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8.1
Using the Distance Formula
In this lesson you will ● ● ●
use the Pythagorean Theorem to help you minimize distance learn how the distance formula and Pythagorean Theorem can help you solve real-world problems use the distance formula to find the equation for a locus of points
In this lesson, you’ll discover and apply a formula for distance.
Investigation: Bucket Race Imagine you are in a race in which you must carry an empty bucket from a starting point A to the edge of a pool, fill the bucket with water, and then carry the bucket to a point B. Point A is 5 m from one end of the pool, point B is 7 m from the other end, and the pool is 20 m long.
Finish B
Start A
7m
5m x
20 x
C 20 m
Let x represent the distance from the end of the pool to a point C on the edge of the pool. Complete Steps 1–4 in your book to find the value of x that gives the shortest possible path from A to C and to B. Here are the answers to Steps 2 and 3.
Step 2 Below are data for several values of x. AC and CB were calculated by using the Pythagorean Theorem, but you can also find them by measuring. Your answers may be slightly different from these depending on how you rounded or on the accuracy of your measurements. x (m)
AC (m)
CB (m)
AC CB (m)
x (m)
AC (m)
CB (m)
AC CB (m)
10.63
23.63
0
5
21.19
26.19
12
13
2
5.39
19.31
24.70
14
14.87
9.22
24.09
4
6.40
17.46
23.87
16
16.76
8.06
24.83
6
7.81
15.65
23.46
18
18.68
7.28
25.96
8
9.43
13.89
23.33
20
20.62
7
27.62
10
11.18
12.21
23.39
Step 3
Method 1: The table above indicates that an x-value of about 8 minimizes the path length, meaning that C should be about 8 m from the end of the pool. (You can get a more accurate answer by trying other values of x that are close to 8.) _______
Method 2: Using the Pythagorean Theorem, AC is 5 2 x 2 and BC is _____________ 7 2 (20 x)2 , so the length, y, of the path can be represented by _______
_____________
y 5 2 x 2 7 2 (20 x)2
Using a calculator graph or table, you can find that the minimum y-value of this function, which is about 23.32, corresponds to an x-value of about 8.33. So, point C should be located about 8.33 m from the end of the pool. Discovering Advanced Algebra Condensed Lessons
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Lesson 8.1 • Using the Distance Formula (continued) The amount of water left in the bucket at the end of the race is an important factor in winning. Imagine that you can carry an empty bucket at a rate of 1.2 m/s and that you can carry a full bucket, without spilling, at a rate of 0.4 m/s. Complete Steps 5 and 6 in your book to find the location of C that minimizes the time it will take you to get from point A to point C to point B. Then, compare your results with those below. distance Use the fact that time _____ rate .
Step 5 x (m)
Time for AC (s)
Time for CB (s)
Total time (s)
0
4.17
52.97
57.14
2
4.49
48.28
52.77
4
5.34
43.66
49.00
6
6.51
39.13
45.64
8
7.86
34.73
42.59
10
9.32
30.52
39.83
12
10.83
26.58
37.41
14
12.39
23.05
35.44
16
13.97
20.16
34.12
18
15.57
18.20
33.77
20
17.18
17.50
34.68
Using the table, the best location for C is 18 m from the end of the pool. Now write expressions for each time to get a more precise answer. The time _______ 5 2 x 2 _______ required to get from A to C is 1.2 , and the time required to get from C to B _____________ Step 6
72 (20 x)2 . Thus, the total time, y, can be represented by the function is ___________ 0.4 _______
_____________
72 (20 x)2 y _________ _______________ 1.2 0.4
5 2
x2
The value of x that minimizes this function is about 17.63. The y-value for this x-value is about 33.75. Thus, to minimize the time, point C should be 17.63 m from the end of the pool. The minimum time will be about 33.75 s. The distance traveled is approximately 25.72 m. This distance is longer than the distance you found in Step 3, but the second leg, CB, is shorter.
Recall that the distance between two points, x 1, y1 and x 2, y2, is given by the ____________________ formula d x 2 x 12 y2 y12 . Review the distance formula by reading Refreshing Your Skills for Chapter 8 through Example B. The distance formula and the Pythagorean Theorem are useful in solving real-world problems. Read Example B of Refreshing Your Skills for Chapter 8 and read Example A of Lesson 8.1 in your book. A locus of points is a set of points that meets a given condition. Example B of Lesson 8.1 shows how to find an equation of a locus of points equidistant from two given points. Work through Example B carefully.
