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2.5 Transformations

Given the graph of a function, graph its transformation under translations, reflections, stretchings, and shrinkings.

Transformations of Functions

The graphs of some basic functions are shown on the following page. Others can be seen on the inside back cover.

These functions can be considered building blocks for many other functions. We can create graphs of new functions by shifting them horizontally or vertically, stretching or shrinking them, and reflecting them across an axis. We now consider these transformations.

Vertical Translations and Horizontal Translations

Suppose that we have a function given by y=f(x) $y = f (x)$ . Let’s explore the graphs of the new functions y=f(x)+b $y = f (x) + b$ and y=f(x)−b $y = f (x) - b$ , for b>0 $b > 0$ .

Consider the functions y=15 x4 $y = \frac{1}{5} x^{4}$ , y=15 x4+5 $y = \frac{1}{5} x^{4} + 5$ , and y=15 x4−3 $y = \frac{1}{5} x^{4} - 3$ and compare their graphs. What pattern do you see? Test it with some other functions.

The effect of adding a constant to or subtracting a constant from f(x) $f (x)$ in y=f(x) $y = f (x)$ is a shift of the graph of f(x) $f (x)$ up or down. Such a shift is called a vertical translation.

Suppose that we have a function given by y=f(x) $y = f (x)$ . Let’s explore the graphs of the new functions y=f(x−d) $y = f (x - d)$ and y=f(x+d) $y = f (x + d)$ , for d>0 $d > 0$ .

Consider the functions y=15 x4 $y = \frac{1}{5} x^{4}$ , y=15(x−3)4 $y = \frac{1}{5} {(x - 3)}^{4}$ , and y=15(x+7)4 $y = \frac{1}{5} {(x + 7)}^{4}$ and compare their graphs. What pattern do you observe? Test it with some other functions.

The effect of subtracting a constant from the x-value or adding a constant to the x-value in y=f(x) $y = f (x)$ is a shift of the graph of f(x) $f (x)$ to the right or to the left. Such a shift is called a horizontal translation.

Example 1

Graph each of the following. Before doing so, describe how each graph can be obtained from one of the basic graphs shown on the preceding pages.

g(x)=x2−6 $g (x) = x^{2} - 6$
h(x)=|x−4| $h (x) = | x - 4 |$
g(x)=x+2−−−−−√ $g (x) = \sqrt{x + 2}$
h(x)=x+2−−−−−√−3 $h (x) = \sqrt{x + 2} - 3$

Solution

To graph g(x)=x2−6 $g (x) = x^{2} - 6$ , think of the graph of f(x)=x2 $f (x) = x^{2}$ . Since g(x)=f(x)−6 $g (x) = f (x) - 6$ , the graph of g(x)=x2−6 $g (x) = x^{2} - 6$ is the graph of f(x)=x2 $f (x) = x^{2}$ , shifted, or translated, down 6 units. (See Fig. 1.)

Figure 1.

Let’s compare some points on the graphs of f and g.

$Points on f : Corresponding points on g (- 3, 9), ⏐ ↓ ⏐ ⏐ (- 3, 3), (0, 0), ⏐ ↓ ⏐ ⏐ (0, - 6), (2, 4) ⏐ ↓ ⏐ ⏐ (2, - 2)$ $\begin{array}{l} Points on f : & (- 3, 9), & (0, 0), & (2, 4) \\ \begin{array}{l} Corresponding \\ points on g \end{array} & \begin{array}{l} ↓ \\ (- 3, 3), \end{array} & \begin{array}{l} ↓ \\ (0, - 6), \end{array} & \begin{array}{l} ↓ \\ (2, - 2) \end{array} \end{array}$

We note that the y-coordinate of a point on the graph of g is 6 less than the corresponding y-coordinate on the graph of f.
To graph h(x)=|x−4| $h (x) = | x - 4 |$ , think of the graph of f(x)=|x| $f (x) = | x |$ . Since h(x)=f(x−4) $h (x) = f (x - 4)$ , the graph of h(x)=|x−4| $h (x) = | x - 4 |$ is the graph of f(x)=|x| $f (x) = | x |$ shifted right 4 units. (See Fig. 2.)

