 5.0363)
3 1.5 0
y (12, 5.0363) (9, 3.998) (6, 3.1737) (3, 2.5194) (0, 2) 3
t 6
9
12
Determine the domain, range, and an asymptote of the function.
The range is the set of real numbers y such that y ≥ 2 . An asymptote for the function is
y=0
.
The y-intercept is the value of y when t = 0, which is the
value of the card when it was sold
.
Reflect
8.
Find a recursive rule that models the exponential growth of y = 1800( 1.11 )t. a 1 = 1800, a n = 1.11a n-1
9.
Find a recursive rule that models the exponential growth of y = 2(1.08) . a 1 = 2, a n = 1.08a n-1
Module 10
© Houghton Mifflin Harcourt Publishing Company
The domain is the set of real numbers t such that t ≥ 0 .
t
462
Lesson 3
DIFFERENTIATE INSTRUCTION IN2_MNLESE389830_U4M10L3.indd 462
Critical Thinking
4/2/14 1:50 AM
To help students understand the difference between exponential growth and linear growth, point out that in exponential growth or decay, the amount added or subtracted in each time period is proportional to the amount already present. For exponential growth, this means that as the amount becomes greater, the amount of increase in each time period also becomes greater. Contrast this to linear growth, in which the amount of increase remains constant.
Modeling Exponential Growth and Decay
462
Your Turn
EXPLAIN 2
10. Write and graph an exponential growth function, and state the domain and range. Tell what the y-intercept represents. Sara sold a coin for $3, and its value increases by 2% each year after it is sold. Find the value of the coin in 8 years. t 8 y = a (1 + r ) y = 3(1.02)
Modeling Exponential Decay
= 3(1.02)
≈ 3.51
t
QUESTIONING STRATEGIES
4 3
y (6, 3.38) (2, 3.12) (4, 3.25) (8, 3.52) (0, 3)
2 1
After 8 years, the coin will be worth approximately $3.51.
What is the domain for a function that represents the exponential growth or decay of a population? Explain. The domain is all non-negative real numbers, because the model applies to all future times (positive values of t) but not past times (negative values of t).
The domain is the set of real numbers t such that t ≥ 0.
t 0
2
4
6
8
The range is the set of real numbers y such that y ≥ 3.
The y-intercept is the value of y when t = 0, which is the value of the coin when it was sold.
Modeling Exponential Decay
Explain 2
Recall that a function of the form y = ab x represents exponential decay when a > 0 and 0 < b < 1. If b is replaced t by 1 - r and x is replaced by t, then the function is the exponential decay model y = a(1 - r) , where a is the initial amount, the base (1 - r) is the decay factor, r is the decay rate, and t is the time interval.
What is the range for a function describing the exponential growth or decay of a population? For a growth function, the range consists of all numbers greater than or equal to the starting value of the population. For a decay function, the range includes all numbers between the starting value and 0.
Example 2
Write an exponential decay function for each situation. Graph each function and state its domain and range. What does the y-intercept represent in the context of the problem?
The population of a town is decreasing at a rate of 3% per year. In 2005, there were 1600 people. Find the population in 2013. Write the exponential decay function for this situation. y = a(1 - r)
t
© Houghton Mifflin Harcourt Publishing Company
= 1600(1 - 0.03) = 1600(0.97)
t
t
Find the value in 8 years. y = 1600(0.97)
t
= 1600(0.97)
8
≈ 1254 After 8 years, the town’s population will be about 1254 people.
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463
Lesson 3
LANGUAGE SUPPORT IN2_MNLESE389830_U4M10L3.indd 463
Connect Vocabulary Remind students that exponential growth refers to an increasing function and exponential decay refers to a decreasing function. While students are probably familiar with the word growth, they may be less familiar with decay. Explain that, in everyday use, it can mean rot, or loss of strength and health. Have students list key words that can indicate growth or decay in descriptions of real-world situations. Examples of key words for growth include increases, goes up, rises, gains. Examples of key words for decay include decreases, goes down, falls, loses value, declines, depreciates.
463
Lesson 10.3
4/2/14 1:50 AM
Create a table of values to graph the function.
(t, y)
t
y
0
1600
8
1254
16
983
24
770
32
604
y (0, 1600)
1600
(8, 1254)
1200
(0, 1600)
(16, 983)
800
(8, 1254)
(24, 770) (32, 604)
400
(16, 983)
t
(24, 770)
0
8
16
24
32
(32, 604)
Determine the domain and range of the function. The domain is the set of real numbers t such that t ≥ 0 The range is the set of real numbers y such that 0 ≤ y ≤ 1600. The y-intercept is the value of y when t = 0, the number of people before it started to lose population.
B
The value of a car is depreciating at a rate of 5% per year. In 2010, the car was worth $32,000. Find the worth of the car in 2013. Write the exponential decay function for this situation. y = a(1 - r)
t
( ) 32,000 ( 0.95 )
t
= 32,000 1 - 0.05 t
=
Find the value in 3 years. y = a(1 + r)
t
( 0.95 ) 32,000 ( 0.95 ) ≈
= 32,000
3
27,436
© Houghton Mifflin Harcourt Publishing Company
=
t
After 3 years, the car’s value will be $ 27,436 . Create a table of values to graph the function.
t
y
0
32,000
1
30,400
2
28,880
3
27,346
(t, y)
(0, 32,000) (1, 30,400) (2, 28,880) (3, 27,346)
Determine the domain and range of the function.
y (1, 30,400) 32,000 (3, 28,346) (0, 32,000) (2, 28,880) 24,000 16,000 8,000 t 0
1
2
3
4
The domain is the set of real numbers t such that t ≥ 0 . The range is the set of real numbers y such that 0 ≤ y ≤ 32,000 .
The y-intercept, 32,000, is the value of y when t = 0, the original value of the car.
Module 10
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Modeling Exponential Growth and Decay
464
Reflect
EXPLAIN 3
11. Find a recursive rule that models the exponential decay of y = 1600(0.97) . a 1 = 1,600, a n = 0.97a n-1 t
Comparing Exponential Growth and Decay
12. Find a recursive rule that models the exponential decay of y = 32,000(0.95) . a 1 = 32,000, a n = 0.95a n-1 t
Your Turn
AVOID COMMON ERRORS
13. The value of a boat is depreciating at a rate of 9% per year. In 2006, the boat was worth $17,800. Find the worth of the boat in 2013. Write an exponential decay function for this situation. Graph the function and state its domain and range. What does the y-intercept represent in the context of the problem?
Some students may forget to convert the percent growth rate to decimal form. Remind them that growth rate must be written as a decimal because the percent sign means “parts out of 100.”
y = a(1 - r) = 17,800(0.91) t
t
y 20,000 (0, 17,800) (2, 14,740) 15,000
7
After 7 years, the boat will be worth approximately $9,198.35. The domain is the set of real numbers t such that t ≥ 0. The range is the set of real numbers y such that 0 ≤ y ≤ 17,800.
QUESTIONING STRATEGIES
(8, 8370.5)
5,000
y = 17,800(0.91) ≈ 9198.35
(6, 10,108)
(4, 12,206)
10,000
t 0
2
4
6
8
The y-intercept is 17,800, the value of y when t = 0, which is the original value of the boat.
When you graph two functions on the same coordinate grid, what does the intersection point of the graphs represent? the time when the two functions are equal in value
Explain 3
Comparing Exponential Growth and Decay
Graphs can be used to describe and compare exponential growth and exponential decay models over time. Example 3
© Houghton Mifflin Harcourt Publishing Company
Use the graphs provided to write the equations of the functions. Then describe and compare the behaviors of both functions.
The graph shows the value of two different shares of stock over the period of 4 years since they were purchased. The values have been changing exponentially.
16 12
y (0, 16) Stock A (1, 12)
8
The graph for Stock A shows that the value of the stock is decreasing as time increases.
4
The initial value, when t = 0, is 16. The value when t = 1 is 12. Since 12 ÷ 16 = 0.75, the function that represents the value of Stock A after t t years is A(t) = 16(0.75) . A(t) is an exponential decay function.
0
Stock B (0, 2) 1
(1, 3) 2
x 3
4
The graph for Stock B shows that the value of the stock is increasing as time increases. The initial value, when t = 0, is 2. The value when t = 1 is 3. Since 3 ÷ 2 = 1.5, the t function that represents the value of Stock B after t years is B(t) = 2(1.5) . B(t) is an exponential growth function. The value of Stock A is going down over time. The value of Stock B is going up over time. The initial value of Stock A is greater than the initial value of Stock B. However, after about 3 years, the value of Stock B becomes greater than the value of Stock A. Module 10
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Lesson 10.3
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Lesson 3
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B
The graph shows the value of two different shares of stocks over the period of 4 years since they were purchased. The values have been changing exponentially. The graph for Stock A shows that the value of the stock is
decreasing
y 100 (0, 100) Stock A 75
as time increases.
The initial value, when t = 0, is 100 . The value when t = 1 is 50 . Since 50 ÷ 100 = 0.5 , the function that
t
represents the value of Stock A after t years is A(t) = 100 0.5 . A(t) is an exponential decay function.
25 (0, 1.5) 0
exponential growth and decay, ensure that students understand the meaning of the real-world values on the graph. Point out that the independent variable represents time, while the meaning of the dependent variable can vary with the situation. Remind students to label the axes of their graphs with the correct quantities and units. Students should also recognize that the y-intercept is the value of the independent variable at time t = 0, which may be called its initial value or starting value.
(1, 50)
50
( )
INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 When modeling real-world examples of Stock B
(1, 3) 1
2
3
4
The graph for Stock B shows that the value of the stock is increasing as time increases. The initial value, when t = 0, is 1.5 . The value when t = 1 is 3 . Since 3 ÷ 100 = 0.5 , the function that represents the value of Stock B after t years is B(t) = 1.5 ( 2 )t. B(t) is an exponential
growth function. The value of Stock A is going down over time. The value of Stock B is going up over time. The initial value of Stock A is greater than the initial value of Stock B. However, after about
3 years,
the value of Stock B becomes greater than the value of Stock A. Reflect
14. Discussion In the function B(t) = 1.5(2) , is it likely that the value of B can be accurately predicted in 50 years? The exponential function grows much more quickly than any stock can reasonably grow, it t
is unlikely that the value of B can be accurately predicted in 50 years by using the function.
15. The graph shows the value of two different shares of stocks over the period of 4 years since they were purchased. The values have been changing exponentially. Use the graphs provided to write the equations of the functions. Then describe and compare the behaviors of both functions. y 150 (0, 150) Stock B: Stock A: Stock A 100 t(0) = 5 t(0) = 150 (1, 45) t(1) = 45 t(1) = 15 50 Stock B (1, 15) 45 ÷ 150 = .3 15 ÷ 5 = 3 t (0, 5) t t 0 1 2 3 ) ) ( ( ) ) ( ( A t = 150 0.3 B t =5 3
© Houghton Mifflin Harcourt Publishing Company
Your Turn
The value of Stock A is going down over time. The value of Stock B is going up over time. The initial value of Stock A is greater than the initial value of Stock B. However, after about 1.5 years, the value of Stock B becomes greater than the value of Stock A. Module 10
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Lesson 3
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Modeling Exponential Growth and Decay
466
Elaborate
ELABORATE
16. If b > 1 in a function of the form y = ab x, is the function an example of exponential growth or an example of exponential decay? If b >1, the function is an example of exponential growth.
QUESTIONING STRATEGIES
17. What is an asymptote of the function y = 35( 1.1 )x ? x An asymptote of the function y = 35(1.1) is y = 0.
Describe three real-world situations that can be described by exponential growth or exponential decay functions. Possible answers: interest earned on an investment, population growth or decline, radioactive decay
18. Essential Question Check-In What equation should be used when modeling an exponential function that models a decrease in a quantity over time? When modeling an exponential function that models a decrease in a quantity over time,
the equation y = a(1 - r) should be used. t
SUMMARIZE How do you write an exponential growth or decay function? The formula for growth is t y = a(1 + r) , and the formula for decay is t y = a(1 - r) , where y represents the final amount, a represents the original amount, r represents the rate of growth expressed as a decimal, and t represents time.
Evaluate: Homework and Practice Graph the function on a graphing calculator, and state its domain, range, end behavior, and an asymptote. 1.
ƒ(x) = 300(1.16)
© Houghton Mifflin Harcourt Publishing Company
ƒ(x) = 65(1.64)
x
ƒ(x) = 800(0.85)
x
Domain: {x| - ∞ < x < ∞} Range: {y| y > 0} End behavior: As x → -∞, y → ∞ and as x → ∞, y → 0 Asymptote: y = 0 4.
Domain: {x| - ∞ < x < ∞} Range:{y| y > 0} End behavior: As x → -∞, y → 0 and as x → ∞, y → ∞ Asymptote: y = 0
ƒ(x) = 57(0.77)
x
Domain: {x| - ∞ < x < ∞} Range: {y| y > 0} End behavior: As x → -∞, y → ∞ and as x → ∞, y → 0 Asymptote: y = 0
Write an exponential function to model each situation. Then find the value of the function after the given amount of time. 5.
Annual sales for a company are $155,000 and increases at a rate of 8% per year for 9 years.
6.
The value of a textbook is $69 and decreases at a rate of 15% per year for 11 years.
y = 69(0.85)
y = 155,000(1.08)
t
= 69(0.85)
y = 155,000(1.08)
9
t
11
= 155,000(1.999)
= 69(0.167)
= $309,845.72
= $11.52
Module 10
IN2_MNLESE389830_U4M10L3.indd 467
Lesson 10.3
2.
Domain: {x| - ∞ < x < ∞} Range: {y| y > 0} End behavior: As x → -∞, y → 0 and as x → ∞, y → ∞ Asymptote: y = 0 3.
467
x
• Online Homework • Hints and Help • Extra Practice
467
Lesson 3
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7.
A new savings account is opened with $300 and gains 3.1% yearly for 5 years.
8.
y = 300(1.031)
y = 7800(0.92)
= 300(1.16)
= 7800(0.61)
t
= 300(1.031)
= 7800(0.92)
5
= $349.47
9.
The value of a car is $7800 and decreases at a rate of 8% yearly for 6 years.
