Section 2.7 Transformations of Functions

Before Starting this Section, Review

  1. 1 Graphs of basic functions (Section 2.6 , page 257)

  2. 2 Symmetry (Section 2.2 , page 188)

  3. 3 Completing the square (Section 1.3 , page 109)

Objectives

  1. 1 Define transformations of functions.

  2. 2 Use vertical or horizontal shifts to graph functions.

  3. 3 Use reflections to graph functions.

  4. 4 Use stretching or compressing to graph functions.

  5. 5 Use a sequence of transformations to graph functions.

Measuring Blood Pressure

Blood pressure is the force of blood per unit area against the walls of the arteries. Blood pressure is recorded as two numbers: the systolic pressure (as the heart beats) and the diastolic pressure (as the heart relaxes between beats). If your blood pressure is “120 over 80,” it is rising to a maximum of 120 millimeters of mercury as the heart beats and is falling to a minimum of 80 millimeters of mercury as the heart relaxes.

The most precise way to measure blood pressure is to place a small glass tube in an artery and let the blood flow out and up as high as the heart can pump it. That was the way Stephen Hales (1677–1761), the first investigator of blood pressure, learned that a horse’s heart could pump blood 8 feet 3 inches, up a tall tube. Such a test, however, consumed a great deal of blood. Jean Louis Marie Poiseuille (1797–1869) greatly improved the process by using a mercury-filled manometer, which allowed a smaller, narrower tube.

In Exercise 124, we investigate Poiseuille’s Law for arterial blood flow.

Transformations

  1. 1 Define transformations of functions.

If a new function is formed by performing certain operations on a given function f, then the graph of the new function is called a transformation of the graph of f. For example, the graphs of y=|x|+2y=|x|+2 and y=|x|−3y=|x|−3 are transformations of the graph of y=|x|;y=|x|; that is, the graph of each is a special modification of the graph of y=|x|.y=|x|.

Vertical and Horizontal Shifts

  1. 2 Use vertical or horizontal shifts to graph functions.

A transformation that changes only the position of a graph but not its shape is called a rigid transformation. We first consider rigid transformations involving vertical shifts and horizontal shifts of a graph.

Example 1 Graphing Vertical Shifts

Let f(x)=|x|, g(x)=|x|+2,f(x)=|x|, g(x)=|x|+2, and h(x)=|x|−3.h(x)=|x|−3. Sketch the graphs of these functions on the same coordinate plane. Describe how the graphs of g and h relate to the graph of f.

Solution

Make a table of values and graph the equations y=f(x), y=g(x),y=f(x), y=g(x), and y=h(x).y=h(x).

Notice that in Table 2.14 and Figure 2.78, each point (x, |x|)(x, |x|) on the graph of f has a corresponding point (x, |x|+2)(x, |x|+2) on the graph of g and (x, |x|−3)(x, |x|−3) on the graph of h. So the graph of g(x)=|x|+2g(x)=|x|+2 is the graph of y=|x|y=|x| shifted 2 units up, and the graph of y=|x|−3y=|x|−3 is the graph of y=|x|y=|x| shifted 3 units down.

Figure 2.78 Vertical shifts of y=|x|y=|x|

TABLE 2.14

x y=|x|y=|x| g(x)=|x|+2g(x)=|x|+2 h(x)=|x|−3h(x)=|x|−3
−5−5 5 7 2
−3−3 3 5 0
−1−1 1 3 −2−2
0 0 2 −3−3
1 1 3 −2−2
3 3 5 0
5 5 7 2

Practice Problem 1

  1. Let

    f(x)=x3, g(x)=x3+1, and h(x)=x3−2.
    f(x)=x3, g(x)=x3+1, and h(x)=x3−2.

    Sketch the graphs of all three functions on the same coordinate plane. Describe how the graphs of g and h relate to the graph of f.

Example 1 illustrates the concept of the vertical (up or down) shift of a graph.

Next, we consider the operation that shifts a graph horizontally.