Figure 2.

Let’s again compare points on the two graphs.

$Points on f : Corresponding points on h : (- 4, 4), ⏐ ↓ ⏐ (0, 4), (0, 0), ⏐ ↓ ⏐ (4, 0), (6, 6) ⏐ ↓ ⏐ (10, 6)$ $\begin{array}{l} Points on f : & (- 4, 4), & (0, 0), & (6, 6) \\ \begin{array}{l} Corresponding \\ points on h : \end{array} & \begin{array}{l} ↓ \\ (0, 4), \end{array} & \begin{array}{l} ↓ \\ (4, 0), \end{array} & \begin{array}{l} ↓ \\ (10, 6) \end{array} \end{array}$

Noting points on f and h, we see that the x-coordinate of a point on the graph of h is 4 more than the x-coordinate of the corresponding point on f.
To graph g(x)=x+2−−−−−√ $g (x) = \sqrt{x + 2}$ , think of the graph of f(x)=x−−√ $f (x) = \sqrt{x}$ . Since g(x)=f(x+2) $g (x) = f (x + 2)$ , the graph of g(x)=x+2−−−−−√ $g (x) = \sqrt{x + 2}$ is the graph of f(x)=x−−√ $f (x) = \sqrt{x}$ , shifted left 2 units. (See Fig. 3.)

Figure 3.
To graph h(x)=x+2−−−−−√−3 $h (x) = \sqrt{x + 2} - 3$ , think of the graph of f(x)=x−−√ $f (x) = \sqrt{x}$ . In part (c), we found that the graph of g(x)=x+2−−−−−√ $g (x) = \sqrt{x + 2}$ is the graph of $f (x) = \sqrt{x}$ shifted left 2 units. Since $h (x) = g (x) - 3$ , we shift the graph of $g (x) = \sqrt{x + 2}$ down 3 units. Together, the graph of $f (x) = \sqrt{x}$ is shifted left 2 units and down 3 units. (See Fig. 4.)

Figure 4.

Now Try Exercises 3 and 15.

Reflections

Suppose that we have a function given by $y = f (x)$ . Let’s explore the graphs of the new functions $y = - f (x)$ and $y = f (- x)$ .

Compare the functions $y = f (x)$ and $y = - f (x)$ by looking at the graphs of $y = \frac{1}{5} x^{4}$ and $y = - \frac{1}{5} x^{4}$ shown on the left below. What do you see? Test your observation with some other functions $y_{1}$ and $y_{2}$ where $y_{2} = - y_{1}$ .

Compare the functions $y = f (x)$ and $y = f (- x)$ by looking at the graphs of $y = 2 x^{3} - x^{4} + 5$ and $y = 2 {(- x)}^{3} - {(- x)}^{4} + 5$ shown on the right below. What do you see? Test your observation with some other functions in which x is replaced with −x.

Given the graph of $y = f (x)$ , we can reflect each point across the x-axis to obtain the graph of $y = - f (x)$ . We can reflect each point of $y = f (x)$ across the y-axis to obtain the graph of $y = f (- x)$ . The new graphs are called reflections of $y = f (x)$ .

The following photographs illustrate reflection.

Example 2

Graph each of the following. Before doing so, describe how each graph can be obtained from the graph of $f (x) = x^{3} - 4 x^{2}$ .