EVALUATE
t
6
= $4729.57
The starting salary at a construction company is fixed at $55,000 and increases at a rate of 1.8% yearly for 4 years.
y = 55,000(1.018)
ASSIGNMENT GUIDE
t
= 55,000(1.018)
4
= 55,000(1.074) = $59,068.21
10. The value of a piece of fine jewelry is $280 and decreases at a rate of 3% yearly for 7 years.
y = 280(0.97) = 280(0.97)
11. The population of a town is 24,000 and is increasing at a rate of 6% per year for 3 years.
y = 24,000(1.06)
t
= 24,000(1.06)
7
= 280(0.808)
t
= 28,584
12. The value of a new stadium is $3.4 million and decreases at a rate of 2.39% yearly for 10 years.
y = 3.4(0.9761)
t
10
= 3.4(0.785)
= $2.67 million
Write an exponential function for each situation. Graph each function and state its domain and range. Determine what the y-intercept represents in the context of the problem.
200,000
13. The value of a boat is depreciating at a rate of 7% per year. In 2004, the boat was worth $192,000. Find the worth of the boat in 2013.
100,000
y = a(1 - r) = 192,000(0.93) t
y (0, 192,000)
150,000
t
50,000
y = 192,000(0.93) ≈ 99,918,93 9
(4, 143,626) (8, 107,440) (12, 80,370) t
After 9 years, the boat will be worth approximately $99,918.93. Domain: {x|0 < x < ∞} Range: {y| y > 0}
0
4
8
12
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©CoolKengzz/Shutterstock
= 3.4(0.9761)
Exercise
IN2_MNLESE389830_U4M10L3.indd 468
COMMON CORE
Mathematical Practices
1–4
1 Recall of Information
MP.5 Using Tools
5–12
1 Recall of Information
MP.4 Modeling
13–20
2 Skills/Concepts
MP.6 Precision
21–24
2 Skills/Concepts
MP.5 Using Tools
1 Recall of Information
MP.5 Using Tools
3 Strategic Thinking
MP.3 Logic
25 26–28
Exercises 1, 3
Explore 2 Describing End Behavior of a Decay Function
Exercises 2, 4
Example 1 Modeling Exponential Growth
Exercises 5, 7, 9, 11, 14, 16, 18, 20, 25–26
Example 2 Modeling Exponential Decay
Exercises 6, 8, 10, 12–13, 15, 17, 19, 27–28
Example 3 Comparing Exponential Growth and Decay
Exercises 21–24
results on a graphing calculator. They can enter the function in the Y = menu and then press GRAPH. Then they can press the TABLE key to get a list of x- and y-values, and can scroll up or down to see more values. This method is especially convenient when seeking the value of the function for more than one input value, because it is an alternative to repeatedly typing the expression into the calculator.
Lesson 3
468
Depth of Knowledge (D.O.K.)
Explore 1 Describing End Behavior of a Growth Function
INTEGRATE MATHEMATICAL PRACTICES Focus on Technology MP.5 Students may want to check their
The y-intercept is 192,000, the original value of the boat in 2004.
Module 10
Practice
3
= 24,000(1.191)
= $226.24
Concepts and Skills
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Modeling Exponential Growth and Decay
468
14. The value of a collectible baseball card is increasing at a rate of 0.5% per year. In 2000, the card was worth $1350. Find the worth of the card in 2013.
VISUAL CUES Focus on visual cues as you discuss how to recognize exponential growth and decay from a graph. Students should recognize that a rising line represents growth and a falling line represents decay. The steeper the line, the greater the rate of growth or decay.
y = a(1 + r) = 1350(1.005) t
t
y = 1350(1.005) ≈ 1440.43 13
After 13 years, the card will be worth approximately $1440.43. Domain: {x|0 < x < ∞} Range: {y| y > 0} The y-intercept is 1350.00, the original value of the card in 2000.
AVOID COMMON ERRORS
500 t 0
15. The value of an airplane is depreciating at a rate of 7% per year. In 2004, the airplane was worth $51.5 million. Find the worth of the airplane in 2013.
CONNECT VOCABULARY
y = a(1 - r) = 51.5(0.93) t
Students may not be familiar with the word depreciate. Explain that to depreciate is to decrease in value. One meaning of the word appreciate is the opposite of depreciate: to increase in value.
30
After 9 years, the airplane will be worth approximately $26.8 million.
20
Domain: {x|0 < x < ∞} Range: {y| y > 0}
10
The y-intercept is 51.5, the original value of the airplane in 2004.
16. The value of a movie poster is increasing at a rate of 3.5% per year. In 1990, the poster was worth $20.25. Find the worth of the poster in 2013.
y = a(1 + r) = 20.25(1.035) t
y = 20.25(1.035)
23
t
≈ 44.67
Domain: {x|0 < x < ∞} Range: {y| y > 0} The y-intercept is 20.25, the original value of the poster in 1990.
469
10
15
y (0, 51.5) (4, 38.5) (6, 33.3)
(10, 24.9)
(8, 28.8)
t 0
50
2
4
6
8
10
y
40
(15, 33.93)
30 (5, 24.05)
After 23 years, the poster will be worth approximately $44.67.
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40 (2, 44.5)
t
9
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50
y = 51.5(0.93) ≈ 26.8
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(15, 1454.90)
(0, 1350) (10, 1419.00) 1,000
Some students may forget to add 1 to the rate of growth in the exponential growth model. Remind students that 1 + r is the growth factor in the model.
469
(5, 1384.10) y
1,500
20 (0, 20.25)
(20, 40.29)
(10, 28.57)
10 t 0
5
10
15
20
25
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17. The value of a couch is decreasing at a rate of 6.2% per year. In 2007, the couch was worth $1232. Find the worth of the couch in 2014. The y-intercept is 20.25, the value of y when t = 0, which is the original value of the poster in 1990.
y = a(1 - r) = 1232(0.938) t
1600
y
1200 (2, 1083.97) 800
t
y = 1232(0.938) ≈ 787.10 7
QUESTIONING STRATEGIES
(0, 1232.00)
The population of a town is decreasing at the rate of 1% a year. You are asked to find the population after four years. Solving the exponential growth function, you get an answer of 1248.77. What answer will you record? Explain. Round the answer to 1249 because there cannot be a fraction of a person.
(4, 953.72) (6, 839.19) (8, 738.30)
400
After 7 years, the couch will be worth approximately $787.10. Domain: {x|0 < x < ∞} Range: {y| y > 0}
0
2
4
6
8
The y-intercept is 1232, the original value of the couch in 2007. 18. The population of a town is increasing at a rate of 2.2% per year. In 2001, the town had a population of 34,567. Find the population of the town in 2018.
y = a(1 + r) = 34,567(1.022) t
50,000
y
(9, 42,046) (3, 36,899)
40,000
t
y = 34,567(1.022) ≈ 50,041
30,000 (0, 34,567)
After 17 years, the town will have about 50,041 people.
20,000
17
Domain: {x|0 < x < ∞} Range: {y| y > 0}
y = a(1 - r) = 131,000(0.946) t
t
y = 131,000(0.946) ≈ 75,194 10
150,000
30,000
20. An account is gaining value at a rate of 4.94% per year. The account held $113 in 2005. What will the bank account hold in 2017? t
y = 113(1.0494) ≈ 201.54 12
200
1
40
2
y
160 (0, 113) 120 80
The y-intercept is 113, the original amount in the account in 2005.
12
15
(4, 104,915)
t
Domain: {x|0 < x < ∞} Range: {y| y > 0}
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(0, 131,000) (2, 117,234)
0
After 12 years, the bank account will hold about $201.54.
470
y
6
3
4
5
(8, 166.19) (4, 137.04) (6, 150.91) (2, 124.44)
© Houghton Mifflin Harcourt Publishing Company
Domain: {x|0 < x < ∞} Range: {y| y > 0} The y-intercept is 131,000, the original value of the house in 2009.
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120,000 (1, 123,926) 90,000 (3, 110,903) 60,000
t
Some students may fail to write percents as decimals before applying the exponential growth model. Encourage them to write out the decimal equivalent of a growth or decay rate before adding it to 1 or subtracting it from 1 in the exponential equation.
t 0
After 10 years, the house will be worth about $75,194.
y = a(1 + r) = 113(1.0494)
AVOID COMMON ERRORS
(12, 44,882) (6, 39,388)
10,000
The y-intercept is 34,567, the original population of the town in 2001.
19. A house is losing value at a rate of 5.4% per year. In 2009, the house was worth $131,000. Find the worth of the house in 2019.
(15, 47,910)
t 0
2
4
6
8
10
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Modeling Exponential Growth and Decay
470
Describe and compare each pair of functions.
PEERTOPEER DISCUSSION
21. A(t) = 13(0.6) and B(t) = 4(3.2) t
Ask students to discuss with a partner how to write an exponential growth or decay model from a verbal description of a real-world situation. They should consider which key words tell whether the situation involves growth or decay; what the starting value a is; and how to use the growth or decay rate r in the equation.
t
The value of A(t) is decreasing. The value of B(t) is increasing. The initial value of A(t) is greater than the initial value of B(t). However, after about .7 units, the value of B(t) becomes greater than the value of A(t).
22. A(t) = 9(0.4) and B(t) = 0.6(1.4) t
t
The value of A(t) is decreasing. The value of B(t) is increasing. The initial value of A(t) is greater than the initial value of B(t). However, after about 2.2 units, the value of B(t) becomes greater than the value of A(t).
23. A(t) = 547(0.32) and B(t) = 324(3) t
t
The value of A(t) is decreasing. The value of B(t) is increasing. The initial value of A(t) is greater than the initial value of B(t). However, after about .2 units, the value of B(t) becomes greater than the value of A(t).
JOURNAL
t t 24. A(t) = 2(0.6) and B(t) = 0.2(1.4)
Have students write a journal entry in which they describe how to graph an exponential growth or decay function, and how to find the equation for a real-world situation involving exponential growth or decay.
The value of A(t) is decreasing. The value of B(t) is increasing. The initial value of A(t) is greater than the initial value of B(t). However, after about 2.7 units, the value of B(t) becomes greater than the value of A(t). 25. Identify the y-intercept of each of the exponential functions. a. 3123(432,543) b. 0 c. 45(54)
x
x
(0, 3123)
d. 76(89, 047, 832)
(0, 0)
e. 1
x
(0, 76) (0, 1)
(0, 45)
© Houghton Mifflin Harcourt Publishing Company
H.O.T. Focus on Higher Order Thinking
26. Explain the Error A student was asked to find the value of a $2500 item after 4 years. The item was depreciating at a rate of 20% per year. What is wrong with the student’s work? 4 25000(0.2) $4 t t The exponential decay function is y = 2500(1 - 0.2) , or y = 2500(0.8) . The student forgot to subtract the rate of depreciation from 1 before solving. 27. Make a Conjecture The value of a certain car can be modeled by the function t y = 18000(0.76) , where t is time in years. Will the value of the function ever be 0?
The value of the function will never be 0 because the right side of the function is a product of positive numbers. Although the value can become extremely close to 0, it can never equal 0. 28. Communicate Mathematical Ideas Explain how a graph of an exponential function may resemble the graph of a linear function.
A graph of an exponential function may appear to be a linear function if only a small part of the graph is shown and the values in that part are changing slowly.
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Lesson Performance Task
CONNECT VOCABULARY Some students may not be familiar with the terms archeologist, artifacts, or radiometric dating. Explain that an archeologist studies prehistoric people and cultures by analyzing items that remain for us to find, such as their bones, weapons, artifacts, and tools. Artifacts are objects that the people made, such as pottery. Radiometric dating, also called radioactive dating, is a method of determining the age of an object based on the rate of decay of an element in the object.
Archeologists have several methods of determining the age of recovered artifacts. One method is radioactive dating. All matter is made of atoms. Atoms, in turn, are made of protons, neutrons, and electrons. An “element” is defined as an atom with a given number of protons. Carbon, for example, has exactly 12 protons. Carbon atoms can, however, have different numbers of neutrons. These are known as “isotopes” of carbon. Carbon-12 has 12 neutrons, carbon-13 has 13 neutrons, and carbon-14 has 14 neutrons. All carbon-based life forms contain these different isotopes of carbon. Carbon-12 and carbon-13 account for over 99% of all the carbon in living things. Carbon-14, however, accounts for approximately 1 part per trillion or 0.0000000001% of the total carbon in living things. More importantly, carbon-14 is unstable and has a half-life of approximately 5700 years. This means that, within the span of 5700 years, one-half of any amount of carbon will “decay” into another atom. In other words, if you had 10 g of carbon-14 today, only 5 g would remain after 5700 years. But, as long as an organism is living, it keeps taking in and releasing carbon-14, so the level of it in the organism, as small as it is, remains constant. Once an organism dies, however, it no longer ingests carbon-14, so the level of carbon-14 in it drops due to radioactive decay. Because we know how much carbon-14 an organism had when it was alive, as well as how long it takes for that amount to become half of what it was, you can determine the age of the organism by comparing these two values.
QUESTIONING STRATEGIES What is the growth factor (the base raised to a power) in the exponential model? Explain. It is 1 - 0.5 = 0.5, because the amount of carbon-14 decreases by half during every time period.
n
Carbon-14 Concentration In Parts Per Quadrillion
C(n) = 1000(0.5)
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©worker/ Shutterstock
Use the information presented to create a function that will model the amount of carbon-14 in a sample as a function of its age. Create the model C(n) here C is the amount of carbon-14 in parts per quadrillion (1 part per trillion is 1000 parts per quadrillion) and n is the age of the sample in half-lives. Graph the model.
1000 800 600 400 200 0
2
4
6
8
Age in Half-Lives
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Have students research how very large numbers such as million, billion, trillion, and quadrillion compare to each other. Then have students rewrite the model C(n) in different units and tell how the graph would change.
Students may find that since 1 trillion, or 10 12, is equal to 1000 billion, or 10 3 ∙ 10 9, 1 part per trillion is the same as 0.001 parts per billion. Therefore, the model could be C(n) = 0.001(0.5)n. The graph would differ only in that the scale of the vertical axis would be in parts per billion rather than parts per quadrillion.
4/2/14 1:49 AM
Scoring Rubric 2 points: Student correctly solves the problem and explains his/her reasoning. 1 point: Student shows good understanding of the problem but does not fully solve or explain his/her reasoning. 0 points: Student does not demonstrate understanding of the problem.
Modeling Exponential Growth and Decay
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LESSON
10.4
Name
Modeling with Quadratic Functions
Class
Date
10.4 Modeling with Quadratic Functions Essential Question: How can you use tables to recognize quadratic functions and use technology to create them?
Common Core Math Standards
Resource Locker
The student is expected to: COMMON CORE
A-CED.A.2
A linear function is a straight line, a quadratic function is a parabola, and an exponential function is a curve that approaches a horizontal asymptote in one direction and curves upward to infinity in the other direction.
Mathematical Practices COMMON CORE
You can determine if the function is linear or exponential when the values of x and y are presented in a table. For a constant change in x-values, if the difference between the associated y-values is constant, then the function is linear. If the ratio of the associated y-values is constant, then the function is exponential.