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8.2
Circles and Ellipses
In this lesson you will ● ● ● ●
review the standard equations for a circle and an ellipse learn the locus definitions of a circle and an ellipse locate the foci of an ellipse learn how the eccentricity of an ellipse is related to the shape of the ellipse
Circles, ellipses, parabolas, and hyperbolas are called conic sections because each can be created by slicing a double cone.
Circle
Ellipse
Parabola
Hyperbola
Each conic section can also be defined as a locus of points. For example, a circle is the set of all points a fixed distance from a given point. Read the text in your book from the definition of a circle through Example B. This material reviews what you learned about circles in previous chapters. To find the equation in Example B, you must first find the point where the circle and tangent line intersect. Make sure you understand each step in the solution. y
Recall from Chapter 4 that you can translate and dilate the unit circle to create an ellipse. In general, if an ellipse has center (h, k), horizontal scale factor a, and vertical scale factor b, then its equation in standard form is
b
x_____ h yk _____ 1 a b
2
a
2
Like a circle, an ellipse can be defined as a locus. However, whereas the locus definition of a circle involves one fixed point (namely, the center), the locus definition of an ellipse involves two fixed points: An ellipse is a locus of points in a plane the sum of whose distances from two fixed points is always a constant. In the diagram, the two fixed points, or foci (the plural of focus), are labeled F1 and F 2. For all points on the ellipse, the distances d 1 and d 2 add up to the same value. Page 455 of your book shows how you can construct an ellipse using a string, a pencil, and two pins. If you have these materials, you may want to try this.
(h, k) x
P d1 F1
d2
d1
d2
d1 P
F2 d2
P
(continued)
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Lesson 8.2 • Circles and Ellipses (continued) The length of half the horizontal axis of an ellipse is the horizontal scale factor. Similarly, the length of half the vertical axis is the vertical scale factor. y y b b
c x
c
a
x
a
The segment that forms the longer dimension of an ellipse, and contains the foci, is the major axis. The segment along the shorter dimension is the minor axis. y Major axis
d2
Major axis
y d1
x
d2
d1
2b
x 2a
Minor axis Minor axis When the major axis is vertical, d1 d2 2b.
When the major axis is horizontal, d1 d2 2a.
If you connect an endpoint of the minor axis to the foci, you form two congruent triangles. Because of the definition of an ellipse (think about the string technique for drawing an ellipse), the distance from a focus to an end of the minor axis is the same as half the length of the major axis. From these facts you can conclude that the distance between the center and a focus, c, is related to a and b through the Pythagorean Theorem. For the ellipse on the left, b 2 c 2 a 2. For the ellipse on the right, which has a vertical major axis, a 2 c 2 b 2. y
y
a F1
c
F1
a
b c
F2
x
b
c
x
a c
b
F2
(continued)
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Lesson 8.2 • Circles and Ellipses (continued) The coordinates of the foci are c units away from the center of the ellipse, (h, k), along the major axis and can be found by adding or subtracting c from the appropriate coordinate of the center.
(h, k c) (h c, k)
(h c, k) (h, k)
c
(h, k)
c (h, k c)
Example C in your book shows how to use the relationships among a, b, and c to locate the foci of an ellipse, and how to sketch an ellipse by hand. Work through Example C carefully. Here is another example. Try to solve it yourself before reading the solution.
EXAMPLE
Locate the foci of the ellipse. y 7
–7
7
x
–7
䊳
Solution
The ellipse has a horizontal major axis, so b 2 c 2 ___ a 2. In this__case, b 4 and 2 2 2 a 6, __ so c 6 4__ 20, and therefore c 20 25 . So the foci are 2 5 , 0 and 25 , 0, or about (4.47, 0) and (4.47, 0).
Investigation: A Slice of Light Read the investigation in your book through Step 2. If you have a flashlight and someone to help you, complete the steps. (continued)
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Lesson 8.2 • Circles and Ellipses (continued) Eccentricity is a measure of how elongated an ellipse is. For an ellipse with a horizontal major axis, eccentricity is _ac . For an ellipse with a vertical major axis, it is _bc . The eccentricity of an ellipse is always between 0 and 1. The closer the eccentricity is to 0, the more circular the ellipse is. The closer it is to 1, the more elongated the ellipse is. y
y
y
6
6
5
–5
5
x
–5
5
x
6
–6
x
–5 –6 Eccentricity 0.18
Eccentricity 0.50
–6 Eccentricity 0.99
If you can, complete Steps 3 and 4 in your book. You should find that when the eccentricity becomes too large, the ellipse becomes a parabola and then a single branch of a hyperbola.