Example 2 Writing Functions for Horizontal Shifts

Let f(x)=x2, g(x)=(x−2)2,f(x)=x2, g(x)=(x−2)2, and h(x)=(x+3)2.h(x)=(x+3)2. A table of values for f, g, and h is given in Table 2.15. The three functions f, g, and h are graphed on the same coordinate plane in Figure 2.80. Describe how the graphs of g and h relate to the graph of f.

Figure 2.80

Horizontal shifts

TABLE 2.15

Solution

First, notice that all three functions are squaring functions.

  1. Each point (x, x2)(x, x2) on the graph of f has a corresponding point (x+2, x2)(x+2, x2) on the graph of g because g(x+2)=(x+2−2)2=x2g(x+2)=(x+2−2)2=x2. So the graph of g(x)=(x−2)2g(x)=(x−2)2 is just the graph of f(x)=x2f(x)=x2 shifted 2 units to the right. Noticing that f(0)=0f(0)=0 and g(2)=0g(2)=0 will help you remember which way to shift the graph. Table 2.15(a) and Figure 2.80 illustrate these ideas.

  2. Each point (x, x2)(x, x2) on the graph of f has a corresponding point (x−3, x2)(x−3, x2) on the graph of h because h(x−3)=(x−3+3)2=x2.h(x−3)=(x−3+3)2=x2. So the graph of h(x)=(x+3)2h(x)=(x+3)2 is just the graph of f(x)=x2f(x)=x2 shifted 3 units to the left. Noticing that f(0)=0f(0)=0 and h(−3)=0h(−3)=0 will help you remember which way to shift the graph. Table 2.15(b) and Figure 2.80 confirm these considerations.

Practice Problem 2

  1. Let

    f(x)=x3, g(x)=(x−1)3, and h(x)=(x+2)3.
    f(x)=x3, g(x)=(x−1)3, and h(x)=(x+2)3.

    Sketch the graphs of all three functions on the same coordinate plane. Describe how the graphs of g and h relate to the graph of f.

Practice Problem 3

  1. Sketch the graph of

    f(x)=√x−2+3.
    f(x)=x−2−−−−−√+3.

Reflections

  1. 3 Use reflections to graph functions.

We now consider rigid transformations that reflect a graph about a coordinate axis.

Comparing the Graphs of y=f(x)y=f(x) and y=−f(x)y=−f(x)

Consider the graph of f(x)=x2,f(x)=x2, shown in Figure 2.82. Table 2.16 gives some values for f(x) and g(x)=−f(x).g(x)=−f(x). Note that the y-coordinate of each point in the graph of g(x)=−f(x)g(x)=−f(x) is the opposite of the y-coordinate of the corresponding point on the graph of f(x). So the graph of y=−x2y=−x2 is the reflection of the graph of y=x2y=x2 about the x-axis. This means that the points (x, x2)(x, x2) and (x,−x2)(x,−x2) are the same distance from, but on opposite sides of, the x-axis. See Figure 2.82.

Figure 2.82 Reflection about the x-axis

TABLE 2.16

x f(x)=x2f(x)=x2 g(x)=−f(x)=−x2g(x)=−f(x)=−x2
−3−3 9 −9−9
−2−2 4 −4−4
−1−1 1 −1−1
0 0 0
1 1 −1−1
2 4 −4−4
3 9 −9−9

Comparing the Graphs of y=f(x)y=f(x) and y=f(−x)y=f(−x)

To compare the graphs of f(x) and g(x)=f(−x),g(x)=f(−x), consider the graph of f(x)=√x.f(x)=x−−√. Then f(−x)=√−x.f(−x)=−x−−−√. See Figure 2.84. Table 2.17 gives some values for f(x) and g(x)=f(−x).g(x)=f(−x). The domain of f is [0, ∞),[0, ∞), and the domain of g is (−∞, 0].(−∞, 0]. Each point (x, y) on the graph of f has a corresponding point (−x, y)(−x, y) on the graph of g. So the graph of y=f(−x)y=f(−x) is the reflection of the graph of y=f(x)y=f(x) about the y-axis. This means that the points (x, √x)(x, x−−√) and (−x, √x)(−x, x−−√) are the same distance from, but on opposite sides of, the y-axis. See Figure 2.84.