MP.5 Using Tools
Language Objective
What if neither the ratio of successive terms nor the first differences are roughly constant? There is a clue in the method for recognizing a linear function. Find the second difference. The second difference is the value obtained by subtracting consecutive first differences. If this number is non-zero, then the function will be quadratic. Examine the graph of the given quadratic function; then construct a table with values for x, y, the first and second differences, and the ratio of consecutive terms.
Explain to a partner how to create a quadratic function to fit data.
ENGAGE
A
Essential Question: How can you use tables to recognize quadratic functions and use technology to create them?
PREVIEW: LESSON PERFORMANCE TASK View the Engage section online. Discuss the photo and how knowing the times that it takes for the object to reach various heights above the ground might help determine a function that models the data. Then preview the Lesson Performance Task.
Graph the function ƒ(x) = x 2 on the given axes.
8
Use the table to complete Steps B, D, and F. y = f(x)
x © Houghton Mifflin Harcourt Publishing Company
You can compare second differences from the table to determine if a relationship is quadratic. Then you can perform a quadratic regression using the values in the table to create a function.
Using Second Differences to Identify Quadratic Functions
Explore
Create equations in two... variables to represent relationships between quantities… Also F-IF.B.4, F-IF.B.5
First Difference
1
2
22 =
3
3 = 9
4
4 2 = 16
16 - 9 = 7
5
5 = 25
25 - 16 =
4
2
2
6
Second Difference
1
4-1=
3
9-4=
5
2 x 5-3= 2 7-5=
-4
-2
0
2
4
2
9 - 7 =2
9
Complete the second column of the table with indicated values of ƒ(x) = x 2.
C
Is there a constant difference between x-values? yes/no
D
Recall that the differences between y-values are called the first differences. Complete the third column of the table with the indicated first differences.
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E
Are the first differences constant (the same)? yes/no
F
The differences between the first differences are called the second differences. Complete the last column of the table with the indicated second differences.
G
Are the second differences constant (the same)? yes/no
H
Complete the table for another quadratic function: ƒ(x) = -3x 2.
y = f(x)
x 1 2 3 4 5
I
-3 ∙ 1 = -3
-3 ∙ 2 2 = -12 -3 ∙ 3 2 = -27 -3 ∙ 4 = -48 2
-3 ∙ 5 2 = -75
First Difference -12 - (-3) = -9
-27 - (-12) = -15
-48 - (-27) = -21 -75 - (-48) = -27
EXPLORE Using Second Differences to Identify Quadratic Functions INTEGRATE TECHNOLOGY
Second Difference
Have students complete the Explore activity in either the book or online lesson.
-15 - (-9) = -6
QUESTIONING STRATEGIES
-21 - (-15) = -6
How do you determine the second differences for a function? Second differences are found by finding the difference between consecutive first differences, which are the differences between y-values for a constant change in x-values.
-27 - (-21) = -6
Is there a constant difference between x-values? yes/no Are the first differences constant (the same)? yes/no Are the second differences constant (the same)? yes/no
Reflect
1. 2.
EXPLAIN 1
Discussion When a table of values with constant x-values leads to constant y-values (first differences), what kind of function does that indicate? (linear/quadratic)
Explain 1
Verify Quadratic Relationships Using Quadratic Regression
The second differences for ƒ(x) = x 2, the parent quadratic function, are constant for values of y when the corresponding differences between x-values are constant. Now, do the reverse. For a given set of data, verify that the second differences are constant and then use a graphing calculator to find a quadratic model for the data. Enter the independent variable into List 1 and the dependent variable into List 2, and perform a quadratic regression on the data. When your calculator performs a quadratic regression, it uses a specific statistical method to fit a quadratic model to the data. As with linear regression, the data will not be perfect. When finding a model, if the first differences are close but not exactly equal, a linear model will still be a good fit. Likewise, if the second differences aren’t exactly the same, a quadratic model will be a good fit if the second differences are close to being the same.
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Verifying Quadratic Relationships Using Quadratic Regression
© Houghton Mifflin Harcourt Publishing Company
When a table of values with constant x-values leads to constant second differences, what kind of function does that indicate? (linear/quadratic)
AVOID COMMON ERRORS Make sure students can correctly substitute values for a, b, and c into the standard form of a quadratic equation, y = ax 2 + bx + c. For example, if a = 3, b = 5, and c = -6, the quadratic equation should be. y = 3x 2 + 5x - 6.
Lesson 4
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Integrate Mathematical Practices
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This lesson provides an opportunity to address Mathematical Practice MP.5, which calls for students to “use tools.” Students use paper and pencil to find the first differences and second differences in order to determine whether a quadratic function can fit given data. Students also use graphing calculators to find a quadratic function that fits a data set, and use that function to solve real-world problems.
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Example 1
INTEGRATE MATHEMATICAL PRACTICES Focus on Technology MP.5 Tell students that to enter values for
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©oksana. perkins/Shutterstock
plotting they must enter values in L1 and L2 by using STAT1:Edit. To find the equation for the quadratic regression, select STAT:CALC:5:QuadReg.
Find a quadratic model for the given situation. Begin by creating a scatter plot of the given data on your graphing calculator, and then find the second differences to verify the data is quadratic. Finally, use a graphing calculator to perform a quadratic regression on the data and graph the regression equation on the scatter plot.
A student is measuring the kinetic energy of a pickup truck as it is travels at various speeds. The speed is given in meters per second, and the kinetic energy is given in kilojoules. Use the given data to find a quadratic model for the data.
Speed x
Kinetic Energy y = K(x)
First Difference
Second Difference
20
410
25
640
230
30
922
282
52
35
1256
334
52
40
1640
384
50
45
2047
436
52
50
2563
487
51
Enter the data into a graphing calculator, placing the x-values into List 1 and the y-values into List 2. Next view a scatter plot of the data points. The calculator window shown is 15 < x < 55 with an x-scale of 5 and 0 < y < 3000 with a y-scale of 500. Next find the first and second differences and fill in the table. The first difference of the first and second y-value is found by evaluating the expression below. K(25) − K(20) 640 − 410 230 Find the next first difference in the same manner. K(30) − K(25) 922 − 640 282 Find the rest of the first differences and fill in the table.
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Peer-to-Peer Activity Have students work in groups of two. Have each student create a data set by roughly sketching a parabola, then selecting 10 points that are close to the parabola to create a data set. Students trade data sets, and use quadratic regression to find and graph a quadratic equation that fits the data. Each student sketches a graph of the function, and compares it to the original sketch made by the partner.
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The first of the second differences is the difference between the values in the third and fourth rows of the first difference column.
QUESTIONING STRATEGIES
282 − 230 = 52
How is performing a quadratic regression on a data set similar to performing a linear regression on a data set? Performing a quadratic regression produces a quadratic equation that fits the data. Performing a linear regression produces an equation of a line that fits the data.
Find the rest of the second differences and fill in the rest of the table. Notice that the second differences are very close to being constant. Use a graphing calculator to find the equation for the quadratic regression. y ≈ 1.026x2 - 0.0548x + 0.3571 Note that the correlation coefficient is very close to 1, so the model is a good fit. Plot the regression equation over the scatter plot.
The table shows the speed of a car in meters per second as it accelerates from a stop at a constant rate, measured every 2 seconds.
Time Height
2
4
6
8
10
5.1
20.4
45.8
81.2
126.1
Create a scatter plot of the data using a graphing calculator. Find the first and second differences and fill out the table.
Time x
Speed y
First Difference
Second Difference
5.1
4
20.4
15.3
6
45.8
25.4
10.1
8
81.2
35.4
10
10
126.1
44.9
9.5
© Houghton Mifflin Harcourt Publishing Company
2
Find the regression equation using a graphing calculator. Report the results to 4 significant digits.
y ≈ 1.236x2 + 0.3114x - 0.52 Based on the correlation coefficient, the model is a good fit. Plot the regression equation over the scatter plot.
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Critical Thinking
4/2/14 1:51 AM
After finding first differences and second differences for a data set, students may be interested in extending the concept to third differences. Point out that a data set with similar first differences can be modeled with a function in the form y = ax + b, and a data set with similar second differences can be modeled with a function in the form y = ax 2 + bx + c. Discuss with students what kind of function would model a data set with similar third differences. For example, have students find the third differences of the data sets generated by y = x 3 and y = x 3 + 4x 2 - 6x + 7.
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Reflect
3.
Discussion Give examples of reasons why the second differences in real-world data won’t necessarily be equal. Sample answer: The second differences will not be exactly equal because of human error, inexactness of measurements, minor differences between theoretical and real-world conditions, and using multiple measuring devices that are not calibrated to the same degree of accuracy.
Your Turn
Find a quadratic model for the given situation. Begin by creating a scatter plot of the given data on your graphing calculator, and then find the second differences to verify the data is quadratic. Finally, use a graphing calculator to perform a quadratic regression on the data and graph the regression equation on the scatter plot. The table shows the height of a soccer ball in feet for every half-second after a goalie dropkicks it.
Time
0.5
1.0
1.5
2.0
2.5
3.0
3.5
Height
54
104
142
173
195
208
216
© Houghton Mifflin Harcourt Publishing Company
4.
Time
Height
0.5
54
First Difference
1.0
104
50
1.5
142
38
Second Difference
-12
2.0
173
31
-7
2.5
195
22
-9
3.0
208
13
-9
3.5
216
8
-5
a = -16.8 b = 120.6 c = -1.143 R = 0.9997 2
y ≈ -16.8x2 + 120.6x - 1.143
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Connect Vocabulary Students may not be familiar with the word regression beyond a mathematical context. Explain that the word regress means to move in reverse and that this describes what happens in the mathematical process. One step is to find data points that are on the graph of a quadratic equation. The reverse action is to find a quadratic equation that fits given data points.
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5.
A company that makes flying discs to use as promotional materials will produce a flying disc of any size. The table shows the cost of 100 flying discs based on the desired size.
Time Height
4
4.5
5
5.5
6
6.5
7
34.99
44.45
54.95
66.50
78.95
92.45
106.99
Size
Cost
4
34.99
4.5
44.99
9
5
54.99
10
1
5.5
66.99
12
2
6
79.99
13
1
92.99
13
0
107.99
15
2
6.5 7
First Difference
EXPLAIN 2 Using Quadratic Relationships to Solve a Real-World Problems
Second Difference
INTEGRATE MATHEMATICAL PRACTICES Focus on Modeling MP.4 When students use calculators to perform quadratic regressions, help them understand how to interpret the value of R 2. It is a measure of how closely data, on average, fit a regression line. Explain to students that the closer R 2 is to 1, the better the function fits the data.
a = 2 b = 2.286 c = -6.081 R 2 = 0.9999 y ≈ -2x2 + 2.286x - 6.081
Explain 2
Using Quadratic Regression to Solve a Real-World Problem
After performing quadratic regression on a given data set, the regression equation can be used to answer questions about the scenario represented by the data. Example 2
© Houghton Mifflin Harcourt Publishing Company
Use a graphing calculator to perform quadratic regression on the data given. Then solve the problem and identify and interpret the domain and range of the function.
The height of a model rocket in feet t seconds after it is launched vertically is shown in the following table. Determine the maximum height the rocket attains.
Time Height
1
2
3
4
5
6
7
8
342
667
902
1163
1335
1459
1584
1864
Enter the data into List 1 and List 2 of a graphing calculator and perform the quadratic regression.
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Then plot the regression function over a scatter plot of the data.
QUESTIONING STRATEGIES
Increase the values of Xmax and Ymax until you can see the maximum value of the function. Then use the maximum function on your graphing calculator to find the maximum height of the rocket.
How do you use a calculator to find the intersection of two functions? To find the point of intersection, select the Calculate menu and then choose the Intersect option.
The model rocket attains a maximum height of approximately 2150 feet 13.5 seconds after launch. The domain of the function will be 0 ≤ t ≤ +∞. Because the independent variable is time, it doesn’t make sense to consider negative time. The range of the function is 0 ≤ y ≤ 2150 because the height of the rocket should never be negative and it will not go higher than its maximum height.
When a rock is thrown into a pond, it makes a series of waves. The area enclosed by the first wave is recorded every second and is shown in the table below. If the rock lands 15 meters from shore, when will the first wave reach the shoreline?
Time
1
2
3
4
5
Area
9.0
35.8
79.8
145.2
225.1
Enter the values into List 1 and List 2 of a graphing calculator and use QuadReg or to find the model. y≈
8.564 x2 +
2.444 x + -2.78
6 319.1
quadratic regression
R2 ≈ 0.999
y ≈ 8.564x2 + 2.444x - 2.78
© Houghton Mifflin Harcourt Publishing Company
Based on the value of R 2, this function will be a close fit for the data. The area enclosed by the wave is a circle, so the wave will reach the shore when the radius 15 of the wave is meters. The area of a circle is given by A = πr 2. A = πr 2
Plot the regression equation as Y 2 and let Y 2 =
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706.9 .
Find the intersection of the two lines.
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2
= π 15 = 255 π ≅ 706.858
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The model intersects the line y = 706.9 at x = 8.9614 . The first wave will reach the 9 seconds .
shoreline in
circular . Once it The function only makes sense while the wave is still the shore reaches , the wave will no longer increase in size in the same way. Therefore, 0 ≤ y ≤ 706.9 . the domain of the function is 0 ≤ x < 9.0 and the range is Your Turn
Use a graphing calculator to perform quadratic regression on the data given. Then solve the problem and identify and interpret the domain and range of the function. 6.
A company needs boxes to package the goods it produces. One product has a standard shape and thickness but comes in a variety of sizes. The sizes are given as integers. The costs of the various sizes in cents are shown in the table. When the packaging cost reaches $2.00, the company will need to add a surcharge. What is the first size that will have the surcharge added?
Size
1
2
3
4
5
6
7
Cost
7.1
13.8
20.1
29.3
50.1
62.3
86.9
y ≈ 1.693x2 - 0.457x + 6.486
R2 ≈ 0.9973
The intersection of y = 1.693x2 - 0.4571x + 6.486 and y = 200 is (10.83, 200). So the first size with the surcharge will be size 11. The function represents the cost of specific sizes so the domain will be integer values of x with x > 0 and the range will be y > 0. 7.
A company sells simple circular wall clocks in a variety of sizes. The production cost of each clock is dependent on the diameter of the clock in inches. The costs of making several sizes of clocks in dollars are given in the table. How big would a clock be that costs $4.00 to make? (Round to the nearest eighth.)