$g (x) = {(- x)}^{3} - 4 {(- x)}^{2}$
$h (x) = 4 x^{2} - x^{3}$

Solution

We first note that

$f (- x) = {(- x)}^{3} - 4 {(- x)}^{2} = g (x) .$

Thus the graph of g is a reflection of the graph of f across the y-axis. (See the figure at left.) If (x, y) is on the graph of f, then (−x, y) is on the graph of g. For example, (2, −8) is on f and (−2, 8) is on g.
We first note that

$\begin{array}{rcll} - f (x) & = & - (x^{3} - 4 x^{2}) \\ = & - x^{3} + 4 x^{2} \\ = & 4 x^{2} - x^{3} \\ = & h (x) . \end{array}$

Thus the graph of h is a reflection of the graph of f across the x-axis. (See the figure at right.) If (x, y) is on the graph of f, then (x, −y) is on the graph of h. For example, (2, −8) is on f and (2, 8) is on h.

Vertical and Horizontal Stretchings and Shrinkings

Suppose that we have a function given by $y = f (x)$ . Let’s explore the graphs of the new functions $y = a f (x)$ and $y = f (c x)$ .

Consider the functions $y = f (x) = x^{3} - x$ , $y = \frac{1}{10} (x^{3} - x) = \frac{1}{10} f (x)$ , $y = 2 (x^{3} - x) = 2 f (x)$ , and $y = - 2 (x^{3} - x) = - 2 f (x)$ and compare their graphs. What pattern do you observe? Test it with some other functions.

Consider any function f given by $y = f (x)$ . Multiplying $f (x)$ by any constant a, where $| a | > 1$ , to obtain $g (x) = a f (x)$ will stretch the graph vertically away from the x-axis. If $0 < | a | < 1$ , then the graph will be flattened or shrunk vertically toward the x-axis. If $a < 0$ , the graph is also reflected across the x-axis.

Consider the functions $y = f (x) = x^{3} - x, y = {(2 x)}^{3} - (2 x) = f (2 x), y = {(\frac{1}{2} x)}^{3} - (\frac{1}{2} x) = f (\frac{1}{2} x)$ , and $y = {(- \frac{1}{2} x)}^{3} - (- \frac{1}{2} x) = f (- \frac{1}{2} x)$ and compare their graphs. What pattern do you observe? Test it with some other functions.

The constant c in the equation $g (x) = f (c x)$ will shrink the graph of $y = f (x)$ horizontally toward the y-axis if $| c | > 1$ . If $0 < | c | < 1$ , the graph will be stretched horizontally away from the y-axis. If $c < 0$ , the graph is also reflected across the y-axis.

Example 3

Shown at left is a graph of $y = f (x)$ for some function f. No formula for f is given. Graph each of the following.

$g (x) = 2 f (x)$
$h (x) = \frac{1}{2} f (x)$
$r (x) = f (2 x)$
$s (x) = f (\frac{1}{2} x)$
$t (x) = f (- \frac{1}{2} x)$

Solution

Since $| 2 | > 1$ , the graph of $g (x) = 2 f (x)$ is a vertical stretching of the graph of $y = f (x)$ by a factor of 2. We can consider the key points(−5, 0), (−2, 2), (0, 0), (2, −4), and (4, 0) on the graph of $y = f (x)$ . The transformation multiplies each y-coordinate by 2 to obtain the key points (−5, 0), (−2, 4), (0, 0), (2, −8), and (4, 0) on the graph of $g (x) = 2 f (x)$ , as shown below.
Since $| \frac{1}{2} | < 1$ , the graph of $h (x) = \frac{1}{2} f (x)$ is a vertical shrinking of the graph of $y = f (x)$ by a factor of $\frac{1}{2}$ . We again consider the key points (−5, 0), (−2, 2), (0, 0), (2, −4), and (4, 0) on the graph of $y = f (x)$ . The transformation multiplies each y-coordinate by $\frac{1}{2}$ to obtain the key points (−5, 0), (−2, 1), (0, 0), (2, −2), and (4, 0) on the graph of $h (x) = \frac{1}{2} f (x)$ . The graph is shown on the left below.
Since $| 2 | > 1$ , the graph of $r (x) = f (2 x)$ is a horizontal shrinking of the graph of $y = f (x)$ . We consider the key points (−5, 0), (−2, 2), (0, 0), (2, −4), and (4, 0) on the graph of $y = f (x)$ . The transformation divides each x-coordinate by 2 to obtain the key points (−2.5, 0), (−1, 2), (0, 0), (1, −4), and (2, 0) on the graph of $r (x) = f (2 x)$ . The graph is shown on the right above.
Since $| \frac{1}{2} | < 1$ , the graph of $s (x) = f (\frac{1}{2} x)$ is a horizontal stretching of the graph of $y = f (x)$ . We consider the key points (−5, 0), (−2, 2), (0, 0), (2, −4), and (4, 0) on the graph of $y = f (x)$ . The transformation divides each x-coordinate by $\frac{1}{2}$ (which is the same as multiplying by 2) to obtain the key points (−10, 0), (−4, 2), (0, 0), (4, −4), and (8, 0) on the graph of $s (x) = f (\frac{1}{2} x)$ . The graph is shown below.
The graph of $t (x) = f (- \frac{1}{2} x)$ can be obtained by reflecting the graph in part (d) across the y-axis.