Figure 2.84 Graphing y=√−xy=−x−−−√

TABLE 2.17

x f(x)=√xf(x)=x−−√ −x−x g(x)=√−xg(x)=−x−−−√
−4−4 Undefined 4 2
−1−1 Undefined 1 1
0 0 0 0
1 1 −1−1 Undefined
4 2 −4−4 Undefined

Example 4 Combining Transformations

Explain how the graph of y=−|x−2|+3y=−|x−2|+3 can be obtained from the graph of y=|x|.y=|x|.

Solution

Start with the graph of y=|x|.y=|x|. Follow the point (0, 0) on y=|x|.y=|x|. See Figure 2.86(a).

Figure 2.86 Transformations of y=|x|y=|x|

  1. Step 1 Shift the graph of y=|x|y=|x| to the right 2 units to obtain the graph of y=|x−2|.y=|x−2|. Add 2 to the x-coordinate of each point. The point (0, 0) moves to (2, 0). See Figure 2.86(b) .

  2. Step 2 Reflect the graph of y=|x−2|y=|x−2| about the x-axis to obtain the graph of y=−|x−2|.y=−|x−2|. Change the y-coordinate of each point to its opposite. The point (2, 0) remains (2, 0). See Figure 2.86(c) .

  3. Step 3 Finally, shift the graph of y=−|x−2|y=−|x−2| up 3 units to obtain the graph of y=−|x−2|+3.y=−|x−2|+3. Add 3 to the y-coordinate of each point. The point (2, 0) moves to (2, 3). See Figure 2.86(d) .

Practice Problem 4

  1. Explain how the graph of y=−(x−1)2+2y=−(x−1)2+2 can be obtained from the graph of y=x2.y=x2.

Example 5 Graphing y=|f(x)|y=|f(x)|

Use the graph of f(x)=(x+1)2−4f(x)=(x+1)2−4 to sketch the graph of y=|f(x)|.y=|f(x)|.

Solution

The graph of f(x)=(x+1)2−4f(x)=(x+1)2−4 is obtained by shifting the graph of y=x2y=x2 left by 1 unit and then shifting the resulting graph down by 4 units. See Figure 2.87(a). The function f has domain (−∞, ∞)(−∞, ∞) and range [−4, ∞).[−4, ∞).

Figure 2.87 Graphing y=|f(x)|y=|f(x)|

We know that

|y|={yif y≥0−yif y<0.
|y|={y−yif y≥0if y<0.

This means that the portion of the graph on or above the x-axis (y≥0)(y≥0) is unchanged and the portion of the graph below the x-axis (y<0)(y<0) is reflected above the x-axis. See Figure 2.87(b). The function | f|| f| has domain (−∞, ∞)(−∞, ∞) and range [0, ∞).[0, ∞).

Practice Problem 5

  1. Use the graph of f(x)=2x−4f(x)=2x−4 to sketch the graph of y=|f(x)|.y=|f(x)|.

Stretching or Compressing

  1. 4 Use stretching or compressing to graph functions.

We now look at transformations that distort the shape of a graph, called nonrigid transformations. We consider the relationship of the graphs of y=af(x)y=af(x) and y=f(bx)y=f(bx) to the graph of y=f(x).y=f(x).

Comparing the Graphs of y=f(x)y=f(x) and y=af(x)y=af(x)

Example 6 Stretching or Compressing a Graph Vertically

Let f(x)=|x|, g(x)=2|x|,f(x)=|x|, g(x)=2|x|, and h(x)=12|x|.h(x)=12|x|. Sketch the graphs of f, g, and h on the same coordinate plane and describe how the graphs of g and h are related to the graph of f.

Solution

The graphs of y=|x|, y=2|x|,y=|x|, y=2|x|, and y=12|x|y=12|x| are sketched in Figure 2.88. Table 2.18 gives some typical function values.