Size
8
8 __12
9
9 __38
9 __12
10
12
Cost
1.07
1.16
1.30
1.32
1.36
1.53
2.23
R2 ≈ 0.9983
The intersection of y = 10.0344x 2 - 0.3996x + 2.071 and y = 4 is (15.28, 4). A clock with 1 inches can be made for $4.00. diameter of 15 _ 4
The function represents the cost of different sizes of clocks, so the domain will be x > 0 and the range will be y > 0.
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Elaborate
ELABORATE
8.
INTEGRATE MATHEMATICAL PRACTICES Focus on Communication MP.3 Explain to students that for a quadratic
between the x-values of the data points. 9.
equation to have a value of R 2 = 1, all the points in the data set must lie exactly on the quadratic function.
graph can be used to find the independent variable associated with a specific value of the dependent variable. 10. Essential Question Check-In When using technology to create a regression model, name two methods for judging the fit of the regression equation. Technology produces the square of the correlation coefficient. The closer that value is to 1,
the better the fit is. Using technology, it is also possible to plot the regression equation
How do you know when to use quadratic regression instead of linear regression? The shape of the scatter plot tells you which kind of regression to perform. Linear regression is appropriate if the data appear to fall along a line. Quadratic regression is appropriate if the data appear to fall along a curve that resembles a quadratic function.
over a scatter plot of the data and gauge the fit by eye.
Evaluate: Homework and Practice Determine if the function represented in the table is quadratic by finding the second differences. © Houghton Mifflin Harcourt Publishing Company
EVALUATE
ASSIGNMENT GUIDE Concepts and Skills
Practice
Explore 1 Using Second Differences to Identify Quadratic Functions
Exercises 1-4
Example 1 Verifying Quadratic Relationships Using Quadratic Regression
Exercises 5-11
Example 2 Using Quadratic Regression to Solve a Real-World Problems
Exercises 12-20
Lesson 10.4
A function modeling a situation can be represented as both a function and a graph. Identify some situations where one representation is more helpful than the other. Possible answers: A graph of a model allows for estimation of the value of the function
for given values of the independent variable while an equation will give an exact value; a
SUMMARIZE THE LESSON
481
Are there any limitations to identifying data that can be modeled by a quadratic function using the method of second differences? Yes, the method of second differences can only be used if there is fairly uniform separation
1.
First Difference
f(x)
1
2
2
4
2
3
8
4
4
16
8
4
5
32
16
8
6
64
32
16
1–4
2
is not quadratic.
Module 10
Exercise
Second Difference
x
The function
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Depth of Knowledge (D.O.K.)
COMMON CORE
Mathematical Practices
1 Recall of Information
MP.2 Reasoning
5
2 Skills/Concepts
MP.2 Reasoning
6–7
2 Skills/Concepts
MP.5 Using Tools
8
2 Skills/Concepts
MP.2 Reasoning
9
2 Skills/Concepts
MP.5 Using Tools
10
2 Skills/Concepts
MP.2 Reasoning
11–14
2 Skills/Concepts
MP.5 Using Tools
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2.
x
First Difference
f(x)
1
3
2
12
9
3
37
15
6
4
48
21
6
5
75
27
6
6
108
33
6
First Difference
Second Difference
The function
is
INTEGRATE MATHEMATICAL PRACTICES Focus on Critical Thinking MP.3 Review the standard form for quadratic
Second Difference
equations, y = ax 2 + bx + c. Students should understand that if a > 0, the function will have no maximum value, and if a < 0, the function will have no minimum value.
quadratic.
3.
x
f(x)
1
9
2
13
4
3
17
4
0
4
21
4
0
5
25
4
0
6
29
4
0
First Difference
Second Difference
The function is not quadratic. 4.
f(x)
1
2
2
18
16
3
48
30
4
92
44
14
5
150
58
14
6
222
72
14
The function
is
Exercise
14
quadratic.
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x
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482
Depth of Knowledge (D.O.K.)
COMMON CORE
Mathematical Practices
15–17
3 Strategic Thinking
MP.4 Modeling
18–19
3 Strategic Thinking
MP.5 Using Tools
20
3 Strategic Thinking
MP.2 Reasoning
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Find the second differences of the given data to verify that the relationship will be quadratic. Then use a graphing calculator to find the quadratic regression equation and R 2 with a precision of 4 digits.
MODELING In graphing the quadratic function that fits the data set, show the points of the data set as well as the function to help students see how well the function fits the data. To graph the function and show the points of the data set, select STAT PLOTS: Plot1; select On, then select the first option under Type, which is the scatter plot option. Then select GRAPH to view the scatter plot.
5.
x
1
7
13
19
25
31
37
43
49
55
61
67
y
93
107
125
148
174
203
237
274
316
361
410
462
Second Difference: 5, 3, 4, 3, 5, 3, 5, 3, 4, 3 y ≈ 0.0531x2 + 1.985x + 90.89 6.
R2 = 1
x
8
11.4
14.8
18.2
21.6
25
28.4
31.8
35.2
38.6
42
45.4
y
24
60
106
161
227
302
387
483
588
703
827
962
Second Difference: 10, 9, 11, 9, 10, 11, 9, 10, 9, 11 y ≈ 0.4273x2 + 2.265x - 21.41 7.
R2 = 1
x
3.6
17.9
32.2
46.5
60.8
75.1
89.4
103.7
118
132.3
146.6
160.9
y
1946
1684
1442
1219
1012
821
648
494
357
237
133
44
Second Difference: 20, 19, 16, 16, 18, 19, 17, 17, 16, 15 y ≈ 0.0423x2 - 19.03x + 2012
© Houghton Mifflin Harcourt Publishing Company
8.
x
9
9.2
9.4
9.6
9.8
y
44
465
844
1180
1475
y ≈ -503.4x2 + 11250x - 60400
9.
10.2
10.4
10.6
10.8
11
11.2
1728
1941
2117
2255
2355
2409
2416
R2 = 1
x
17
20.1
23.2
26.4
29.5
32.7
35.8
38.9
42.1
45.2
48.4
51.5
y
1000
995
974
936
882
814
729
627
510
376
225
58
Second Difference: -16, -17, -16, -14, -17, -17, -15, -17, -17, -16 y ≈ -0.8172x2 - 28.73x - 747.5
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Second Difference: -42, -43, -41, -42, -40, -37, -38, -38, -46, -47
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R2 = 1
R2 = 1
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Find a quadratic model for the given situation. Begin by creating a scatter plot of the given data on your graphing calculator, and then find the second differences to verify the data is quadratic. Finally, use a graphing calculator to perform a quadratic regression on the data.
AVOID COMMON ERRORS Students may incorrectly assume that quadratic regression will produce a function that passes through most of the points in the data set. Remind students that the quadratic function that fits the data set should be close to the points in the data set, but probably will not pass through all the points.
10. The table shows the height of an arrow in feet x seconds after being loosed at a target down range by an archery student.
Time
0.25
0.5
0.75
1.0
1.25
1.5
1.75
Height
12.0
23.7
32.5
39.3
44.6
47.5
47.8
Second Difference: -2.9, -2, -1.5, -2.4, -2.6 y ≈ -17.05x2 + 57.97x - 1.314
R2 = 0.9998
11. The table shows the cost of cleaning a lap pool based on the number of lanes it has.
Number of Lanes
6
8
10
12
14
16
Cleaning Cost
30
95
263
518
875
1299
Second Difference: 103, 87, 102, 67 y ≈ 11.39x2 - 122.8x - 353.6
R2 = 0.9999 © Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Butter45/ Shutterstock
Use a graphing calculator to find a quadratic model for the given data. 12.
x
2.9
3.9
4.7
5.6
6.9
7.7
8.5
y
8
14
23
29
40
53
70
y ≈ 1.189x2 - 3.132x + 8.335
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13.
x
4.7
6.7
8.5
10.1
12.8
y
32
17
-27
-94
-193
y ≈ -4.645x2 + 50.89x - 113.4 14.
14.3
15.9
-321
-499
R2 = 0.9907
x
-2.7
-1.9
-0.9
0
0.9
1.9
2.7
y
-13
-8
-6
-4
-5
-8
-13
y ≈ -1.184x2 + 0.0384x - 4.180
R2 = 0.9842
Use a graphing calculator to perform quadratic regression on the given data. Then solve the problem and identify and interpret the domain and range of the function. 15. The revenue of a company based on the price of its product is in the table below. How much should the company sell the product for to maximize revenue?
Cost
1
2
2.75
4
4.5
6
8
8.4
9
Revenue
90
228
303
384
406
396
282
229
135
y ≈ -1843x2 + 191.4x - 83.36
R2 = 0.9961
The maximum value of the function is 413.77. If the company sells its product for $5.19, it will maximize its revenue. The model is for selling price and profit, so the domain will be x > 0 and the range will be -∞ < y < ∞.
© Houghton Mifflin Harcourt Publishing Company
16. A scuba diver brought an air-filled balloon 150 feet underwater to the bottom of a lake. The diver conducts an experiment to measure the surface area of the balloon while ascending back to the surface. The results of the measurements are shown in the table. How far will the balloon have risen when it has doubled in surface area?
Distance from Bottom Surface Area
40
60
80
100
120
35.4
39.5
45.0
52.8
65.4
R2 = 0.9925
Find the intersection of the regression equation and y = 57.2. The surface area of the balloon will have doubled when it has risen 106.5 feet from the bottom. The function models the surface area of a real-world object based on its distance from the bottom of a lake, so the domain is x > 0 and the range is y > 0.
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y ≈ 0.0021x2 + 0.035x - 29.64
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17. The height of a ski jumper with respect to the low point of the ramp in meters is measured every 0.3 seconds. The results are given in the table. If the skier lands at a point 30 meters below the reference point, how long was the skier in the air?
Time
0.3
0.6
0.9
1.2
1.5
1.8
2.1
2.4
Height
11.2
14.06
16.32
18.47
19.48
20.52
21.01
21.01
y ≈ -2.627x2 + 11.74x - 7.942
R2 = 0.9991
Find the intersection of the regression equation and y = -30. The skier was in the air for 6.6 seconds. The function models height based on a reference point after an event begins, so the domain will be x > 0 and the range is -∞ < y < ∞.
H.O.T. Focus on Higher Order Thinking
Use the table for Exercises 18 and 19. x
1
1.25
1.5
1.75
2
3
4
5
6
7
8
9
y
2
2.378
2.828
3.364
4
8
16
32
64
128
256
512
18. What If? If you perform a quadratic regression on the data, will the value of R2 be close to 1? Justify your answer. © Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Tetra Images/Alamy
The value of R 2 for a quadratic regression model is about 0.95, but the value of R 2 for an exponential regression is nearly 1, so an exponential regression model is better for this data set. 19. Communicate Mathematical Ideas Perform a quadratic regression on the data; then perform a quadratic regression using only the first four data points. Explain the difference in R 2 values between the two models.
R2 = 0.9493 Full Set: y ≈ 13.84x2 - 86.13x + 104.6 2 R2 = 1 Four points: y ≈ 0.632x + 0.0788x + 1.290 The value of R2 for the full set reflects the entire graph, so even though a model could be very close for small values of x, any model over a large domain will not be close at all.
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20. Multi-Part A trebuchet is a catapult that was used in the Middle Ages to hurl projectiles during a siege. It is now used in various regions of the United States to throw pumpkins. Teams build trebuchets to compete to see who can throw a pumpkin the farthest. On the practice field, one team has measured the height of its pumpkin after it is launched at 1-second intervals. The results are displayed in the table below.
JOURNAL Have students write the steps for using quadratic regression to find a quadratic equation that models a data set. Students should include which calculator options to use in order to identify and graph a quadratic equation that fits the data.
Time Height
1
2
3
4
5
6
7
8
152
265
377
441
470
450
396
342
a. Find a quadratic function that models the data. y ≈ -17.36x2 + 182.9x - 18.45 R2 = 0.9925 b. Determine the flight time of the pumpkin. The pumpkin will travel until its height is 0. This occurs 10.43 seconds after launch. c.
If the pumpkin travels horizontally at a speed of 120 feet per second, how far does it travel before it hits the ground? 10.43 ⋅ 120 = 1251.6 feet.
d. At the official competition, the trebuchet is situated on a slight rise 10 feet above the targeting area. How far will the pumpkin travel in the competition, assuming the height relative to the base of the trebuchet is modeled by the same function and it moves with the same horizontal speed?
Find the zero of the sum of the regression function and 10 to find the time and then multiply by the horizontal speed.
© Houghton Mifflin Harcourt Publishing Company
The new zero is 10.49 seconds after launch. So the pumpkin will travel 10.49 ⋅ 120 = 1258.8 feet.
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Lesson Performance Task
MODELING Relate the negative values in the data table to the distances below the base of the lighthouse. Draw a sketch of the lighthouse with its base at point 0 and its top at point 200. The water level should be at –300. Ask: After 5 seconds have elapsed, has the object fallen into the water? No, it has not; it is at –200, which is 100 feet above the water.
A student stands at the top of a lighthouse that is 200 feet tall. The base of the lighthouse is an additional 300 feet above the ocean below, and the student has a clear shot to the water below to examine the claims made by Galileo. But, this being the 21st century, the student also has a sophisticated laser tracker that continually tracks the exact height of the dropped object from the ground as well as the length of time elapsed from the drop. At the end of the trial, the student gets sample data in the form of a table.
Time
Height above Ground
0
200
0.5
196
1
184
1.5
164
2
136
2.5
100
3 3.5 4
AVOID COMMON ERRORS Some students may subtract in the wrong order when finding the first differences or the second differences. Remind students that to find a first difference, they need to pick a height and then subtract the previous height. A second difference is a first difference minus the difference before it.
56 4 -56 -124
5
-200
Examine the data and determine the relationship between time and height. Then find the function that models the data. (Hint: Negative values represent when the object passes the base of the lighthouse.)
The common ratio and first difference are not constant. The second difference is constant so the data is quadratic. The formula modeling the height of the ball above ground at time t is h(t) = -16t 2 + 200.
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Have students find out more about Galileo’s experiment, dropping two cannonballs of different weights from the Leaning Tower of Pisa. Students may discover that Galileo’s experiment may actually have been a thought experiment, and that someone else later dropped the cannonballs from the tower. Have students research how air resistance affects the rate of fall of, for example, a hammer and a feather. Then have students investigate how astronauts on the moon showed that a hammer and a feather dropped together from the same height above the surface struck the ground at the same time. There was no air resistance because the moon has no atmosphere.
4/2/14 1:51 AM
Scoring Rubric 2 points: Student correctly solves the problem and explains his/her reasoning. 1 point: Student shows good understanding of the problem but does not fully solve or explain his/her reasoning. 0 points: Student does not demonstrate understanding of the problem.