Now Try Exercises 59 and 61.

Example 4

Use the graph of $y = f (x)$ shown at left to graph $y = - 2 f (x - 3) + 1$ .

Solution

Now Try Exercise 63.

Summary of Transformations of $y = f (x)$

Vertical Translation: $y = f (x) \pm b$

For $b > 0$ :

the graph of $y = f (x) + b$ is the graph of $y = f (x)$ shifted up b units;
the graph of $y = f (x) - b$ is the graph of $y = f (x)$ shifted down b units.

Horizontal Translation: $y = f (x \mp d)$

For $d > 0$ :

the graph of $y = f (x - d)$ is the graph of $y = f (x)$ shifted right d units;
the graph of $y = f (x + d)$ is the graph of $y = f (x)$ shifted left d units.

Reflections

Across the x-axis:

The graph of $y = - f (x)$ is the reflection of the graph of $y = f (x)$ across the x-axis.

Across the y-axis:

The graph of $y = f (- x)$ is the reflection of the graph of $y = f (x)$ across the y-axis.

Vertical Stretching or Shrinking: $y = a f (x)$

The graph of $y = a f (x)$ can be obtained from the graph of $y = f (x)$ by

stretching vertically for $| a | > 1$ , or
shrinking vertically for $0 < | a | < 1$ .

For $a < 0$ , the graph is also reflected across the x-axis.

Horizontal Stretching or Shrinking: $y = f (c x)$

The graph of $y = f (c x)$ can be obtained from the graph of $y = f (x)$ by

shrinking horizontally for $| c | > 1$ , or
stretching horizontally for $0 < | c | < 1$ .

For $c < 0$ , the graph is also reflected across the y-axis.

Visualizing the Graph

Match the function with its graph. Use transformation graphing techniques to obtain the graph of g from the basic function $f (x) = | x |$ shown at top left.

$g (x) = - 2 | x |$
$g (x) = | x - 1 | + 1$
$g (x) = - | \frac{1}{3} x |$
$g (x) = | 2 x |$
$g (x) = | x + 2 |$
$g (x) = | x | + 3$
$g (x) = - \frac{1}{2} | x - 4 |$
$g (x) = \frac{1}{2} | x | - 3$
$g (x) = - | x | - 2$

Answers on page A-10

2.5 Exercise Set

Describe how the graph of the function can be obtained from one of the basic graphs on p. 134. Then graph the function.

