Math 233 - Spring 2009
Chapter 8 - Quadratic Functions 8.1 8.1.1
Solving Quadratic Equations by Completing the Square Square Roots Revisited
Recall in the last chapter, we studied square roots and found that for any square roots there were two possible roots: √ √ 9 = 3, 9 = −3 At that time we defined the square root to be the positive or principle square root. However, when solving equations we want to find all possible solutions so for example: EX 1. Solve: x2 = 9 We square root both sides and we get: x = 3 however, we also have that x = −3 is a solution. √ In General if x2 = a where a is a real number then x = ± a √ REMARK 1. The notation ± is read “plus or minus” and is understood to mean that + a and √ − a are both solutions to the equation. EX 2. Solve: 1. t2 + 3 = 27
2. x2 + 11 = 0
3. (t + 4)2 + 27 = 0
√ REMARK 2. √ • Note that the statement t = −4±3i 3 means there are two possible solutions: √ t = −4 + 3i 3 and t = −4 + 3i 3. • In past chapters we solved x2 = 9 by rewriting it: x2 − 9
=
0
(x + 3)(x − 3)
=
0
and we know that x = 3 or x = −3. However for the first example above t2 = 24 this method doesn’t work because t2 − 24 doesn’t factor. That will be the case for most of our polynomials in this chapter. • We will develop methods to solve quadratic equations that do not factor. 1
8.1.2
Perfect Square Trinomials
We will learn a process called comleting the squares. In order to do this, we need to understand perfect square trinomials. Consider: x2 + 4x + 4
=
(x + 2)(x + 2)
=
(x + 2)2
x2 − 4x + 4
=
(x − 2)(x − 2)
=
(x − 2)2
x2 + 6x + 9
=
(x + 3)(x + 3)
=
(x + 3)2
x2 − 6x + 9
=
(x − 3)(x − 3)
=
(x − 3)2
EX 3. Let’s examine this further. Consider x2 + 6x + 9:
We will use this property a lot, if x2 + bx + c is a perfect square then c = ( 2b )2 . 8.1.3
Completing the Square
Steps: 1. Divide by the leading coefficient. (We want the coefficient of the square term to be 1). 2. Isolate the constant term on one side of the equation. 3. Take the coefficient of the first degree term, divide it by two, and square the resulting number. This is what we add to both sides of the equation. 4. Replace the trinomial with the square of a binomial. 5. Square root both sides. 6. Solve and check. EX 4. Solve by completing the square. 1. x2 + 10x = −21
2. r2 + 9r + 8 = 0
2
3. −x2 = −3x − 10
4. x2 − 10x + 37 = 0
5. −3a2 + 3a = −18
8.2 8.2.1
The Quadratic Formula Derivation
Recall the standard form of a quadratic equation is ax2 + bx + c = 0 where a, b, c are real numbers. EX 5. Identify a, b, and c 1. x2 − 3x + 4
2. 2.3x2 − 7.2
3. − 23 x2 + 14 x
Let’s solve ax2 + bx + c = 0 by completing the square.
3
(derivation continued) The quadratic formula:
√
b2 − 4ac 2a This is an extremely useful formula for solving quadratic equations. x=
8.2.2
−b ±
Using the Quadratic Formula
Let’s look at some examples. EX 6. Solve using the quadratic formula. 1. x2 + 5x − 6 = 0
2. −25x2 = 10x + 1
3. m2 − 12 m +
3 4
=0
4. f (x) = 3x2 + 5x Find all values of x for which f (x) = −3
4
8.2.3
Working Backwards
Suppose we are given the solutions to a quadratic equation but not the equation itself. Can we derive the equation? EX 7. Determine an equation that has the given solutions. 1. Solutions: -3, 2
2. Solutions: 2 + i, 2 − i
REMARK 3. Why would we care about this? In an application suppose we know that our system follows a quadratic equation. Example: Suppose we know the temperature follows a quadratic function and we know temperature is zero at times 3 and 2
8.2.4
Number of Real Solutions
The discriminant of a quadratic equation is b2 − 4ac the part under the square root of the quadratic formula. Solutions of a Quadratic Equation For an equation of the form ax2 + bx + c = 0 where a 6= 0: • If b2 − 4ac > 0, the equation has two distinct real number soltuions. • If b2 − 4ac = 0, the equation has a single real number solution. • if b2 − 4ac < 0, the equation has no real number solutions. EX 8. 1. For the equation x2 − 5x − 14 find the discriminant, the number of real number solutions and check your answer by finding the solutions.