Modeling with Quadratic Functions
488
LESSON
10.5
Name
Comparing Linear, Exponential, and Quadratic Models
Essential Question: How can you determine whether a given data set is best modeled by a linear, quadratic, or exponential function?
F-LE.A.1b
Recall that you learned to characterize the end behavior of a function by recognizing what the behavior of the function is as x approaches positive or negative infinity. Look at the three graphs to see what the function does as x approaches negative infinity.
Recognize ... changes at a constant rate per unit interval relative to another. Also F-IF.B.6, F-LE.A.1, F-LE.A.3
Linear
Mathematical Practices MP.5 Using Tools
4
-4
-2
2
PREVIEW: LESSON PERFORMANCE TASK
© Houghton Mifflin Harcourt Publishing Company
ENGAGE
4
y
6
6
4
4
2
-2 -4
Exponential
y
y
x
Explain to a partner how to determine whether a data set is best modeled by a linear, quadratic, or exponential function.
You can (a) look at the graph of the function and consider its shape and its end behavior; (b) use a table of data and determine first differences, second differences, and ratios of consecutive function values; or (c) perform regressions on the data and compare the fit of the data.
Quadratic
2
Language Objective
Essential Question: How can you determine whether a given data set is best modeled by a linear, quadratic, or exponential function?
Resource Locker
Exploring End Behavior of Linear, Quadratic, and Exponential Functions
Explore
The student is expected to:
COMMON CORE
Date
10.5 Comparing Linear, Exponential, and Quadratic Models
Common Core Math Standards COMMON CORE
Class
2 x
-4
0
-2
2
x -4
4
-2
0
2
4
infinity, ƒ (x) approaches infinity , and A For the linear function, ƒ(x), as x approaches approaches negative infinity as x approaches negative infinity, ƒ (x)
.
infinity, g (x) B For the quadratic function, g(x), as x approaches approaches infinity as x approaches negative infinity, g (x)
approaches infinity , and .
C For the exponential function, h(x), as x approaches infinity, h(x)
and as x approaches negative infinity, h(x) approaches zero .
Module 10
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IN2_MNLESE389830_U4M10L5.indd 489
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© Houghto
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Publishin
Watch for the hardcover student edition page numbers for this lesson.
6
y
Lesson 5 489 Module 10
0L5 489 30_U4M1
ESE3898
IN2_MNL
489
Lesson 10.5
4/9/14
7:12 PM
07/04/14 7:05 PM
Examine the end behavior and the rate of the change of the three function types by filling in the values of the table.
D
E
Linear
Quadratic
Exponential
x
L(x) = 5x - 2
Q(x) = 5x 2 - 2
E(x) = 5 x - 2
1
3
3
3
Exploring End Behavior of Linear, Quadratic, and Exponential Functions
2
8
18
23
3
13
43
123
INTEGRATE TECHNOLOGY
4
18
78
623
5
23
123
3123
6
28
178
15,623
Have students complete the Explore activity in either the book or online lesson.
7
33
243
78,123
8
38
318
390,623
QUESTIONING STRATEGIES How can you tell what happens to a function as x approaches infinity? As x approaches negative infinity? Evaluate the function for larger and larger values of x. Evaluate the function for smaller and smaller values of x, such as –10, –100, –1000.
Use first differences to find the growth rate over each interval and determine which function ultimately grows fastest.
Linear
Quadratic
Exponential
x
L(x + 1) - L(x)
Q(x + 1) - Q(x)
E(x + 1) - E(x)
1
5
15
20
2
5
25
100
3
5
35
500
4
5
45
2500
5
5
55
12,500
The fastest growing function of the three is the
exponential
.
© Houghton Mifflin Harcourt Publishing Company
F
Reflect
1.
What is the end behavior of y = 7x + 12? As x approaches infinity, y approaches infinity. As x approaches negative infinity, y
approaches negative infinity. 2.
What is the end behavior of y = 5x 2 + x + 2? As x approaches infinity, y approaches infinity. As x approaches negative infinity, y
approaches infinity. 3.
EXPLORE
Fill in the missing values of the table
What is the end behavior of y = 3 x - 5? As x approaches infinity, y approaches infinity. As x approaches negative infinity, y
approaches -5. Module 10
490
Lesson 5
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Integrate Mathematical Practices
This lesson provides an opportunity to address Mathematical Practice MP.5, which calls for students to “use tools.” Students compare the end behavior of the graphs of linear, quadratic, and exponential equations. Students construct tables to identify the best model for a particular data set, and use calculators to perform regressions to identify a function that best fits a data set.
Comparing Linear, Exponential, and Quadratic Models
490
4.
EXPLAIN 1
Make a Conjecture Does an increasing exponential function always grow faster than an increasing quadratic function? Will the growth rate of an increasing exponential function eventually exceed that of an increasing quadratic function? The exponential will not always grow faster initially, because the average growth in an interval depends on the parameters (for example Q (x) = 5x 2 grows faster than
Justifying a Quadratic Model as More Appropriate Than a Linear Model
E (x) = 2 x from 0 to 1). However, the exponential will always catch up and grow faster than the quadratic.
AVOID COMMON ERRORS
Explain 1
Some students may think that a quadratic function will not model a data set if, as x increases, y always increases, or if, as x increases, y always decreases. Use sketches to show that a quadratic function can model such a data set. For example, y = -x 2 might model the data set that begins (1, –1), (2, –4), (3, –9).
Justifying a Quadratic Model as More Appropriate Than a Linear Model
The first step in modeling data is selecting an appropriate functional form. If you are trying to decide between a quadratic and a linear model, for example, you may compare interval rates of change or the end behavior. First and second differences are useful for identifying linear and quadratic functions if the data points have equally spaced x-values. Example 1
© Houghton Mifflin Harcourt Publishing Company
Examine the data sets provided and determine whether a quadratic or linear model is more appropriate by examining the graph, the end behavior, and the first and second differences. x
f (x)
0
3
y 8
1
1.5
2
1
6
3
1.5
4
4
3
2
5
5.5
6
x 0
9
2
4
6
8
Shape: The graph of the data appears to follow a curved path that starts downward and turns back upward. End Behavior: The path appears to increase without end as x approaches infinity and to increase without end as x approaches negative infinity. Based on the apparent curvature and end behavior, the function is quadratic.
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Small Group Activity Have students work in groups of four. Have each student write the equation of a linear function, a quadratic function, or an exponential function, then create a data set of points that are close to the graph of the function. Then, each student graphs a data set selected by another in the group, decides which type of function is modeled by the graph, and uses a regression to identify the function. Finally, students check with the student who wrote the original function to see whether they chose the right kind of function.
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Lesson 10.5
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Interval Behavior:
f (x)
x 0
3
1
1.5
2 3
First differences
Second differences
-1.5
1
-0.5
1
1
0.5
1
1.5
1.5
1
4
3
2.5
1
5
5.5
3.5
6
9
QUESTIONING STRATEGIES What are two things that differentiate a data set that is better modeled by a quadratic function from a data set that is better modeled by a linear function? The graph of the first data set appears to follow a curve, while the graph of the second appears to follow a line. The first set does not have constant first differences but does have constant second differences, while the second set has constant first differences.
The first differences increase as x increases, while the second differences are constant, which is characteristic of a quadratic function.
B
x
f (x)
0
8
1
6.75
6
2
5
4
3
2.75
5
-3.25
6
8
y
2 x
-7
0
2
4
6
8
Plot the data on the graph. Shape: The graph of the data appears to follow a curved | straight, downward | upward path. End Behavior: The path appears to
decrease
as x increases and to
increase
as x decreases. © Houghton Mifflin Harcourt Publishing Company
The curvature is more consistent with a quadratic than a line, but the apparent end behavior is not. Fill in the first and second differences to discuss internal behavior. Interval Behavior:
x
f (x)
0
8
1
6.75
2
5
3
2.75
4
0
5
-3.25
6
-7
First differences -1.25
-1.75
Second differences -0.5 -0.5
-2.25
-0.5
-3.25
-0.5
-2.75
-3.75
-0.5
The first differences increase as x increases, while the second differences are constant, which is characteristic of a quadratic function. Module 10
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Curriculum Integration Discuss why having a function that models a data set can be useful. For example, having such a model can help people make projections into the future, which may help them plan and make policy decisions. Have students brainstorm issues for which data might be useful in making such decisions.
Comparing Linear, Exponential, and Quadratic Models
492
Reflect
5.
Was the end behavior helpful in determining that the function in Example 1B was a quadratic? Explain. No, it would have been misleading to assume the small range of data supplied
demonstrated the same behavior as the overall function. All quadratic functions change vertical direction at the vertex of the parabola and have the same end behavior for increasing and decreasing x. However, a data set corresponding to a quadratic function need not include the vertex, and the apparent end behavior can be misleading. 6.
Discussion Can you always tell that a function is quadratic by looking at a graph of it? No, it depends on how much of the function is plotted. A quadratic function can appear linear or exponential depending on how many data points are shown on the graph.
Your Turn
7.
4
y
2 x 0
2
4
6
8
-2 -4
The shape appears curved upward. f (x) appears to increase without end as x approaches infinity and to flatten out as x approaches negative infinity.
© Houghton Mifflin Harcourt Publishing Company
The function appears to be quadratic based on the curvature, although the apparent end behavior is not consistent with either a linear or a quadratic function. Interval behavior: x
f (x)
First Differences
Second Differences
0
-4
0.2
0.4
1
-3.8
0.6
0.4
2
-3.2
1
0.4
3
-2.2
1.4
0.4
4
-0.8
1.8
0.4
5
1
2.2
6
3.2
The function has increasing first differences and constant second differences so it is a quadratic function. Module 10
493
Lesson 5
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Connect Vocabulary Make sure that students understand what it means for a variable to approach infinity and to approach negative infinity. For example, to say that as x approaches infinity, f(x) approaches infinity means that as x gets larger and larger, f(x) gets larger and larger without any limit. If you pick a number N, no matter how large, you can find a number x so that f(x) > N. In other words, there is no limit to how large f(x) gets as x increases.
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Justifying a Quadratic Model as More Appropriate Than an Exponential Model
Explain 2
EXPLAIN 2
Previously, you learned to model data with an exponential function. How do you choose between a quadratic and an exponential function to model a given set of data? Graph the given data points and compare the trend of the data with the general shape and end behavior of the parent quadratic and exponential functions. Use the results to decide if the function appears to be quadratic or exponential. Then examine the first and second differences and the ratios of the function using the function values corresponding to x-values separated by a constant amount.
Justifying a Quadratic Model as More Appropriate Than an Exponential Model
Properties of f (x) = x 2 and g (x) = b x f (x) = x 2
g (x) = b x
AVOID COMMON ERRORS
End behavior as: x approaches infinity
f (x) approaches infinity
g (x) approaches infinity
x approaches negative infinity
f (x) approaches infinity
g (x) decreases to zero
Example 2
Sometimes students confuse quadratic functions with exponential functions because both have exponents. Remind students that in order for a function to be an exponential function, it must be of the form, f (x ) = ab x, with the variable x as an exponent.
Determine if the function represented in the given table is quadratic or exponential. Plot the given points and analyze the graph. Draw a conclusion if possible. Then find the first and second differences and ratios and either verify your conclusion or determine the family of the function.
x
f (x)
8
-3
3
6
-2
1.5
-1
1
(-2, 1.5) (-1, 1)
1.5
1
3
2
5.5
3
9
-4
-2
(3, 9) (2, 5.5)
4 2 0
(1, 3) (0, 1.5) 2
x 4
Graph ƒ(x) on the axes provided by plotting the given points and connecting them with a smooth curve. The data appears to be parabolic.
© Houghton Mifflin Harcourt Publishing Company
0
(-3, 3)
y
Also, as x approaches infinity, ƒ(x) appears to increase without end, and as x approaches negative infinity, ƒ(x) appears to increase without end. It appears that ƒ(x) is a quadratic function. Now find the first and second differences and the ratio of the values of ƒ(x). Module 10
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Comparing Linear, Exponential, and Quadratic Models
494
QUESTIONING STRATEGIES
f (x)
x
Suppose that you are trying to determine whether a quadratic function or an exponential function is a more appropriate model for a data set. If the difference between successive x values of the data set is a constant c, how do you find the ratios? Find the ratio of consecutive function values, that is, the ratio of f(x + c) to f(x) for each x in the data set (except the last one).
First Difference
Second Difference
Ratio
-3
3
-2
1.5
-1.5
-1
1
-0.5
1
0.67
1.5
0.5
1
1.5
1
3
1.5
1
2
2
5.5
2.5
1
1.83
3
9
3.5
1
1.64
0
0.5
The second differences are constant so the function is quadratic as predicted.
B
x
f (x)
-2
First Difference
Second Difference
Ratio
4
0
3.5
-0.5
2
2
-1.5
-1
4
-0.5
-2.5
-1
-0.25
6
-4
-3.5
-1
8
8
-8.5
-4.5
-1
2.125
0.875 0.571
Graph ƒ(x) on the axes provided by plotting the given points and connecting them with a smooth curve.
© Houghton Mifflin Harcourt Publishing Company
The data appears to be either quadratic or exponential . As x approaches infinity, ƒ(x)
Now find the first and second differences and the ratio of the values of f (x).
IN2_MNLESE389830_U4M10L5.indd 495
Lesson 10.5
(2, 2) x
-2
0 -2 -4
2
4
6
(4, -0.5) (6, -4)
The second differences are constant so the function is quadratic .
Module 10
495
could be either quadratic or exponential .
y (0, 3.5)
2
appears to decrease without end .
is not defined . As x approaches negative infinity, ƒ(x) It appears that ƒ(x)
(-2, 4)
4
495
Lesson 5
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Your Turn
Determine if the function represented in the given table is quadratic or exponential. Plot the given points and analyze the graph. Draw a conclusion if possible. Then find the first and second differences and ratios and either verify your conclusion or determine the family of the function. 8.
x
f (x)
First Difference
Second Difference
Ratio
y
4
-3
-5
-2
-3.11
1.89
-1
-1.44
1.67
-0.22
0.46
0
0
1.44
-0.22
0
-2
1
1.22
1.22
-0.22
Undefined
-4
2
2.22
1
-0.22
1.82
3
3
0.78
-0.22
1.35
4
3.56
0.56
-0.22
1.19
5
3.89
0.33
-0.22
1.09
6
4
0.11
-0.22
1.03
(6, 4)
(3, 3)
(4, 3.56)
2
0.62
(1, 1.22)
(0, 0) 0
-2
x
2 4 (-1, -1.44)
6
(-3, -5)
Based on the graph, the data could be either quadratic or exponential. Since f (x) has not been defined for x > 6, the end behavior of the function as x approaches infinity cannot be determined. As x approaches negative infinity, f (x) appears to decrease without end. No conclusions can be drawn about f (x) from the graph. The second differences are constant. Therefore, the function is quadratic. 9.
f (x)
-2
8.25
First Difference
Second Difference
Ratio 8
0
6
-2.25
2
4.25
-1.75
0.5
0.71
4
3
-1.25
0.5
0.71
6
2.25
-0.75
0.5
0.75
8
2
-0.25
0.5
0.89
0.73
y
6 (0, 6) f(x) 4
(4, 3)
2 0
(2, 4.25) (6, 2.25) (8, 2)
2
4
6
x
8
Based on the graph, the data could be either quadratic or exponential. Since f (x) has not been defined for x > 8, the end behavior of the function as x approaches infinity cannot be determined. As x approaches negative infinity, f (x) appears to decrease without end.