1. $f (x) = {(x - 3)}^{2}$
2. $g (x) = x^{2} + \frac{1}{2}$
3. $g (x) = x - 3$
4. $g (x) = - x - 2$
5. $h (x) = - \sqrt{x}$
6. $g (x) = \sqrt{x - 1}$
7. $h (x) = \frac{1}{x} + 4$
8. $g (x) = \frac{1}{x - 2}$
9. $h (x) = - 3 x + 3$
10. $f (x) = 2 x + 1$
11. $h (x) = \frac{1}{2} | x | - 2$
12. $g (x) = - | x | + 2$
13. $g (x) = - {(x - 2)}^{3}$
14. $f (x) = {(x + 1)}^{3}$
15. $g (x) = {(x + 1)}^{2} - 1$
16. $h (x) = - x^{2} - 4$
17. $g (x) = \frac{1}{3} x^{3} + 2$
18. $h (x) = {(- x)}^{3}$
19. $f (x) = \sqrt{x + 2}$
20. $f (x) = - \frac{1}{2} \sqrt{x - 1}$
21. $f (x) = \sqrt[3]{x} - 2$
22. $h (x) = \sqrt[3]{x + 1}$

Describe how the graph of the function can be obtained from one of the basic graphs on p. 134.

23. $g (x) = | 3 x |$
24. $f (x) = \frac{1}{2} \sqrt{x}$
25. $h (x) = \frac{2}{x}$
26. $f (x) = | x - 3 | - 4$
27. $f (x) = 3 \sqrt{x} - 5$
28. $f (x) = 5 - \frac{1}{x}$
29. $g (x) = | \frac{1}{3} x | - 4$
30. $f (x) = \frac{2}{3} x^{3} - 4$
31. $f (x) = - \frac{1}{4} {(x - 5)}^{2}$
32. $f (x) = {(- x)}^{3} - 5$
33. $f (x) = \frac{1}{x + 3} + 2$
34. $g (x) = \sqrt{- x} + 5$
35. $h (x) = - {(x - 3)}^{2} + 5$
36. $f (x) = 3 {(x + 4)}^{2} - 3$

The point (−12, 4) is on the graph of $y = f (x)$ . Find the corresponding point on the graph of $y = g (x)$ .

37. $g (x) = \frac{1}{2} f (x)$
38. $g (x) = f (x - 2)$
39. $g (x) = f (- x)$
40. $g (x) = f (4 x)$
41. $g (x) = f (x) - 2$
42. $g (x) = f (\frac{1}{2} x)$
43. $g (x) = 4 f (x)$
44. $g (x) = - f (x)$

Given that $f (x) = x^{2} + 3$ , match the function g with a transformation of f from one of A–D.

45. $g (x) = x^{2} + 4$
46. $g (x) = 9 x^{2} + 3$
47. $g (x) = {(x - 2)}^{2} + 3$
48. $g (x) = 2 x^{2} + 6$
1. $f (x - 2)$
2. $f (x) + 1$
3. $2 f (x)$
4. $f (3 x)$

Write an equation for a function that has a graph with the given characteristics.

49. The shape of $y = x^{2}$ , but upside-down and shifted right 8 units
50. The shape of $y = \sqrt{x}$ , but shifted left 6 units and down 5 units
51. The shape of $y = | x |$ , but shifted left 7 units and up 2 units
52. The shape of $y = x^{3}$ , but upside-down and shifted right 5 units
53. The shape of $y = 1 / x$ , but shrunk horizontally by a factor of 2 and shifted down 3 units
54. The shape of $y = x^{2}$ , but shifted right 6 units and up 2 units
55. The shape of $y = x^{2}$ , but upside-down and shifted right 3 units and up 4 units
56. The shape of $y = | x |$ , but stretched horizontally by a factor of 2 and shifted down 5 units
57. The shape of $y = \sqrt{x}$ , but reflected across the y-axis and shifted left 2 units and down 1 unit
58. The shape of $y = 1 / x$ , but reflected across the x-axis and shifted up 1 unit

A graph of $y = f (x)$ follows. No formula for f is given. In Exercises 59–66, graph the given equation.