Figure 2.88

Vertical stretch and compression

TABLE 2.18

x f(x)=|x|f(x)=|x| g(x)=2|x|g(x)=2|x| h(x)=12|x|h(x)=12|x|
−2−2 2 4 1
−1−1 1 2 1212
0 0 0 0
1 1 2 1212
2 2 4 1

The graph of y=2|x|y=2|x| is the graph of y=|x|y=|x| vertically stretched (expanded) by multiplying each of its y-coordinates by 2. It is twice as high as the graph of |x||x| at every real number x. The result is a taller V-shaped curve. See the red graph in Figure 2.88.

The graph y=12|x|y=12|x| is the graph of y=|x|y=|x| vertically compressed (shrunk) by multiplying each of its y-coordinates by 12.12. It is half as high as the graph of |x||x| at every real number x. The result is a flatter V-shaped curve. See the green graph in Figure 2.88.

Practice Problem 6

  1. Let f(x)=√xf(x)=x−−√ and g(x)=2√x.g(x)=2x−−√. Sketch the graphs of f and g on the same coordinate plane and describe how the graph of g is related to the graph of f.

Comparing the Graphs of y=f(x)y=f(x) and y=f(bx)y=f(bx)

Given a function y=f(x),y=f(x), we explore the effect of the constant b in graphing the function y=f(bx).y=f(bx). Consider the graphs of y=f(x)y=f(x) and y=f(2x)y=f(2x) in Figure 2.90(a). Multiplying the independent variable x by 2 compresses the graph y=f(x)y=f(x) horizontally toward the y-axis. Because the value 2x is twice the value of x, a point on the x-axis will be only half as far from the origin when y=f(2x)y=f(2x) has the same y value as y=f(x).y=f(x). See Figure 2.90(a).

Figure 2.90

Horizontal stretch or compression

Side Note

Notice that replacing the variable x with x±c,x±c, or bx in y=f(x)y=f(x) results in a horizontal change in the graphs. However, replacing the y value in y=f(x)y=f(x) by f(x)±d,f(x)±d, or af(x), results in a vertical change in the graph of y=f(x)y=f(x).

Now consider the graphs of y=f(x)y=f(x) and y=f(12x)y=f(12x) in Figure 2.90(b). Multiplying the independent variable x by 1212 stretches the graph of y=f(x)y=f(x) horizontally away from the y-axis. The value 12x12x is half the value of x so that a point on the x-axis will be twice as far from the origin for y=f(12x)y=f(12x) to have the same y-value as in y=f(x).y=f(x). See Figure 2.90(b).

Example 7 Stretching or Compressing a Function Horizontally

The graph of the function y=f(x)y=f(x) whose equation is not given is shown in Figure 2.92. Sketch the following graphs.

Figure 2.92

  1. f(12x)f(12x)

  2. f(2x)

  3. f(−2x)f(−2x)

Solution

Note that the domain of f is [−2, 2][−2, 2] and its range is [−1, 3][−1, 3].

  1. To graph y=f(12x),y=f(12x), we stretch the graph of y=f(x)y=f(x) horizontally by a factor of 2. In other words, we transform each point (x, y) in Figure 2.92 to the point (2x, y). The x-intercept 2 changes to 4 but the y-intercept is the same. See Figure 2.93(a).

    Figure 2.93

  2. To graph y=f(2x),y=f(2x), we compress the graph of y=f(x)y=f(x) horizontally by a factor of 12.12.

    Therefore, we transform each point (x, y) in Figure 2.92 to (12x, y)(12x, y). The x-intercept 2 changes to 1, but the y-intercept is the same. See Figure 2.93(b).

  3. To graph y=f(−2x),y=f(−2x), we reflect the graph of y=f(2x)y=f(2x) in Figure 2.93(b) about the y-axis. So we transform each point (x, y) in Figure 2.93(b) to the point (−x, y)(−x, y). The x-intercept 1 changes to −1−1, but the y-intercept is the same. See Figure 2.93(c).

Practice Problem 7

  1. The graph of a function y=f(x)y=f(x) is given in the margin. Sketch the graphs of the following functions.

    1. f(12x)f(12x)

    2. f(2x)

Multiple Transformations in Sequence

  1. 5 Use a sequence of transformations to graph functions.