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8.3
Parabolas
In this lesson you will ● ● ●
learn the locus definition of a parabola find the focus and directrix of a parabola based on its equation use the locus definition of a parabola to construct a parabola using patty paper
In previous chapters you learned that a parabola is a transformation of the graph with the standard equation y x 2. A parabola can also be defined as a locus of points.
P
A parabola is a locus of points P in a plane whose distance from a fixed point, called the focus, is the same as the distance from a fixed line called the directrix. That is, d 1 d 2. In the diagram, F is the focus and l is the directrix.
d1 d1 P F
If the directrix of a parabola is a horizontal line, then the parabola is vertically oriented. If the directrix is a vertical line, then the parabola is horizontally oriented.
The parabola at bottom is vertically oriented with vertex (0, 0), focus (0, f ), and directrix y f. By the locus definition, you know that d 1 d 2. That _________________ _________________ 2 2 is, (x 0) (y f ) (x x)2 (y f )2 . You can use a little algebra 1 2 to rewrite this equation as x 2 4f y, or y __ x . So when the equation for a 4f parabola is in either form, you can find the distance, f, from the vertex to the focus.
ᐉ
y d2 d1 x F(f, 0) Directrix
If a parabola is horizontally oriented, with vertex (0, 0), then its focus is inside the curve at a point, (f, 0), as shown at right. Because the directrix is the same distance from the vertex as the focus, its equation is x f. The text on page 464 of your book shows how you can use this information, along with the locus definition, to derive the equation y 2 4f x for the parabola. Read the derivation carefully. So, when the equation for a parabola is in the form y 2 4f x, you know that the distance from the vertex to the focus is f, one-fourth the coefficient of x.
d2
d2
y
d1 (0, f )
(x, y) d2
(0, 0) (x, f )
x y f
Read the example in your book carefully, and then read the example on next page. Try to answer each part on your own before reading the solution. (continued)
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Lesson 8.3 • Parabolas (continued) EXAMPLE
Consider the parent equation, y x 2, of a vertically oriented parabola. a. Write the equation of the image of the graph after a horizontal dilation by a factor of 2, a vertical dilation by a factor of 0.5, and then a translation left 3 units. b. Where is the focus of y x 2 ? Where is the directrix? c. Where are the focus and directrix of the transformed parabola?
䊳
Solution
Recall the transformations of functions that you studied in Chapter 4. a. Begin with the parent equation and perform the transformations. y x2
Original equation.
y 2 ___ __x 0.5
Dilate horizontally by a factor of 2 and vertically by a factor of 0.5.
2
y x3 2 ___ _____ 0.5
2
Translate left 3 units.
b. Use the general form, x 2 4f y. The coefficient of y is 4f in the general form and 1 in the equation x 2 y. So, 4f 1, or f _14 . Thus, the focus is 0, _14 and the directrix is y _14 . y x3 2 ____ c. First, rewrite the equation __ 0.5 2 as
8y (x The coefficient of y is 8, so 4f 8, or f 2. The focus and the directrix will both be 2 units from the vertex, which is (3, 0), in the vertical direction. Therefore, the focus is (3, 2) and the directrix is the line y 2. 3)2.
y 5 8y (x 3)2
(–3, 2) –5
5
x
y 2
–5
The box on page 466 of your book summarizes the standard form of the equation for both vertically oriented and horizontally oriented parabolas. Read this material carefully.
Investigation: Fold a Parabola y
Follow the directions in your book to construct a parabola using patty paper, and find its equation. Here is an example: Suppose this parabola was overlaid on graph paper and traced, with focus (3, 0) and directrix x 1. The general form of the parabola is y 2 4f x. The distance from the focus to the vertex is 1, so f 1 and the general equation of the parabola is y 2 4x. However, the parabola is translated right 2 units, so the final equation is y 2 4(x 2).
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4
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8
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8.4
Hyperbolas
In this lesson you will ● ● ●
learn the locus definition for a hyperbola use the asymptotes of a hyperbola to help you sketch the curve locate the foci of a hyperbola using the horizontal and vertical scale factors
A hyperbola is a locus of points P in a plane the difference of whose distances from two fixed points is always a constant. That is, ⏐d 2 d 1⏐ is always a constant. In the diagram below, the two fixed points, F1 and F2, are called foci. The points where the two branches of the hyperbola are closest to each other are called vertices. The center of a hyperbola is the point midway between the vertices. F1
d1
d1 P
Vertex Center d2
d1
P
d2
Vertex
P
d2
F2
Notice that the constant difference, ⏐d 2 d 1⏐, is equal to the distance between the vertices. The distance from P to F1 is d1. F1 d1 The distance from P to F2 is d2.
d2
P
d2 d1
The distance between the vertices is equal to the constant difference d2 d1 .