5
2. For the following, determine the number of real number solutions without actually solving. (a) 3x2 − 2x + 8 = 0
(b) 4x2 − 4x + 1 = 0
The Geometry We can visualize what is going on here if we consider the function f (x) = ax2 +bx+ c. Then our solutions (quadratic formula) are where the function f (x) = 0 (the x-intercepts). We’ve mentioned that the graphs of quadratic functions are parabolas, we have the following graphical possibilities:
8.3
Applications and Problem Solving
Quadratic functions appear frequently in applications. Let’s look at a couple of contrived word problems: EX 9. 1. A motorboat travels 16 miles with a 6-mph current and then turns around and returns to the starting point traveling against the current. If the entire trip takes 2 hours, find the speed of the motorboat in still water.
6
2. Dan and Bill can paint a living room in 4 hours working together. Dan can paint the room 1 hour faster than Bill. How long will it ake each of them working alone to paint the living room.
8.4
Quadratic Form
Definition 1. An equation that can be written in the form au2 + bu + c = 0 for a 6= 0, where u is an algebraic expression, is called quadratic form. We will be doing substitution to solve equations that are almost in quadratic form. For example: Equation Quadratic in Form 4
2
y −y −6=0 2(x + 5)2 − 5(x + 5) − 12 = 0 2
1
x 3 + 4x 3 − 3 = 0
Substition u=y
2
u=x+5 1
u = x3
EX 10. Solve 1. y 4 − 5y 2 − 14 = 0
2. x4 − 13x2 + 36 = 0
3. x4 + 4x2 − 45 = 0
7
Equation with Substitution u2 − u − 6 = 0 2u2 − 5u − 12 = 0 u2 + 4u − 3 = 0
4. 6(a + 1)2 − 7(a + 1) − 3 = 0
5. Find the x-intercepts of the graph of the function f (x) = 4x−2 − 9x−1 + 2
2
1
6. x 3 + x 3 − 12 = 0
7. 2m −
√
m − 21 = 0
8
8.5 8.5.1
Graphing Quadratic Functions Characteristics
The graphs of quadratic functions are called parabolas. There are several characteristics of a parabola that we will use to draw the graph. For the quadratic function f (x) = ax2 + bx + c We have the following properties: • Opening: A parabola can either open upwards or downwards. – When a > 0 it opens upwards. 6
-
– When a < 0 it opens downwards. 6
-
• Axis of Symmetry: A parabola is symmetric about a given vertical line. This line is called the axis of symmetry. The equation for the axis of symmetry is given by: x=−
b 2a
6
-
• Vertex: The vertex is the lowest point of a parabola that opens upward or the highest point for a parabola that opens downward.
9
6
6
-
-
or REMARK 4. The vertex is also the minimum or maximum value of the function (depending on whether it opens up or down). – At what value does the vertex occur? b . – Answer: Always occur at the axis of symmetry. Thus when x = − 2a
– Remember the vertex is a point so we should list it as a point, the formula is: (− or (−
b b , f (− )) 2a 2a
b 4ac − b2 , ) 2a 4a
• y-intercept: Occurs when the parabola passes through the y-axis. Occurs when x = 0. Thus y-intercept = (0, f (0)). • x-intercepts: Occur when the parabola passes through the x-axis. Must solve f (x) = 0. Recall that for a parabola there can be zero, one, or two x-intercepts depending on discriminant. – If b2 − 4ac > 0, the equation has two distinct x-intercepts. – If b2 − 4ac = 0, the equation has a single x-intercept. – if b2 − 4ac < 0, the equation has no x-intercepts. 8.5.2
Graphing Quadratic Functions
Let’s now apply all of these ideas to graph some parabolas. EX 11. Do a - e for the given equations. a. Determine whether the parabola opens upward or downward. b. Find the y-intercept. c. Find the vertex d. Find the x-intercepts if any. e. Draw the graph.
10
1. For the equation y = −x2 + 2x + 3 6
-
2. For the equation f (x) = x2 − 3x + 4 6
-
8.5.3
Maximum and Minimum Problems
Recall that the vertex gives us the maximum value if the parabola opens downard and the minimum value if the parabola opens upward. For example: y = ax2 + bx + c For a > 0 minimum value 6
For a < 0 maximum value 6
-
-
11
We can use these facts to solve interesting application problems. EX 12. 1. Consider the following rectangle. (a) Find an equation for the area, A(x) (b) Find the value for x that gives the largest (maximum) area. (c) Find the maximum area.
2. If the perimeter of a rectangle is 60 feet, find the dimensions that will give the greatest area.
12