© Houghton Mifflin Harcourt Publishing Company
x
No conclusions can be drawn about f (x) from the graph. The second differences are constant. Therefore, the function is quadratic.
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Comparing Linear, Exponential, and Quadratic Models
496
Explain 3
EXPLAIN 3
Selecting an Appropriate Model Given Linear, Exponential, or Quadratic Data
It is important to be able to choose among a variety of models when solving real-world problems.
Selecting an Appropriate Model Given Linear, Exponential, or Quadratic Data
Example 3
Decide which type of function is best represented by each of the following data sets. Then perform the following steps: 1. Graph the data on a scatterplot and draw a fit curve. 2. Identify which function the data appear to represent. 3. Predict the function’s end behavior as x approaches infinity.
INTEGRATE MATHEMATICAL PRACTICES Focus on Technology MP.5 As students work with finding a model that
4. Use a function table to calculate the first differences, second differences, and ratios. 5. Perform the appropriate regression on a graphing calculator. Plot the regression line and data together to evaluate the fit of regression. 6. Answer any additional questions.
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©videnko/ Shutterstock
Demographics The data table describes the average lifespan in the United States over time. Year
Average Lifespan (years)
1900
47.3
1910
50.0
1920
54.1
1930
59.7
1940
62.9
1950
68.2
1960
69.7
1970
70.8
1980
73.1
1990
75.4
What will be the estimated average lifespan in 2000? Graph the scatterplot and an approximate line of fit to determine the best function to use for this data set. The data set appears to best fit a linear function. The end behavior of the data is that as x approaches infinity, ƒ (x) approaches infinity. Complete the function table for first differences, second differences, and ratios.
Average Lifespan (years)
fits the data, discuss why it is important to identify the type of function before performing a regression using a graphing calculator. Students should understand that the calculator will perform different calculations depending on the type of regression.
75
y
60 45 30 15 x 0
20
40
60
80
100
Year Since 1900
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Year
Average Lifespan (years)
First Difference
Second Difference
Ratio
1900
47.3
2.7
1.4
50.0 ____ = 1.06
1910
50.0
4.1
1.5
54.1 ____ = 1.08
-2.4
59.7 ____ = 1.10
2.1
62.9 ____ = 1.05
-3.8
68.2 ____ = 1.08
-0.4
69.7 ____ = 1.02
1920 1930 1940 1950
54.1 59.7 62.9 68.2
5.6 3.2 5.3 1.5
AVOID COMMON ERRORS Remind students that the value for r, the correlation coefficient, will be close to 1 if the function is a good fit for the data set. Students who select an inappropriate model for the data can use a low value for r to determine that they have selected an incorrect model.
47.3 50.0 54.1 59.7 62.9 68.2
1960
69.7
1.1
1.2
70.8 ____ = 1.02
1970
70.8
2.3
0
73.1 ____ = 1.03
1980
73.1
2.3
1990
75.4
69.7 70.8
75.4 ____ = 1.03 73.1
Since the ratios are dropping, it is possible that the data set can be modeled by an exponential regression. Since the average of the second differences is around 0, however, it is most likely that the data set should be modeled by a linear regression. Perform the linear regression by first creating a data table by using the STAT function on a calculator. Go to STAT, move over to CALC, type 4, and press ENTER to perform the regression.
© Houghton Mifflin Harcourt Publishing Company
Press ZOOM and 9 to fit the data. Plot the line from the regression to test its fit. The linear regression is a good fit for the data set. To find the estimated average lifespan in 2000, use the equation y = 3.23x + 48.57 and substitute 10 for x. y = 3.23x + 48.57 y = 3.23(10) + 48.57 = 32.3 + 48.57 = 80.87 The average lifespan will be 80.87 years in the year 2000.
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Comparing Linear, Exponential, and Quadratic Models
498
B
QUESTIONING STRATEGIES
Biology The data table lists the whooping crane population over time. Whooping Crane
When determining the first differences, second differences, and the ratio between x-values, what must be true about the x-values in the data set in order to select an appropriate model? To select an appropriate model, the differences between successive x-values must be the same.
Year
Population
1940
22
1950
34
1960
33
1970
56
1980
76
1990
146
2000
177
2010
281
How many whooping cranes will exist in 2020? Graph the scatterplot and an approximate line of fit to determine the best function to use for this data set. y Whooping Crane Population
250 200 150 100 50 x © Houghton Mifflin Harcourt Publishing Company
0
30
40
50
60
The data set appears to best fit an linear/quadratic/exponential function. The end behavior of this data is that as x approaches infinity, ƒ(x), approaches infinity.
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20
Year Since 1940
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499
10
499
Lesson 5
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Complete the function table for first differences, second differences, and ratios.
Year
First Difference
Population
12
Second Difference
Ratio
-13
34 ___ = 1.55
1940
22
1950
34
1960
33
23
-3
1970
56
20
50
1980
76
70
-39
1990
146
31
73
2000
177
104
2010
281
-1
24
22
33 ___ = 0.97 34 56 ___ = 1.70 33 76 ___ = 1.36 56 146 ____ = 1.92 76 177 ____ = 1.21 146 281 ____ = 1.59 177
Since the ratios are changing/not changing and the second difference does/does not have an average that is close to 0, an linear/quadratic/ exponential regression should be used. Perform the exponential regression by first creating a data table by using the STAT function on a calculator. Go to STAT, move over to CALC, type 0, and press ENTER to perform the regression. Press ZOOM and 9 to fit the data. Plot the line from the regression to test its fit. The linear/quadratic/exponential regression is a good/poor fit for the data set. To find the estimated number of whooping cranes in 2020, use the
y = 20.05(1.44)
x
y = 20.05(1.44)
18
x
and substitute
8
for x.
© Houghton Mifflin Harcourt Publishing Company
equation y = 20.05(1.44)
= 20.05(18.49) =
370.7
= 370
Since it is unrealistic to round up in this situation, the number of whooping cranes in 2020 will be 370 .
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Comparing Linear, Exponential, and Quadratic Models
500
Your Turn
Decide which type of function is best represented by each of the following data sets. Then perform the following steps: 1. Graph the data on a scatterplot and draw a fit curve. 2. Identify which function the data appear to represent. 3. Predict the function’s end behavior as x approaches infinity. 4. Use a function table to calculate the first differences, second differences, and ratios. 5. Perform the appropriate regression on a graphing calculator. Plot the regression line and data together to evaluate the fit of regression. 6. Answer any additional questions.
© Houghton Mifflin Harcourt Publishing Company
Year
% of People
1910
21.2
1920
24.2
1930
30.8
1940
32.5
1950
32.8
1960
32.3
1970
31.4
1980
30.0
35
y
28 % of People
10. Population The data table describes the percentage of people living in central cities in the United States over time. What percentage of people will be living in central cities in the United States in 2000?
21 14 7 x 0
20
40
60
80
Year Since 1910
Graph the scatterplot and an approximate line of fit to determine the best function to use for this data set. The function appears to best fit a quadratic function. The end behavior of the data is as x approaches infinity, f(x) approaches negative infinity. Since the ratios increase quickly and then decrease quickly, a quadratic function should probably be used for this data set. The quadratic regression is a good fit for the data set. To find the percentage of people living in central cities in the United States in 2000, use the equation y = -0.61x 2 + 5.46x + 20.89 and substitute 9 for x. y = -0.61x 2 + 5.46x + 20.89
= -0.61(9) + 5.46(9) + 20.89 2
= -49.09 + 49.14 + 20.89 = 20.62
By the year 2000, 20.62% of people will be living in central cities in the United States.
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11. Automobiles The data table describes the car weight versus horsepower for automobiles produced in 2012. If a car weighed 6500 pounds in 2012, how much horsepower should the car have?
Car Weight (pounds)
Horsepower in 2012
2000
70
2500
105
3000
145
3500
179
4000
259
4500
338
5000
400
5500
557
6000
556
Graph the scatterplot and an approximate line of fit to determine the best function to use for this data set. y 2012 Horsepower
500 400 300 200 100 x 0
1000
2000
3000
4000
5000
Car Weight (pounds)
The end behavior of the data is that as x approaches infinity, f (x) approaches infinity. The changing ratios suggest that the data set is best described by a quadratic or linear function. However, the average of the second difference is close to 0, so a linear regression should be used. The linear regression is a good fit for the data set. To find the horsepower a car weighing 6500 pounds should have in 2012, use the equation y = 0.13x - 239.31 and substitute 6500 for x.
y = 0.13x - 239.31
= 0.13(6500) - 239.31
© Houghton Mifflin Harcourt Publishing Company
The data set appears to best fit a linear function.
= 845.00 - 239.31 = 605.69
It makes sense to round in this problem since horsepower is typically expressed in terms of whole numbers, so a car weighing 6500 pounds will have 606 horsepower in 2012. Module 10
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Elaborate
ELABORATE
12. In general, what are three possible end behaviors of exponential, linear, and quadratic graphs as x increases without bound? When do these end behaviors occur? f (x) can increase without bound, which occurs when a > 0 in an growing exponential or
INTEGRATE MATHEMATICAL PRACTICES Focus on Critical Thinking MP.3 Discuss with students the information that
quadratic function and when m > 0 in a linear function. f (x) can decrease without bound, which occurs when a < 0 in a quadratic function and when m < 0 in a linear function.
can be determined about a model to fit a data set based solely upon the shape of the graph. Guide students to see that the shape of the graph is useful but that making a function table and finding first differences, second differences, and ratios can also provide important information when selecting the model to fit a data set.
13. What do function tables tell you that graphs don’t ? How can this information be used to help you select a model? Function tables can be used to find first and second differences as well as the ratio
between terms to help you distinguish linear, quadratic, and exponential models when the data do not show the end behavior.
14. When does a graph help determine an appropriate model better than examining first and second differences and ratios? When the data are very scattered and do not lie exactly on any model function, the
SUMMARIZE THE LESSON
and general shape observed over the whole set of data will offer a clearer picture of the preferred model.
15. Can two different models be created that represent the same set of data? It is possible to create two different models that represent the same set of data. For
example, in a model best described by a linear function, the model can also be described © Houghton Mifflin Harcourt Publishing Company
How do you select an appropriate model to fit a data set? Use the shape of the graph to determine whether the function seems to be a linear, quadratic, or exponential equation. If more than one model may fit, use first differences, second differences, and the ratio of consecutive function values to select a model.
differences and ratios can fail to show a discernable pattern. Instead, the end behavior
by an exponential model with a very small growth rate.
16. Essential Question Check-In How can a graph be used to determine whether a given data set is best modeled by a linear, quadratic, or exponential function? A graph can be used to determine whether a given data set is best modeled by a linear,
quadratic, or exponential function by identifying its general shape and end behavior.
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Evaluate: Homework and Practice 1.
• Online Homework • Hints and Help • Extra Practice
For the two function ƒ (x) = 2x + 1 and g (x) = 2 x + 1 which function has the greatest average rate of change over the interval from 0 to 1? What about the interval from 2 to 3?
f (1) - f (0) 2+1-1 __ = _=2 1-0
1
2 + 1 - (1 + 1 ) g(1) - g(0) __ = __ = 1 1-0
1
f (x) increases faster from 0 to 1.
2.
EVALUATE
6 + 1 - (4 + 1 ) f (3) - f( 2) __ = __ = 2 1 3-2 8 + 1 - (4 + 1 ) g(3) - g(2) __ __ = =4
ASSIGNMENT GUIDE
1 3-2 g(x) increases faster from 2 to 3.
Plot the data and describe the observed shape and end behavior. Does it appear linear or quadratic? Calculate first and second differences and identify the type of function.
x
f(x)
0
-0.79
3.6
-2
1
2.81
1.6
-2
2
4.41
-0.4
-2
3
4.01
-2.4
-2
4
1.61
-4.4
5
-2.79
First Differences
Second Differences
4
The shape appears curved downward.
2
The function appears to be quadratic.
x -2
The function has constant second differences and decreasing first differences, so it is quadratic.
Exercise
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2
4
6
-2 -4
Practice
Explore Exploring End Behavior of Linear, Quadratic, and Exponential Functions
Exercises 1, 10
Example 1 Justifying a Quadratic Model as More Appropriate Than a Linear Model
Exercises 2–5
Example 2 Justifying a Quadratic Model as More Appropriate Than an Exponential Model
Exercises 6–8
Example 3 Selecting an Appropriate Model Given Linear, Exponential, or Quadratic Data
Exercises 9, 11
INTEGRATE MATHEMATICAL PRACTICES Focus on Reasoning MP.2 Discuss with students whether it is possible for a data set to have constant values for the first differences and the second differences. Guide students to understand that a linear function will have constant values for first differences and second differences. (The second differences will all be 0.)
Lesson 5
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Depth of Knowledge (D.O.K.)
0
© Houghton Mifflin Harcourt Publishing Company
f(x) appears to decrease without end as x approaches infinity and to decrease without end as x approaches negative infinity.
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Concepts and Skills
COMMON CORE
Mathematical Practices
1
2 Skills/Concepts
MP.2 Reasoning
2
2 Skills/Concepts
MP.4 Modeling
3
2 Skills/Concepts
MP.4 Modeling
4
2 Skills/Concepts
MP.4 Modeling
5
2 Skills/Concepts
MP.4 Modeling
6
2 Skills/Concepts
MP.4 Modeling
7
2 Skills/Concepts
MP.4 Modeling
8
2 Skills/Concepts
MP.4 Modeling
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3.
KINESTHETIC EXPERIENCE Have students sketch the graphs of a linear function, a quadratic function, and an exponential function. Have students trace the paths of the functions using their index fingers or colored pencils. Have them focus on the shape of each graph and on how the linear function changes compared to the quadratic function or the exponential function.