59. $g (x) = - 2 f (x)$
60. $g (x) = \frac{1}{2} f (x)$
61. $g (x) = f (- \frac{1}{2} x)$
62. $g (x) = f (2 x)$
63. $g (x) = - \frac{1}{2} f (x - 1) + 3$
64. $g (x) = - 3 f (x + 1) - 4$
65. $g (x) = f (- x)$
66. $g (x) = - f (x)$

A graph of $y = g (x)$ follows. No formula for g is given. In Exercises 67–70, graph the given equation.

67. $h (x) = - g (x + 2) + 1$
68. $h (x) = \frac{1}{2} g (- x)$
69. $h (x) = g (2 x)$
70. $h (x) = 2 g (x - 1) - 3$

The graph of the function f is shown in figure (a) below. In each of Exercises 71–78, match the function g with one of the graphs (a)–(h) that follow. Some graphs may be used more than once and some may not be used at all.

71. $g (x) = f (- x) + 3$
72. $g (x) = f (x) + 3$
73. $g (x) = - f (x) + 3$
74. $g (x) = - f (- x)$
75. $g (x) = \frac{1}{3} f (x - 2)$
76. $g (x) = \frac{1}{3} f (x) - 3$
77. $g (x) = \frac{1}{3} f (x + 2)$
78. $g (x) = - f (x + 2)$

For each pair of functions, determine if $g (x) = f (- x)$ .

79.

$\begin{array}{rcl} f (x) & = & 2 x^{4} - 35 x^{3} + 3 x - 5, \\ g (x) & = & 2 x^{4} + 35 x^{3} - 3 x - 5 \end{array}$
80.

$\begin{array}{rcl} f (x) & = & \frac{1}{4} x^{4} + \frac{1}{5} x^{3} - 81 x^{2} - 17, \\ g (x) & = & \frac{1}{4} x^{4} + \frac{1}{5} x^{3} + 81 x^{2} - 17 \end{array}$

A graph of the function $f (x) = x^{3} - 3 x^{2}$ is shown below. Exercises 81–84 show graphs of functions transformed from this one. Find a formula for each function.

Skill Maintenance

Determine algebraically whether the graph is symmetric with respect to the x-axis, the y-axis, and the origin. [2.4]

85. $y = 3 x^{4} - 3$
86. $y^{2} = x$
87. $2 x - 5 y = 0$

Solve. [1.5]

88. Federal Tax Rules. The number of pages of U.S. federal tax rules that explain the tax code and regulations totaled 74,608 in 2014 (for tax year 2013). This number was an increase of 84.2% over the number of pages in 1995 (for tax year 1994). (Source: Wolters Kluwer, CCH: 2014) Find the number of pages of federal tax rules in 1995.
89. Guns with Airline Passengers. In 2013, the Transportation Security Administration found 1828 guns with travelers preparing to board an airplane. This number was 418 less than twice the number of guns discovered in 2010. (Source: Transportation Security Administration data by Northwestern University Medill National Security Journalism Initiative) How many guns were found with airline travelers in 2010?
90. Acres of Pumpkins. In 2012, 16,200 acres of pumpkins were harvested in Illinois. This amount was about 54.5% of the total number of acres of pumpkins harvested in Michigan, Ohio, and Illinois. (Source: U.S. Department of Agriculture) Find the total number of acres of pumpkins harvested in Michigan, Ohio, and Illinois.

Synthesis

Use the following graph of the function f for Exercises 91 and 92.

91. Graph: $y = | f (x) |$ .
92. Graph: $y = f (| x |)$ .

Use the following graph of the function g for Exercises 93 and 94.

93. Graph: $y = g (| x |)$ .
94. Graph: $y = | g (x) |$ .
95. If (−1, 5) is a point on the graph of $y = f (x)$ , find b such that (2, b) is on the graph of $y = f (x - 3)$ .
96. The graph of $f (x) = | x |$ passes through the points (−3, 3), (0, 0), and (3, 3). Transform this function to one whose graph passes through the points (5, 1), (8, 4), and (11, 1).

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Table of Contents for 2.5 Transformations

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Table of Contents for
2.5 Transformations