When graphing requires more than one transformation of a basic function, we perform transformations in the following order:

  1. Horizontal shift

  2. Stretch or compression

  3. Reflection

  4. Vertical shift

To graph y=af(bx−c)+dy=af(bx−c)+d from the graph of y=f(x),y=f(x), here is one possible order of transformations:

Graph Change each graph point (x, y) on the graph in the previous step to
Step 0 Original graph y=f(x)y=f(x) (x, y)
Step 1 Horizontal Shift y=f(x−c)y=f(x−c) (x+c, y)(x+c, y)
Step 2 Stretch/Compression

  1. y=f(|b|x−c)y=f(|b|x−c)

  2.  y=|a| f(|b|x−c) y=|a| f(|b|x−c)

(x|b|, y)(x|b|, y)

(x, |a|y)(x, |a|y)
Step 3 Reflections

  1. y=|a| f(bx−c)y=|a| f(bx−c) for b<0b<0

  2. y=af(bx−c)y=af(bx−c) for a<0a<0

(−x, y)(−x, y)

(x,−y)(x,−y)
Step 4 Vertical shift y=af(bx−c)+dy=af(bx−c)+d (x, y+d)(x, y+d)

Example 8 Using Multiple Transformations to Graph a Function

Sketch the graph of the function f(x)=−2(x−1)2+3.f(x)=−2(x−1)2+3.

Solution

Begin with the graph of the basic function g(x)=x2.g(x)=x2. Then apply the necessary transformations in a sequence of steps. The result of each step is shown in Figure 2.94. We follow the point (2, 4) in Figure 2.94 on page 273 from the graph of y=x2y=x2.

Figure 2.94

Multiple transformations

  1. Step 0

    y=x2Identify a related function whose graph is familiar.In this case, use y=x2. See Figure 2.94(a).
    y=x2Identify a related function whose graph is familiar.In this case, use y=x2. See Figure 2.94(a).

  2. Step 1

    y=(x−1)2Replace x with x−1; shift the graph of y=x2,1 unit to the right. See Figure 2.94(b).
    y=(x−1)2Replace x with x−1; shift the graph of y=x2,1 unit to the right. See Figure 2.94(b).

  3. Step 2

    y=2(x−1)2Multiply by 2; stretch the graph of y=(x−1)2vertically by a factor of 2. See Figure 2.94(c).
    y=2(x−1)2Multiply by 2; stretch the graph of y=(x−1)2vertically by a factor of 2. See Figure 2.94(c).

  4. Step 3

    y=−2(x−1)2Multiply by −1. Reflect the graph of y=2(x−1)2about the x-axis. See Figure 2.94(d).
    y=−2(x−1)2Multiply by −1. Reflect the graph of y=2(x−1)2about the x-axis. See Figure 2.94(d).

  5. Step 4

    y=−2(x−1)2+3Add 3. Shift the graph of y=−2(x−1)2,up 3 units. See Figure 2.94(e).
    y=−2(x−1)2+3Add 3. Shift the graph of y=−2(x−1)2,up 3 units. See Figure 2.94(e).

Practice Problem 8

  1. Sketch the graph of the function

    f(x)=3√x+1−2.
    f(x)=3x+1−−−−−√−2.

Example 9 Using a Sequence of Transformations

Use the graph of y=f(x)y=f(x) in Figure 2.95(i) to graph y=−13f(4−2x)+5.y=−13f(4−2x)+5.

Solution

Figure 2.95 Graph of y=−13f(4−2x)+5y=−13f(4−2x)+5

Practice Problem 9

  1. Use the graph of y=f(x)y=f(x) in Figure 2.96 to graph y=12f(2x−1)+3.y=12f(2x−1)+3.

    Figure 2.96

Section 2.7 Exercises

Concepts and Vocabulary

  1. The graph of y=f(x)−3y=f(x)−3 is found by vertically shifting the graph of y=f(x)y=f(x), 3 units                           .

  2. The graph of y=f(x+5)y=f(x+5) is found by horizontally shifting the graph of y=f(x)y=f(x), 5 units to the                           .