F2 (continued)
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Lesson 8.4 • Hyperbolas (continued) Example A in your book derives the equation x 2 y 2 1 for the unit hyperbola, shown below. Read this example carefully, following along with a pencil and paper. y 4
(x, y)
2 (–
2, 0) –4
( 2, 0) 2
–2
4
x
–2 –4
Each branch of a hyperbola approaches two lines called asymptotes. Asymptotes are lines that a graph approaches as x- or y-values increase in the positive or negative direction. The asymptotes for the unit hyperbola are y x and y x . Notice that these asymptotes pass through the vertices of a square with corners at (1, 1), (1, 1), (1, 1), and (1, 1). If you zoom out on the graph of a hyperbola, eventually it looks like the letter X. The lines that form the X are the asymptotes. The graph of y 2 x 2 1 is also a hyperbola. This hyperbola has the same shape as the graph of x 2 y 2 1, but it is oriented vertically.
y 4 2 –4
–2
2
4
x
–2 –4
The standard form of the equation of a hyperbola centered at the origin is x
__a
2
y 2 y 2 x 2 __ 1 or __ __ 1 a b b
where a is the horizontal scale factor and b is the vertical scale factor. Example B in your book shows how to graph a hyperbola by first sketching its asymptotes. Read this example very carefully. The foci of a hyperbola are the same distance from the center as the corners of the asymptote rectangle. In the diagram below, this distance is the length of the hypotenuse of a right triangle with legs of lengths a and b. You can use the Pythagorean formula, a 2 b 2 c 2, to locate the foci. On the next page is an example. y (0, 5) a b –5
c 5
x
(0, ⫺5) (continued)
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Lesson 8.4 • Hyperbolas (continued) Sketch a graph of _2x 2 y 2 1 and give the coordinates of the foci.
EXAMPLE 䊳
Solution
From the equation, you can see that this is a horizontally oriented hyperbola with a horizontal scale factor of 2. Start by sketching the asymptotes. To do this, sketch a rectangle centered at the origin that measures 2 2, or 4 units horizontally and 2 1, or 2 units vertically, and then draw lines through opposite vertices. Or use the equations of the asymptotes, y _12 x. The vertices of the hyperbola are at (2, 0) and (2, 0). Use this information to sketch the hyperbola.
y
1 y ⫽ _2 x
2 –4
4 –2
x 1 y ⫽ ⫺ _2 x
To locate the foci, use the relationship a 2 b 2 c__2. In this case, a 2 and _______ __ 2 2 b __1, so c 2 __ 1 5 . So, the foci are 5 units from the center at 5 , 0 and 5 , 0, or about (2.24, 0) and (2.24, 0).
Investigation: Passing By Read the investigation in your book, and make sure you understand the procedure. Here are some sample data. Complete the investigation using these data, and then compare your answers with those below. Time (s)
Distance (m)
Time (s)
Distance (m)
Time (s)
Distance (m)
0
5.000
2.6
1.605
5.2
1.978
0.2
4.720
2.8
1.525
5.4
2.251
0.4
4.381
3.0
1.291
5.6
2.478
0.6
4.016
3.2
1.022
5.8
2.748
0.8
3.688
3.4
0.901
6.0
3.099
1.0
3.558
3.6
0.882
6.2
3.284
1.2
3.302
3.8
0.926
6.4
3.533
1.4
3.078
4.0
0.966
6.6
3.820
1.6
2.709
4.2
1.056
6.8
4.116
1.8
2.410
4.4
1.240
7.0
4.379
2.0
2.249
4.6
1.387
7.2
4.695
2.2
1.969
4.8
1.568
7.4
4.955
2.4
1.770
5.0
1.673
Step 1
Here is a graph of the data:
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Lesson 8.4 • Hyperbolas (continued) The asymptotes y 1.3(x 3.8) and y 1.3(x 3.8) work well.