Plot the data that fits on the grid and describe the observed shape and end behavior. Does it appear to be linear or quadratic? Calculate first and second differences and identify the type of function.
x
f(x)
-2
-4
0.7
1.4
-1
-3.3
2.1
1.4
0
-1.2
3.5
1.4
1
2.3
4.9
1.4
2
7.2
6.3
3
13.5
First Differences
Second Differences
The shape appears slightly curved upward.
The function appears to be quadratic because of the curvature, but the observable end behavior is not consistent with a quadratic function.
2 x -4
-2
© Houghton Mifflin Harcourt Publishing Company
0
2
4
6
8
-2 -4
The function has constant second differences and decreasing first differences, so it is quadratic. 4.
y
4
f(x) appears to increase without end as x approaches infinity and to decrease without end as x approaches negative infinity.
Plot the data on the grid and describe the observed shape and end behavior. Does it appear linear or quadratic? Calculate first and second differences and identify the type of function.
x
f(x)
0
7.6
-1.35
-0.3
1
6.25
-1.65
-0.3
2
4.6
-1.95
-0.3
3
2.65
-2.25
4
0.4
First Differences
Second Differences
The shape appears straight. f(x) appears to decrease without end as x approaches infinity and to increase without end as x approaches negative infinity. The function appears to be linear. The function has constant second differences and decreasing first differences, so it is quadratic.
8
y
6 4 2 x 0
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Depth of Knowledge (D.O.K.)
COMMON CORE
Mathematical Practices
9
3 Strategic Thinking
MP.5 Using Tools
10
3 Strategic Thinking
MP.3 Logic
11
3 Strategic Thinking
MP.3 Logic
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5.
Plot the data and describe the observed the shape and end behavior. Does it appear linear or quadratic Calculate first and second differences and identify the type of function?
AVOID COMMON ERRORS
x
f(x)
First Differences
Second Differences
1
-7.2
3.4
0.4
2
-3.8
3.8
0.4
3
0
4.2
0.4
4
4.2
4.6
5
8.8
The shape appears straight.
8
f(x) appears to increase without end as x increases and to decrease as x decreases.
Sometimes students confuse quadratic functions with exponential functions because both have exponents. Emphasize to students that, when they perform regressions using a calculator, it is important to differentiate between quadratic and exponential functions. Remind students that in order for a function to be an exponential function, it must be of the form f (x ) = ab x, with the variable x as an exponent.
y
4 x
The function appears to be linear.
-8
The function has constant second differences and increasing first differences, so it is quadratic.
-4
0
4
8
-4 -8
6.
Plot the given points. Describe the general shape and end behavior of the graph. Draw a conclusion about the function, if possible. Then complete the table and use the differences to to verify your conclusions.
x
f(x)
First Difference
Second Difference
Ratio
-2
-5.75
————
————
————
0
1
6.75
————
-0.17
2
6.25
5.25
-1.5
6.25
10
3.75
-1.5
1.6
6
12.25
2.25
-1.5
1.23
8
13
0.75
-1.5
1.06
10
12.25
-0.75
-1.5
0.94
12
10
-2.25
-1.5
0.82
-3.75
-1.5
0.63
14
6.25
The data appears to be parabolic. Also, as x approaches infinity, f(x) appears to decrease without end, and as x approaches negative infinity, f(x) appears to decrease without end. It appears that f(x) is a quadratic function. The second differences are constant. Therefore, the function is quadratic.
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y 12 (4, 10) 8 4 (0, 1) 0 -4
(8, 13)
(10, 12.25) (12, 10) (6, 12.25) (14, 6.25) (2, 6.25) f(x) 4
8
© Houghton Mifflin Harcourt Publishing Company
4
x
12
(-2, -5.75) Lesson 5
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7.
JOURNAL Have students compare and contrast the clues they can use to identify a function as linear, quadratic, or exponential. Students should include information about the shape of the graph and its end behavior, as well as first differences, second differences, and the ratio of consecutive function values.
Plot the given points. Describe the general shape and end behavior of the graph. Draw a conclusion about the function, if possible. Then complete the table and use the differences to to verify your conclusions.
x -2
f(x)
First Difference
Second Difference
Ratio
9
————
————
————
0
5.5
-3.5
————
0.61
2
3
-2.5
1
0.55
4
1.5
-1.5
1
0.5
6
1
-0.5
1
0.67
Based on the graph, the data could be either quadratic or exponential. Since f(x) has not been defined for x > 6, the end behavior of the function as x approaches infinity cannot be determined.
(-2, 9)
8 f(x)
The second differences are constant. Therefore, the function is quadratic.
© Houghton Mifflin Harcourt Publishing Company
8.
x
f(x)
First Difference
Second Difference
Ratio
-5
-3
————
————
————
-2
-2
1
————
0.67
1
1
3
2
-0.5
4
6
5
2
6
7
13
7
2
2.17
-4
0
Based on the graph, the data could be either quadratic or exponential.
(2, 3) (6, 1) x
(4, 1.5) 2
4
y
Since f(x) has not been defined for x < -5, the end behavior of the function as x approaches negative infinity cannot be determined.
4
As x approaches infinity, f(x) appears to increase without end.
0
The second differences are constant. Therefore, the function is quadratic. Module 10
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Plot the given points. Describe the general shape and end behavior of the graph. Draw a conclusion about the function, if possible. Then complete the table and use the differences to to verify your conclusions.
No conclusions can be drawn about f(x) from the graph.
507
(0, 5.5)
4
As x approaches negative infinity, f(x) appears to increase without end. No conclusions can be drawn about f(x) from the graph.
6
y
507
2 -4 -2 (-2, -2) -2
(4, 6) f(x) (1, 1) x 2
(-5, -3)
Lesson 5
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H.O.T. Focus on Higher Order Thinking
9.
INTEGRATE TECHNOLOGY
Critical Thinking A set of data was modeled by using the quadratic, linear, and exponential forms of regression. Nearly identical statistically significant r 2-values were produced in all three situations. Which type of regression model should be used? Explain. If a quadratic, linear, and exponential regression model were to produce the same answer, the linear model should be used since it is preferable to use the simplest answer that fits the data. There is no evidence that a quadratic or exponential is a better fit than the linear.
Have students plot the data values and the quadratic function on the same coordinate plane to see how the function fits the data.
INTEGRATE MATHEMATICAL PRACTICES Focus on Modeling MP.4 Have students create a scatter plot of the data
10. Explain the Error To determine if the data represents a linear model, Louise looked at the difference in y-values: 110, 110, 110, 110, 110.
x
7
9
12
14
18
y
150
260
370
480
590
and perform a linear regression, then a quadratic regression of the data. Ask questions such as: Which model best describes the data set? Explain why. The quadratic model best describes the data set because the points appear to lie on a parabola rather than on a straight line. Why does neither model represent the data very accurately? The weight of a dog does not increase endlessly over time as indicated by the linear function, nor does it decrease endlessly over time as indicated by the quadratic function.
She decided that since the differences between the y-values are all the same, a linear model would be appropriate. Explain her mistake.
Although the y-values are separated by the same difference, this difference does not correspond to an equal change in x-values. Louise should have checked that the x-values changed by a constant interval before finding the difference between y-values. 11. Critical Thinking Suppose that the following r 2-values were produced from an unknown set of data.
r 2-values Linear
Quadratic
Exponential
0.15
0.11
0.13
What type of regression model should be chosen for this data set? Explain.
Lesson Performance Task The table shows general guidelines for the weight of a Great Dane at various ages. Create a function modeling the ideal weight for a Great Dane at any age. Justify your choice of models. How well do you think your model will do when the puppy is one or two years old?
Age (months)
Weight (kg)
2
12
4
23
6
33
8
40
10
45
The first differences are too dissimilar and the ratios are decreasing. Use quadratic regression to find a model, even though the second differences are not very similar. w(t) = -0.268t 2 + 7.364t - 1.8, where w is the weight of the puppy and t is the age of the puppy. The model should be fairly accurate when t = 12 because it is fairly close to the domain of the data set. At t = 24, the model will not be very accurate because the end behavior of the weight of an animal is a set value and for a quadratic function with a negative leading term, as t increases without end, w(t) decreases without end.
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© Houghton Mifflin Harcourt Publishing Company
Since all three types of regression models have insignificant r 2-values, none of them should be chosen for this data set.
Lesson 5
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Have students research a table of typical weights for Great Danes. Have students use data from their research to extend or modify the data table given in the Lesson Performance Task. Ask students to evaluate how the new data would change their answers to the Task. Students may find that most weight charts are given in pounds, so they will need to convert to kilograms in order to extend the data in the table. Students will also find that the weight of a Great Dane reaches its maximum and then remains there throughout the dog’s maturity.
4/2/14 1:53 AM
Scoring Rubric 2 points: Student correctly solves the problem and explains his/her reasoning. 1 point: Student shows good understanding of the problem but does not fully solve or explain his/her reasoning. 0 points: Student does not demonstrate understanding of the problem.
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MODULE
10
MODULE
STUDY GUIDE REVIEW
10
Linear, Exponential, and Quadratic Models
Study Guide Review
Essential Question: How can you use linear, exponential, and quadratic models to solve real-world problems?
ASSESSMENT AND INTERVENTION
KEY EXAMPLE
(Lesson 10.3)
A comic book is sold for $3, and its value increases by 6% each year after it is sold. Write an exponential growth function to find the value of the comic book in 25 years. Write the exponential growth function for this situation. y = a(1 + r)
Assign or customize module reviews.
t
= 3(1 + 0.06)
= 3(1.06)
MODULE PERFORMANCE TASK
t
Key Vocabulary
correlation (correlación) exponential function (función exponencial) exponential regression (regresión exponencial) line of best fit (línea de mejor ajuste) linear regression (regresión lineal) quadratic regression (regresión cuadrática)
t
Find the value in 25 years. y = 3(1.06)
t
= 3(1.06)
25
≈ 12.88
After 25 years, the comic book will be worth approximately $12.88.
COMMON CORE
Mathematical Practices: MP.1, MP.2, MP.4, MP.5, MP.6, MP.7 A-CED.A.2, S-ID.B.6a, F-LE.A.1b, F-BF.A.1
KEY EXAMPLE
• How can more than one function be used to model data? Students can research piecewisedefined functions, in which a different function is given for particular domain intervals.
© Houghton Mifflin Harcourt Publishing Company
SUPPORTING STUDENT REASONING Students should begin by focusing on how to find the number of survivors for each year. They can then do research, or you can provide them with specific information. Here is some of the information they may ask for.
The table below shows the cost of shipping a box that has a volume of x cubic feet. Volume
Cost ($)
1
$6.15
2
$8.00
3
$13.90
4
$23.75
23.75 - 13.9 = 9.85
9.85 - 5.9 = 3.95
5
$38.25
38.25 - 23.75 = 14.5
14.5 - 9.85 = 4.65
a ≈ 2.09
b ≈ -4.54 c ≈ 8.65
R ≈ 0.9999 2
SAMPLE SOLUTION A Type I survivorship curve shows a high survival in early and middle life, followed a rapid decline in survival in later life.
(Lesson 10.4)
Find the second differences of the given data to verify that the relationship will be quadratic. Use a graphing calculator to find the quadratic regression equation for the data and R 2 to 4 significant digits.
First Difference
Second Difference
8 - 6.15 = 1.85 13.9 - 8 = 5.9
5.9 - 1.85 = 4.05
The second differences are close to being constant. Plug the volumes, as the x-values, and the costs, as the y-values, into the graphing calculator and run a quadratic regression. The quadratic regression equation is y ≈ 2.09x 2 - 4.54x + 8.65.
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Study Guide Review
SCAFFOLDING SUPPORT
IN2_MNLESE389830_U4M10MC 509
A Type II survivorship curve shows an approximately constant mortality rate, regardless of age. A Type III survivorship curve shows the greatest mortality early in life, with relatively low rates of death for the survivors.
509
Module 10
• Students should understand that to find the number of survivors, they will need to subtract the number of deaths from the previous year’s survivors. For the first year, they will need to subtract from the initial population of 1000 goats. • If students are having trouble identifying the function that fits the goat data, encourage them to consider the scatter plot in two parts, one for the domain 0 ≤ x ≤ 10 and one for the domain 10 < x ≤ 20.
4/12/14 10:21 AM
EXERCISES State whether each situation is best represented by an exponential or linear function. Then write an exponential or linear function for the model and state whether the model is increasing or decreasing. (Lessons 10.1, 10.3) 1.
A customer borrows $950 at 6% interest compounded annually. t exponential; C(t) = 950(1.06) ; increasing
2.
The population of a town is 8548 people and decreases by 90 people each year.
SAMPLE SOLUTION CONTINUED Calculate the number of survivors from the goat data:
linear; P(t) = 8548 - 90t ; decreasing 3.
The table below shows the height of a baseball in inches x seconds after it was thrown. Fill in the table with the first differences and second differences. Then, use a graphing calculator to find the quadratic regression equation for the data. (Lesson 10.4)
Time
First Difference
Height
Second Difference
Age
Survivors
Age
Survivors
1
988
11
862
2
975
12
848
3
966
13
802
4
955
14
740
5
943
15
688
6
932
16
596
7
923
17
495
0
60
0.25
59
59 - 60 = -1
0.5
57
57 - 59 = -2
-2 + 1 = -1
8
912
18
362
0.75
52
52 - 57 = -5
-5 + 2 = -3
9
901
19
203
1
44
44 - 52 = -8
-8 + 5 = -3
10
890
20
0
1.25
35
35 - 44 = -9
-9 + 8 = -1
Make a scatter plot of the data:
Number of Survivors
y ≈ -18.29x 2 + 2.86x + 59.86
MODULE PERFORMANCE TASK
The data table presents the results of a survivorship study for a population of 1000 goats. What type of survivorship curve most closely matches the goat data? Can you find a good mathematical model for these data, using either a linear, quadratic, or exponential function, or a combination?
y
100 90 80 70 60 50 40 30 20 10
Type I
Type II
Type III
0
20
40 60 80 Age (years)
Age (years)
1
2
3
4
5
6
7
8
9
10
Number of Deaths During Year
12
13
9
11
12
11
9
11
11
11
Age (years)
11
12
13
14
15
16
17
18
19
20
Number of Deaths During Year
28
14
46
62
52
92
101
133
159
203
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x 100
© Houghton Mifflin Harcourt Publishing Company
A survivorship curve shows the number of surviving members of a population over time from a given set of births. The graph shows the three types of survivorship curves that commonly occur. Which types of functions would appear to best model each type of curve?
Number of Survivors
What Model Fits a Survivorship Curve?