  3. The graph of y=f(bx)y=f(bx) is a horizontal compression of the graph of y=f(x)y=f(x) if b                           .

  4. The graph of y=f(−x)y=f(−x) is found by reflecting the graph of y=f(x)y=f(x) about the                           .

  5. True or False. The graph of y=f(x)y=f(x) and y=f(−x)y=f(−x) cannot be the same.

  6. True or False. The graphs of y=f(x)y=f(x) and y=f(x)+1y=f(x)+1 cannot be the same.

  7. True or False. You get the same graph by shifting the graph of y=x2y=x2, 2 units up, reflecting the shifted graph about the x-axis or reflecting the graph of y=x2y=x2 about the x-axis, and then shifting the reflected graph up 2 units.

  8. True or False. Combining horizontal and vertical shifts of a graph preserves the shape of the original graph.

Building Skills

In Exercises 9–22, describe the transformations that produce the graphs of g and h from the graph of f.

  1. f(x)=√xf(x)=x−−√

    1. g(x)=√x+2g(x)=x−−√+2

    2. h(x)=√x−1h(x)=x−−√−1

  2. f(x)=|x|f(x)=|x|

    1. g(x)=|x|+1g(x)=|x|+1

    2. h(x)=|x|−2h(x)=|x|−2

  3. f(x)=x2f(x)=x2

    1. g(x)=(x+1)2g(x)=(x+1)2

    2. h(x)=(x−2)2h(x)=(x−2)2

  4. f(x)=1xf(x)=1x

    1. g(x)=1x+2g(x)=1x+2

    2. h(x)=1x−3h(x)=1x−3

  5. f(x)=√xf(x)=x−−√

    1. g(x)=√x+1−2g(x)=x+1−−−−−√−2

    2. h(x)=√x−1+3h(x)=x−1−−−−−√+3

  6. f(x)=x2f(x)=x2

    1. g(x)=−x2g(x)=−x2

    2. h(x)=(−x)2h(x)=(−x)2

  7. f(x)=|x|f(x)=|x|

    1. g(x)=−|x|g(x)=−|x|

    2. h(x)=|−x|h(x)=|−x|

  8. f(x)=√xf(x)=x−−√

    1. g(x)=2√xg(x)=2x−−√

    2. h(x)=√2xh(x)=2x−−√

  9. f(x)=1xf(x)=1x

    1. g(x)=2xg(x)=2x

    2. h(x)=12xh(x)=12x

  10. f(x)=x3f(x)=x3

    1. g(x)=(x−2)3+1g(x)=(x−2)3+1

    2. h(x)=−(x+1)3+2h(x)=−(x+1)3+2

  11. f(x)=√xf(x)=x−−√

    1. g(x)=−√x+1g(x)=−x−−√+1

    2. h(x)=√−x+1h(x)=−x−−−√+1

  12. f(x)=〚x〛f(x)=〚x〛

    1. g(x)=〚x−1〛+2g(x)=〚x−1〛+2

    2. h(x)=3〚x〛−1h(x)=3〚x〛−1

  13. f(x)=3√xf(x)=x−−√3

    1. g(x)=3√x+1g(x)=x−−√3+1

    2. h(x)=3√x+1h(x)=x+1−−−−−√3

  14. f(x)=3√xf(x)=x−−√3

    1. g(x)=23√1−x+4g(x)=21−x−−−−−√3+4

    2. h(x)=−3√x−1+3h(x)=−x−1−−−−−√3+3

In Exercises 23–34, match each function with its graph (a)–(l).

  1. y=−|x|+1y=−|x|+1

  2. y=−√−xy=−−x−−−√

  3. y=√x2y=x2−−√

  4. y=12|x|y=12|x|

  5. y=√x+1y=x+1−−−−−√

  6. y=2|x|−3y=2|x|−3

  7. y=1−2√xy=1−2x−−√

  8. y=−|x−1|+1y=−|x−1|+1

  9. y=(x−1)2y=(x−1)2

  10. y=−x2+3y=−x2+3

  11. y=−2(x−3)2−1y=−2(x−3)2−1

  12. y=3−√1−xy=3−1−x−−−−−√

In Exercises 35–62, graph each function by starting with a function from the library of functions and then using the techniques of shifting, compressing, stretching, and/or reflecting.