Step 2
The asymptotes have slopes _ab , so _ab 1.3. The center of the hyperbola is (3.8, 0) and, according to the sample data, the vertex is approximately (3.8, 0.9). 0.9 So the value of b, the vertical scale factor, is 0.9, and the value of a is __ 1.3 , or approximately 0.7. The equation for the hyperbola is therefore y
___ 0.9
2
x 3.8 2 1 _______ 0.7
The distances from the center to the foci are determined by 0.9 2 0.7 2 c 2, so c 1.14 and the foci are (3.8, 1.14) and (3.8, 1.14). Calculating d 2 d 1 for two different points gives approximately 1.80 units. The differences in the distances are the same. This is the locus definition of a hyperbola. Step 3
y
4 2 –4
–2
d1 d1 d2
d2
2
6
8
10
x
–2 –4
The text in the Equation of a Hyperbola box on page 474 of your book gives the standard equation for a hyperbola. Example C in your book asks you to write an equation for a given hyperbola. Try to do this yourself before reading the solution. (Hint: You will need to write an equation containing b and then use a point on the curve to solve for b.)
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8.5
The General Quadratic
In this lesson you will ● ● ● ●
convert quadratic equations from general form to standard form solve quadratic equations for y so they can be graphed on a calculator find all the possible ways two conic sections can intersect find the intersection points of two conic sections
Circles, parabolas, ellipses, and hyperbolas are called quadratic curves, or seconddegree curves, because the highest power of any variable in their equations is 2. The equations for any of these curves can be written in the general quadratic form Ax 2 Bxy Cy 2 Dx Ey F 0 where A, B, and C are not all zero. For all the curves you have seen so far in this chapter, B is equal to 0 (that is, there is no xy-term). If B is not equal to 0, the curve is rotated (it is not oriented horizontally or vertically). To graph a quadratic equation given in general form by hand, it is helpful to first rewrite it in standard form. And, to graph the equation on your calculator, you must first solve it for y. In this lesson you will practice converting general quadratic equations to these forms. Read Example A in your book. The equation is relatively easy to work with because it has no x-, y-, or xy-terms. If an equation has these terms, then you must use the process of completing the square to rewrite it in standard form. This is demonstrated in Example B of your book. Read that example, and then read the example below.
EXAMPLE A 䊳
Solution
Describe the shape determined by the equation 25x 2 4y 2 150x 16y 109 0. Complete the square to convert the equation to standard form. 25x 2 4y 2 150x 16y 109 0
Original equation.
25x 2 150x 4y 2 16y 109 Group x-terms together and y-terms together and move constants to the other side.
25x 2 6x 4y 2 4y 109 Factor out the coefficients of x 2 and y 2. 25x 2 6x 9 4y 2 4y 4 109 25(9) 4(4) Complete the square for both x and y. Add the same values to the right side of the equation.
25(x 3)2 4(y 2)2 100 (y 2) (x 3) _______ _______ 1 2
2
4
25
x3 _____ 2
2
Write the equation in perfect-square form.
y2 2 _____ 1 5
Divide both sides by 100. Write the equation in standard form.
The equation is that of a horizontally oriented hyperbola with center (3, 2), horizontal scale factor 2, and vertical scale factor 5. (continued) Discovering Advanced Algebra Condensed Lessons
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Lesson 8.5 • The General Quadratic (continued) In the previous example, you could have used clues from the general form of the equation to predict that the graph would be a hyperbola before you converted the equation to standard form. Because the equation has an x 2-term and a y 2-term and no xy-term, the graph must be an ellipse, a hyperbola, or a circle. The coefficient of x 2 is positive and the coefficient of y 2 is negative, so the graph is a hyperbola. The equation in Example C in your book has a y 2-term but no x 2-term and no xy-term. This indicates that its graph is a parabola. The second part of Example C shows you how to use the quadratic formula to solve the equation for y. Work through Example C with a pencil and paper.
Investigation: Systems of Conic Equations There are four conic sections: circles, ellipses, parabolas, and hyperbolas. The diagrams on page 485 of your book show that an ellipse and a hyperbola can intersect in 0, 1, 2, 3, or 4 points. There are nine other possible pairs of two conic sections: Ellipse and ellipse Ellipse and parabola Ellipse and circle Parabola and parabola Parabola and hyperbola Parabola and circle Hyperbola and hyperbola Hyperbola and circle Circle and circle For each pair, consider all the possible numbers of intersection points, and make a sketch for each possibility. When you are done, compare your answers to those below. Circle and circle: 0, 1, 2, infinitely many Ellipse and ellipse: 0, 1, 2, 3, 4, infinitely many Parabola and parabola: 0, 1, 2, 3, 4, infinitely many Hyperbola and hyperbola: 0, 1, 2, 3, 4, infinitely many All other combinations: 0, 1, 2, 3, or 4
To find the points where two conic sections intersect, first graph the curves to see the number of intersection points and their approximate locations. Then, use algebra to find the exact intersection points. This technique is illustrated in Example D in your book. On the next page is another example. You will need to fill in some of the details in the solution on your own. (continued)
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Lesson 8.5 • The General Quadratic (continued) EXAMPLE B 䊳
Solution
y2 2 2 Find the points of intersection of x 2 __ 9 1 and (y 2) x 1.