0
DISCUSSION OPPORTUNITIES • What species could be modeled by each of the survival curve types and why?
y
x 2
4
6
8
10
12
14
16
18
20
Age (years)
The data look like a Type I curve. Looking at the first ten years, the data appear linear. Regression gives the model: y = -10.873x + 998.45 For the second ten years, the data appear quadratic. Regression gives the model:
Study Guide Review
IN2_MNLESE389830_U4M10MC 510
1000 900 800 700 600 500 400 300 200 100
y = -10.345x 2 + 227.96x - 403.1 4/12/14 10:21 AM
So the model is: ⎧ -10.873x + 998.45, 0 ≤ x ≤ 10 y=⎨ -10.345x 2 + 227.96x - 403.1, 10 < x ≤ 20 ⎩
Assessment Rubric 2 points: Student correctly solves the problem and explains his/her reasoning. 1 point: Student shows good understanding of the problem but does not fully solve or explain. 0 points: Student does not demonstrate understanding of the problem.
Study Guide Review 510
Ready to Go On?
Ready to Go On?
ASSESS MASTERY
10.1–10.5 Linear, Exponential, and Quadratic Models
Use the assessment on this page to determine if students have mastered the concepts and standards covered in this module.
• Online Homework • Hints and Help • Extra Practice
The table shows numbers of books read by students in an English class over a summer and the students’ grades for the following semester.
1.
ASSESSMENT AND INTERVENTION
Books
0
0
0
0
1
1
1
2
2
3
5
8
10
14
20
Grade
64
68
69
72
71
74
76
75
79
85
86
91
94
99
98
Find an equation for the line of best fit. Calculate and interpret the correlation coefficient. Then use your equation to predict the grade of a student who read 7 books. (Lesson 10.1)
line of best fit: y ≈ 1.707x + 72.44 ; correlation coefficient is 0.9 which shows a strong, positive correlation; predicted grade: 84 The height of a plant, in inches, x weeks after it was planted is given in the table below. Use a graphing calculator to write a quadratic regression equation for the data set given. About how many weeks did it take the plant to reach a height of 40 inches? (Lesson 10.4)
2.
Access Ready to Go On? assessment online, and receive instant scoring, feedback, and customized intervention or enrichment.
Weeks Height
• Reading Strategies • Success for English Learners • Challenge Worksheets Assessment Resources
20
25
15
31
55
87
127
Graph the data represented in the given table. Determine if the function represented in the given table is best represented by a linear, exponential, or quadratic function. Explain your answer. (Lesson 10.5)
© Houghton Mifflin Harcourt Publishing Company
Differentiated Instruction Resources
15
y = 0.16x + 0.8x + 7; about 12 weeks
ADDITIONAL RESOURCES • Reteach Worksheets
10
2
3.
Response to Intervention Resources
5
x
-3
-2
-1
0
1
2
3
f(x)
5.5
3.5
1.5
1
0.6
0.3
0.1
6
y
The data are best represented by an exponential function. The curvature of the function that best fits the data shows that it is either exponential or quadratic. The end behavior, which approaches 0 as x increases, indicates that the data are better represented by an exponential function.
4 2 x -4
-2
0
2
4
-2
ESSENTIAL QUESTION
• Leveled Module Quizzes
How can you determine if a function is linear, quadratic, or exponential?
4.
Possible Answer: A linear function grows by equal first differences over equal intervals; for a quadratic function, the second differences are equal over equal intervals. An exponential function changes by equal factors over equal intervals. Module 10
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511
Module 10
Study Guide Review
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Common Core Standards
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Content Standards Mathematical Practices
Lesson
Items
10.1
1
S-ID.B.6, S-ID.C.9
MP.6
10.4
2
A-CED.A.2, F-IF.B.4, F-IF.B.5
MP.7
10.5
3
F-LE.A.1
MP.6
MODULE MODULE 10 MIXED REVIEW
MIXED REVIEW
Assessment Readiness
Assessment Readiness
1. Consider each data set and if it is best represented by a linear, exponential, or quadratic model. Choose True or False for each statement. ⎫
⎧
A. ⎨⎩(-5, 2), (-3, 6), (-1, 10), (1, 14), (3, 18)⎬⎭ is best represented by a linear model.
True
False
True
False
True
False
ASSESSMENT AND INTERVENTION
⎫
⎧
B. ⎨⎩(-2, 12), (-1, 6), (0, 3), (1, 1.5), (2, 0.75)⎬⎭ is best represented by a quadratic model. ⎫
⎧
C. ⎨⎩(-5, 4), (-4, 1), (-3, 0), (-2, 1), (-1, 4)⎬⎭ is best represented by an exponential model.
Assign ready-made or customized practice tests to prepare students for high-stakes tests.
2. Does the given equation have 2 real solutions? Select Yes or No for each equation. A. 6x 2 + 15 = 0 B. 8x 2 - 50 = 0
C. 3x + 4x - 10 = 0 2
Yes
No
Yes
No
Yes
No
ADDITIONAL RESOURCES
3. Consider data represented by the following points:
Assessment Resources
⎧ ⎫ ⎨(-2, 0.25), (-1, 1.5), (0, 4), (1, 15), (2, 60)⎬. Determine if the data are best represented ⎩ ⎭
• Leveled Module Quizzes: Modified, B
by a linear, exponential, or quadratic function. Explain your answer.
The data are best represented by an exponential model. The first differences are not constant, so the data should not be represented by a linear model. As x increases, the function appears to increase without end. As x decreases, the function appears to approach 0.
Possible answer: The value of c is 8. For the equation to have 1 real solution, it must be a perfect square trinomial, so b 2 - 4ac = 8 2 - 4(2)c = 0, so 8c = 64 and c = 8.
Module 10
AVOID COMMON ERRORS Item 3 Some students have a hard time deciding between quadratic and exponential representations because they seem to have similar shapes when using only a few data points. Remind students to look for second differences. When the second differences are the same, the function must be quadratic.
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4. The equation 2x 2 + 8x + c = 0 has one real solution. What is the value of c? Explain how you found the value of c.
COMMON CORE
10
Study Guide Review
512
Common Core Standards
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Content Standards Mathematical Practices
Lesson
Items
10.4, 10.5
1
F-LE.A.1, F-LE.A.3
MP.4
9.3
2*
A-REI.B.4
MP.2
10.4, 10.5
3
F-LE.A.1, F-LE.A.3
MP.3
8.2
4*
A-SSE.A.2
MP.3
* Item integrates mixed review concepts from previous modules or a previous course.
Study Guide Review 512
UNIT
4
UNIT 4 MIXED REVIEW
Assessment Readiness
MIXED REVIEW
Assessment Readiness
1. A quadratic equation has the solutions –3 and 6. Can the quadratic equation be the given equation?
ASSESSMENT AND INTERVENTION
Yes
No
B. (6x - 1) (x + 3) + = 0
Yes
No
C. - 3x (x - 6) = 0
Yes
No
2. Factor and solve each equation. Does the equation have a solution of x = –5? A. 3x 2 + 14x - 5 = 0 Yes No
Assign ready-made or customized practice tests to prepare students for high-stakes tests.
ADDITIONAL RESOURCES Assessment Resources
B. x 2 + 3x - 40 = 0
Yes
No
C. x 2 - 3x - 40 = 0
Yes
No
3. Consider the equation 4x 2 - 20 = 0. Is the given statement True or False? A. The equation has 2 solutions. True False B. A solution of the equation is - √20 . True False
• Leveled Unit Tests: Modified, A, B, C • Performance Assessment
C.
A solution of the equation is √― 5.
(
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AVOID COMMON ERRORS Item 6 Since the first difference is the same, and the second difference is also shown and the same, some students will have a hard time choosing between linear and quadratic. Remind students that if a difference is consistently zero, the previous difference should be considered the correct difference.
A. (2x + 6) (x - 6) = 0
―
False
)
B. x = - 5
Yes
No
C. x = __23
Yes
No
5. The equation ax 2 + 12x + c = 0 has one solution. Can a and c equal each of the following values? A. a = 4, c = 9 Yes No B. a = 9, c = 16 C. a = 36, c = 1
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Yes Yes
No No
513
Common Core Standards
Items
Unit 4
True
4. Solve 2x + __23 (x + 5) = 0. Is the given value a solution of the equation? A. x = - __13 Yes No
Unit 4
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• Online Homework • Hints and Help • Extra Practice
Content Standards Mathematical Practices
1*
A-APR.B.3
MP.7
2
A-SSE.B.3a
MP.2
3
A-REI.A.2
MP.1
4*
A-APR.B.3
MP.2
5
A-SSE.A.2
MP.2
6
F-LE.A.3
MP.5
* Item integrates mixed review concepts from previous modules or a previous course.
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6. The table given has been filled out for the function g(x). The values of g(x) are not shown. Is g(x) a linear or quadratic function? Justify your answer. –2
x
0
2
4
PERFORMANCE TASKS There are three different levels of performance tasks:
6
g(x) First difference
–4
Second difference
–4 0
–4 0
* Novice: These are short word problems that require students to apply the math they have learned in straightforward, real-world situations.
–4 0
It’s a linear function because the first difference is constant. 7. The area of a square table top can be represented by (9x 2 - 30x + 25) ft 2. The perimeter of the table top is 34 feet. What is the value of x? Explain how you solved this problem. 4.5; Each side of the table is 8.5 ft long. I factored (9x 2 - 30x + 25) which
** Apprentice: These are more involved problems that guide students step-by-step through more complex tasks. These exercises include more complicated reasoning, writing, and open ended elements.
is a perfect square trinomial. Each side of the table is also (3x - 5) ft long. I
solved 3x - 5 = 8.5 and found that x = 4.5.
***Expert: These are open-ended, nonroutine problems that, instead of stepping the students through, ask them to choose their own methods for solving and justify their answers and reasoning.
8. Solve 4x 2 + 8x = -3 . Which of the following solution methods did you use: factoring, completing the square, or the quadratic formula? Why? Show your work. Possible answer: I completed the square because 4 is a perfect square. 4x 2 + 8x = - 3 4x 2 + 8x + 4 = - 3 + 4
(2x + 2) 2 = 1
SCORING GUIDES
2x + 2 = 1 or 2x + 2 = -1
Item 9 (2 points) Award the student 2 points for correct answer of 2 inches.
2x = - or 2x = -3 x = - 0.5 or x = -1.5
9. Abigail has a rectangular quilt with dimensions 36 inches by 48 inches. She decides to sew a border on the quilt, so that the total area of the quilt is 1900 square inches. What will be the width of the border? 2 inches
Unit 4
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Performance Tasks
514
COMMON CORE
Common Core Standards
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Items
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Content Standards Mathematical Practices
7*
A-CED.A.1
MP.4
8
A-REI.B.4b
MP.3
* Item integrates mixed review concepts from previous modules or a previous course.
Unit 4
514
10. The table shows the average weight of a particular variety of sheep at various ages. A. None of the three models—linear, quadratic, or exponential— fits the data exactly. Which of these is the best model for the data? Explain your choice.
SCORING GUIDES Item 10 (6 points) a. 1 point for reasonable choice 1 point for explanation
B. What would you predict for the weight of a sheep who is 1 year old? C. Do you think you could use your model to find the weight of a sheep at any age? Why or why not?
b. 1 point for reasonable prediction
Sheep Age (mo)
Weight (lb)
2
36
4
69
6
99
8
120
10
135
A. Possible answer: quadratic: the second differences are approximately constant at -6.
c. 1 point for correct answer 2 points for explanation
B. about 143 lbs C. No; this quadratic model will begin to decrease although the sheep’s weight will continue to grow until it eventually remains constant.
Item 11 (6 points)
11. Examine the two models that represent annual tuition for two colleges. A. Describe each model as linear, quadratic, or exponential. B. Write a function rule for each model. C. Both models have the same values for 2004. What does this mean?
a. 1 point for correct models b. 1 point for correct function for College 1 1 point for correct function for College 2
D. Why do both models have the same value for year 1?
c. 1 point for correct explanation
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d. 2 points for correct explanation
Tuition at College 2 ($)
0
2000.00
2000.00
1
2200.00
2200.00
2
2400.00
2420.00
3
2600.00
2662.00
4
2800.00
2928.20
B. College 1: y = 200x + 2000 x College 2: y = 2000(1.1)
C. Both have the same tuition ($2000) in 2000.
D. For College 1, $200 is added each year, so 2000 + 200 = 2200. For College 2, 10% is added each year, so 2000 + (0.1)(2000) = 2200.
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Tuition at College 1 ($)
A. College 1: linear; College 2: exponential
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math in careers
MATH IN CAREERS
competitive Diver Franco and Grace are competitive divers. Grace dives from a 20-meter cliff into the water, with an initial upward speed of 3.2 m/s. Franco dives from a springboard that is 10 meters above the water surface with an initial upward speed of 4.2 m/s.
Competitive Diver In this Unit Performance Task, students can see how a competitive diver uses mathematics on the job.
The height in meters of an object thrown into the air with an initial vertical velocity of v meters per second and initial height of h 0 can be modeled by h(t) = -5t 2 + vt + h 0.
For more information about careers in mathematics as well as various mathematics appreciation topics, visit the American Mathematical Society http://www.ams.org
Assume this model can be applied to both divers. a. Write a function h Grace(t) that models the height of Grace’s dive.
b. Write a function h Franco(t) that models the height of Franco’s dive.
c. Graph both functions on the same coordinate grid. Label each function. d. What is the domain and range of each function in terms of the situation? Explain. Round values to the nearest tenth.
SCORING GUIDES
e. Compare the maximum heights and the time that elapses before each diver hits the water.
a. h Grace (t) = -5t + 3.2t + 20
Task (6 points)
c.
a. 1 point for correct function
2
b. h Franco (t) = -5t 2 + 4.2t + 10
b. 1 point for correct function
20
c. 1 point for correct graph Grace
d. 1 point for correct ranges and domains 1 point for explanation
Height (m)
15
10 Franco
e. 1 point for correct values and comparisons 5
0.5
1
1.5
2
Time (s)
d. for Grace: D: 0 ≤ t ≤ 2.3; R: 0 ≤ h ≤ 20.5 for Franco: D: 0 ≤ t ≤ 1.9; R: 0 3 h 3 10.9
The minimum value of the domain for both divers is zero, because the dives begin at t = 0. The maximum of the domain is the time when each diver hits the water. The minimum value of the range for both divers is 0, because h = 0 represents the height when the divers hit the water. The maximum value of the range is the maximum height each diver reaches.
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0
e. Grace: maximum height: 20.5 m; time for dive: 2.3 s Marco: maximum height: 10.9 m; time for dive: 1.9 s
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