  1. f(x)=x2−2f(x)=x2−2

  2. f(x)=x2+3f(x)=x2+3

  3. g(x)=√x+1g(x)=x−−√+1

  4. g(x)=√x−4g(x)=x−−√−4

  5. f(x)=|x|+2f(x)=|x|+2

  6. f(x)=|x|−1f(x)=|x|−1

  7. f(x)=x3+2f(x)=x3+2

  8. f(x)=x3−1f(x)=x3−1

  9. f(x)=1x+1f(x)=1x+1

  10. f(x)=1x−2f(x)=1x−2

  11. f(x)=(x−3)3f(x)=(x−3)3

  12. f(x)=(x+2)3f(x)=(x+2)3

  13. f(x)=√x−1f(x)=x−1−−−−−√

  14. f(x)=√x+2f(x)=x+2−−−−−√

  15. h(x)=|x+1|h(x)=|x+1|

  16. h(x)=|x−2|h(x)=|x−2|

  17. f(x)=(x+1)3f(x)=(x+1)3

  18. f(x)=(x−3)3f(x)=(x−3)3

  19. f(x)=1x−3f(x)=1x−3

  20. f(x)=1x+2f(x)=1x+2

  21. f(x)=√−xf(x)=−x−−−√

  22. f(x)=−√xf(x)=−x−−√

  23. f(x)=−x2f(x)=−x2

  24. f(x)=−x3f(x)=−x3

  25. f(x)=2x2f(x)=2x2

  26. f(x)=13x2f(x)=13x2

  27. f(x)=2|x|f(x)=2|x|

  28. f(x)=13|x|f(x)=13|x|

In Exercises 63–74, graph each function by starting with a function from the library of functions and then combining shifting and reflecting techniques.

  1. f(x)=(x−2)2+1f(x)=(x−2)2+1

  2. f(x)=(x−3)2−5f(x)=(x−3)2−5

  3. f(x)=5−(x−3)2f(x)=5−(x−3)2

  4. f(x)=2−(x+1)2f(x)=2−(x+1)2

  5. f(x)=√x+1−3f(x)=x+1−−−−−√−3

  6. f(x)=√x−2+1f(x)=x−2−−−−−√+1

  7. f(x)=√1−x+2f(x)=1−x−−−−−√+2

  8. f(x)=−√x+2−3f(x)=−x+2−−−−−√−3

  9. f(x)=|x−1|−2f(x)=|x−1|−2

  10. f(x)=−|x+3|+1f(x)=−|x+3|+1

  11. f(x)=1x−1+3f(x)=1x−1+3

  12. f(x)=2−1x+2f(x)=2−1x+2

In Exercises 75–82, graph each function by starting with a function from the library of functions and then combining shifting, compressing, stretching, and/or reflecting techniques.

  1. f(x)=2(x+1)2−1f(x)=2(x+1)2−1

  2. f(x)=13(x+1)2+2f(x)=13(x+1)2+2

  3. f(x)=2−12(x−3)2f(x)=2−12(x−3)2

  4. f(x)=1−3(x−3)2f(x)=1−3(x−3)2

  5. f(x)=2√x+1−3f(x)=2x+1−−−−−√−3

  6. f(x)=√2x−2+1f(x)=2x−2−−−−−√+1

  7. f(x)=−2|x−1|+2f(x)=−2|x−1|+2

  8. f(x)=−12|3−x|−1f(x)=−12|3−x|−1

In Exercises 83–94, write an equation for a function whose graph fits the given description.

  1. The graph of f(x)=x3f(x)=x3 is shifted 2 units up.

  2. The graph of f(x)=√xf(x)=x−−√ is shifted 3 units left.

  3. The graph of f(x)=|x|f(x)=|x| is reflected about the x-axis.

  4. f(x)=√xf(x)=x−−√ is reflected about the y-axis.