______
Solving both equations for y gives y 31 x 2 and ______ y 2 1 x 2 . (Be sure to verify this.) Graphing the equations shows there are three points of intersection. One appears to be (0, 3). Using the graph you can find that the other two points are approximately (0.98, 0.6) and (0.98, 0.6). To find the intersection points algebraically, solve the y2 first equation for x 2 to get x 2 1 __ 9 , and then substitute y2 __ 2 1 9 for x in the second equation. (y 2)2 x 2 y2 (y 2)2 1 __ 9 y2 y 2 4y 4 1 __ 9 10 y 2 4y 2 ___ 9
1
Original second equation.
1
2
y Substitute 1 __ for x 2. 9
1
Expand the squared binomial.
0
Combine like terms. ____________
10 (2) 4 42 4 ___ 9 ___________________ y 10 2 ___
Use the quadratic formula.
y 0.6 or y 3
Evaluate.
9
To find the corresponding values for x, substitute both values into ______ y2 x 1 __ 9 , which is obtained from the first equation.
___________
(3) x 1 9
(0.6)2 x 1 _______ 0.980 9 _________ 2
_____ 0
The points of intersection are about (0.980, 0.6), (0.980, 0.6), and (0, 3). Sometimes using graphs to find approximate points of intersection is sufficient. Read Example E in your book.
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8.6
Introduction to Rational Functions
In this lesson you will ● ● ● ● ●
model real-world data with a rational function examine transformations of f (x) _1x , the parent function for the inverse variation curve rewrite equations for rational functions to see how they are related to y _1x write an equation for the graph of a rational function use rational expressions to solve a problem involving acid solutions
In the investigation you will see how the length of a pole is related to how much weight it can support. Before you do the investigation, read the introduction on page 490 of your book and look at curves A, B, and C. Which curve do you think most closely resembles the relationship between the length of the pole and the “breaking mass”—that is, the minimum mass that will cause the pole to break?
Investigation: The Breaking Point Read the materials list, the Procedure Note, and Step 1 of the investigation in your book. If you have the materials, collect 10 or 15 data values on your own, and use your data to complete the investigation. If you do not have the materials, use these sample data. The results below are based on the sample data. Length (cm) x
Mass (number of pennies) y
Length (cm) x
Mass (number of pennies) y
16
6
12
7
16
5
11
8
15
7
10
9
15
6
9
10
14
6
8
11
13
7
7
13
13
6
6
16
12
8
The graph does not appear to be linear. It is a curve that decreases quickly at first, and then more slowly.
Step 2
Step 3
90 One possible equation is y __ x.
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Lesson 8.6 • Introduction to Rational Functions (continued) The relationship between length and mass in the investigation is an inverse variation. The parent function for an inverse variation curve is f (x) _1x . This is the simplest rational function. p (x)
A rational function is any function that can be written in the form f (x) ___ , q (x) where p (x) and q (x) are both polynomials. The denominator polynomial cannot equal the constant 0. Notice that the graph of y _1x , shown at right, is a hyperbola rotated 45° with vertices (1, 1) and (1, 1). The x- and y-axes are the asymptotes. The function has no value at x 0 because 10 is undefined. As the x-values get closer to zero, from the left, the y-values become more and more negative. As the x-values get closer to zero from the right, the y-values become more and more positive. As the x-values approach the extreme values on both the left and right ends, the graph approaches the horizontal axis. The features of the graph of y _1x are described in more detail on page 492 of your book.
y 5
–5
5
1 _1 Functions such as y _1x 4, y ____ x 1 , and y 2 x are transformed rational functions. Use your calculator to experiment with various transformations of y _1x .
x
–5
Example A in your book shows how a rational function can be rewritten so that it is clear how it is related to the parent function, y _1x . Read this example carefully. Then try to solve the problem in the example below before reading the solution.
EXAMPLE
䊳
Solution
3x 11 Describe the function y ______ x 3 as a transformation of the parent function, 1 _ y x . Then sketch a graph.
Because the numerator and the denominator have the same degree, you can use division to rewrite the expression. _______3 ) x 3 3x 11 3x 9 2 2 y 3 _____ x3
The parent function has been vertically dilated by a factor of 2, then translated left 3 units and up 3 units. The asymptotes have been translated as well. The vertical asymptote has been translated left 3 units to x 3, and the horizontal asymptote has been translated up 3 units to y 3.
y 7
–7
7
x
–7 (continued)
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Lesson 8.6 • Introduction to Rational Functions (continued) Suppose that you are given the graph of a transformed rational function and you must find its equation. You can identify the translations by looking at the locations of the asymptotes: A horizontal asymptote of y k indicates a vertical translation of k units, and a vertical asymptote of x h indicates a horizontal translation of h units. To identify the scale factors, pick a point, such as a vertex, whose coordinates you would know after the translation. Then, find a point on the dilated graph that has the same x-coordinate. The ratio of the vertical distances from the horizontal asymptotes to those two points is the vertical scale factor. An example is shown below. You will find another example on page 493 of your book. y 7
–7
7
x
An undilated rational function with these translations would have vertices (⫺4, 2) and (⫺2, 4), 1 horizontal unit and 1 vertical unit from the center. Because the distance is now 2 vertical units from the center, include a vertical scale factor of 2 to get the equation 2 y ⫽ 3 ⫹ _____ x⫹3
–7
The problem in Example B in your book uses rational expressions to model a situation involving an acid solution. Work through the example carefully.
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8.7
Graphs of Rational Functions
In this lesson you will ● ● ●
predict holes and asymptotes in the graph of a rational function based on its equation write equations for graphs of rational functions convert rational functions from one form to another
The graphs of rational functions have a variety of shapes. Some examples are shown in the lesson introduction in your book. Notice that the graphs have either asymptotes or holes at values where the functions are undefined. In this lesson you will see how you can predict asymptotes, holes, and other features of a graph based on its equation.
Investigation: Predicting Asymptotes and Holes Complete Step 1 in your book, and then compare your results with those below. a. B. The graph has a vertical asymptote at x 2, which is the value that makes the denominator 0. As x gets close to 2 from the left, the y-values are negative numbers with increasingly extreme values. As x gets close to 2 from the right, the y-values become larger and larger positive numbers. b. D. The graph has a hole at x 2, which is the value that makes both the numerator and denominator 0. For any x-value except 2, the function reduces to y 1, so the graph looks like y 1 everywhere but at that point. c. A. The graph has a vertical asymptote at x 2, which is the value that makes the denominator 0. As x gets close to 2 from either side, the y-values become larger and larger positive numbers. d. C. The graph has a hole at x 2, which is the value that makes both the numerator and denominator 0. For any x-value except 2, the function reduces to y x 2, so the graph looks like y x 2 everywhere but at that point. Now, use what you learned in Step 1 to complete Steps 2 and 3. Then, compare your results with those below. Step 2 1 _1 a. y ____ x 1 . This is the graph of y x translated left 1 unit (so the vertical 1 _ asymptote of y x , namely the y-axis, has been translated left 1 unit to x 1). 2(x 1) b. y ______ x 1 . The function is undefined for x 1 but reduces to y 2 for all other x-values. Thus, the graph looks like the graph of y 2, with a hole at x 1. 1 c. y ______ . The function is undefined for x 1. As the x-values (x 1)2 approach 1 from either direction, (x 1)2 becomes a smaller and 1 smaller positive number, so ______ becomes a larger and larger (x 1)2 positive number. Therefore, x 1 is an asymptote. (x 1)2 d. y ______ x 1 . The function is undefined for x 1 but reduces to y x 1 for all other x-values. Thus, the graph looks like the graph of y x 1 with a hole at x 1.
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Lesson 8.7 • Graphs of Rational Functions (continued) Vertical asymptotes occur at zeroes that appear only in the denominator or zeros that appear more times in the denominator than in the numerator. If an asymptote occurs at x a, then the equation will have (x a) as a factor of its denominator. Holes occur at values that make both the numerator and denominator 0 provided there is not a vertical asymptote at these values. To write an equation for a graph with a hole at x a , imagine the graph has no hole and xa write an equation for it. Then, multiply the result by ____ x a. Step 3
The graph has vertical asymptote x 2. When a factor occurs in both the numerator and denominator, but occurs more times in the denominator, it indicates a vertical asymptote rather than a hole. Step 4
Think carefully about what the clues in the equation of a rational function tell you about its graph. Then, read Example A in your book. Example B shows how factoring the numerator and denominator of a rational function can help you determine features of its graph. Work through that example, and then read the example below.
EXAMPLE 䊳
Solution
x2 x 6 Describe the features of the graph of y ________ . x 2 5x 6 (x 2)(x 3)
Factoring the numerator and denominator gives y __________ . There is (x 2)(x 3) a hole at x 3 because it is a zero occurring the same number of times in both the numerator and denominator. If x 2, the denominator (but not the numerator) is 0, so there is a vertical asymptote at x 2. If x 2, the numerator (but not the denominator) is 0, so there is an x-intercept at x 2. If x 0, then y 1. This is the y-intercept. To find any horizontal asymptotes, consider what happens to the y-values as the x-values get very far from 0. x
10,000
1,000
100
100
1,000
10,000
y
0.99960
0.99601
0.96078
1.04082
1.00401
1.00040
The table shows that the y-values get closer and closer to 1 as x gets farther from 0. So y 1 is a horizontal asymptote. The graph of the function confirms these features. y 6
–4
6
x
–4
Example C in your book shows how to convert a function from a form that shows the transformations of y _1x to rational function form. Read this example carefully. 134
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Operations with Rational Expressions
8.8 In this lesson you will ●
operate with rational expressions
Operating with rational expressions is very similar to operating with whole number fractions. For example, to add or subtract two rational expressions, you must first rewrite the expressions so that they have a common denominator. To review adding fractions, read the text on page 505 of your book. Work through Example A in your book, and then add the rational expressions in the example below. After you have found the sum, compare your work with the solution.
EXAMPLE A
Add rational expressions to rewrite the right side of the equation as a single rational expression in factored form. 7 2x 5 y _____________ ______ x3 (x 4)(x 3)
䊳
Solution
The least common denominator is (x 4)(x 3). 2x 5 7 ______ y _____________ x3 (x 4)(x 3) 7 2x 5 y _____________ ______ x3 (x 4)(x 3)
Original equation.
(x 4)
_______ (x 4)
Multiply the second fraction by an equivalent of 1 to get a common denominator.
7 2x 2 3x 20 y _____________ _____________ (x 4)(x 3) (x 3)(x 4)
Multiply the numerator of the second fraction.
2x 2 3x 13 y _____________ (x 3)(x 4)
Add the numerators.
The numerator cannot be factored with rational roots. To subtract rational expressions, you must also find a common denominator. This is demonstrated in Example B in your book. Work through Example B carefully. The text between Examples B and C in your book reviews how to multiply and divide fractions. Read that text if you need to. You use the same process to multiply and divide rational expressions. When you multiply and divide rational expressions, factor all the expressions first. This will make it easy to reduce common factors and identify features of the graph. Work through Example C. Then, find the product in the example on the next page. After you have found the product, compare your results with the solution. (continued)
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Lesson 8.8 • Operations with Rational Expressions (continued) EXAMPLE B 䊳
Solution
x 2 7x 6 Multiply __________ x 2 5x 6
2x 2x . ________ x1 2
(x 6)(x 1) _____________
Factor any expressions that you can.
(x 6)(x 1)2x (x 1) _____________________
Combine the two expressions.
(x 6)(x 1)2x (x 1) _____________________
Reduce common factors.
2x
Rewrite.
2x (x 1) _________ (x 6)(x 1) (x 1) (x 6)(x 1)(x 1) (x 6)(x 1)(x 1)
(x 6)(x 1) Note that the graphs of y __________ (x 6)(x 1)
2x (x 1)
and y 2x will look the same, _______ (x 1) (x 6)(x 1) 2x (x 1) _______ will have holes at x 6, except that the graph of y __________ (x 6)(x 1) (x 1) x 1, and x 1.
Read Example D in your book, and then find the quotient in the example below.
EXAMPLE C
x 2 8x 9 __________ x2 Divide ____________ . x 2 2x 3 ___________ x 2 7x 18
䊳
Solution
x 2 8x 9 ___________ x2
x 7x 18 ____________ x 2 2x 3
(x 9)(x 1) _____________ (x 2)
2
(x 9)(x 2)
_____________ (x 3)(x 1)
Invert the fraction in the denominator and multiply.
Factor all expressions.
(x 9)(x 1)(x 9)(x 2) _________________________
Multiply.
(x 9)(x 9) _____________
Reduce all common factors.
(x 2)(x 3)(x 1) (x 3)
You can check your answer by comparing the graphs of x 2 8x 9 __________ (x 9)(x 9) x2 and y _____________ y ____________ 2 (x 3) x 2x 3 ___________ x 2 7x 18
The graphs should be identical except for holes.
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