  5. The graph of f(x)=x2f(x)=x2 is shifted 3 units right and then 2 units up.

  6. The graph of f(x)=x2f(x)=x2 is shifted 2 units left and then reflected about the x-axis.

  7. The graph of f(x)=√xf(x)=x−−√ is shifted 3 units left, reflected about the x-axis, and then shifted 2 units down.

  8. The graph of f(x)=√xf(x)=x−−√ is shifted 2 units down, reflected about the x-axis, and then compressed vertically by a factor of 12.12.

  9. The graph of f(x)=x3f(x)=x3 is shifted 4 units left, stretched vertically by a factor of 3, reflected about the y-axis, and then shifted 2 units up.

  10. The graph of f(x)=x3f(x)=x3 is reflected about the x-axis, shifted 1 unit up, shifted 1 unit left, and then reflected about the y-axis.

  11. The graph of f(x)=|x|f(x)=|x| is shifted 4 units right, compressed horizontally by a factor of 2, reflected about the x-axis, and then shifted 3 units down.

  12. The graph of f(x)=|x|f(x)=|x| is shifted 2 units right, reflected about the y-axis, stretched horizontally by a factor of 2 and then shifted 3 units down.

In Exercises 95–108, graph the function y=g(x)y=g(x), given the following graph of y=f(x).y=f(x).

  1. g(x)=f(x)−1g(x)=f(x)−1

  2. g(x)=f(x)+3g(x)=f(x)+3

  3. g(x)=−f(x)g(x)=−f(x)

  4. g(x)=f(−x)g(x)=f(−x)

  5. g(x)=f(2x)

  6. g(x)=f(12x)

  7. g(x)=f(x+1)

  8. g(x)=f(x−2)

  9. g(x)=2f(x)

  10. g(x)=12f(x)

  11. g(x)=f(x−1)+2

  12. g(x)=−f(−2x)−1

  13. g(x)=f(1−2x)

  14. g(x)=f(3−2x)

In Exercises 109–118, graph (a) y=g(x) and (b) y=|g(x)|, given the following graph of y=f(x).

  1. g(x)=f(x)+1

  2. g(x)=−2f(x)

  3. g(x)=f(12x)

  4. g(x)=f(−2x)

  5. g(x)=f(x−1)

  6. g(x)=f(2−x)

  7. g(x)=−2 f(x+1)+3

  8. g(x)=−f(−x+1)−2

  9. g(x)=2f(1−2x)−3

  10. g(x)=−f(−2−2x)+1

Applying the Concepts

In Exercises 119–128, let f be the function that associates the employee number x of each employee of the ABC Corporation with his or her annual salary f(x) in dollars.

  1. Across-the-board raise. Each employee was awarded an across-the-board raise of $800 per year. Write a function g(x) to describe the new salary.

  2. Percentage raise. Suppose each employee was awarded a 5% raise. Write a function h(x) to describe the new salary.

  3. Across-the-board and percentage raises. Suppose each employee was awarded a $500 across-the-board raise and an additional 2% of his or her increased salary. Write a function p(x) to describe the new salary.

  4. Across-the-board percentage raise. Suppose the employees making $30,000 or more received a 2% raise, while those making less than $30,000 received a 10% raise. Write a piecewise function to describe these new salaries.

  5. Health plan. The ABC Corporation pays for its employees’ health insurance at an annual cost (in dollars) given by

    C(x)=5,000+10√x−1,

    where x is the number of employees covered.

    1. Use transformations of the graph of y=√x to sketch the graph of y=C(x).

    2. Assuming that the company has 400 employees, find its annual outlay for the health coverage.

  6. Poiseuille’s Law. For an artery of radius R, the velocity v (in mm/min) of blood flow at a distance r from the center of the artery is given by

    v=c(R2−r2), 0≤r≤R,

    where c is a constant.

    In an artery for Angie, c=1000 and R=3 mm. Find the velocity of blood flow in this artery at

    1. The center of the artery.

    2. The inner linings of the artery.

    3. Midway between the center and the inner linings.

  7. Demand. The weekly demand for paper hats produced by Mythical Manufacturers is given by

    x(p)=109,561−(p+1)2,

    where x represents the number of hats that can be sold at a price of p cents each.

    1. Use transformations on the graph of y=p2 to sketch the graph of y=x(p).

    2. Find the price at which 69,160 hats can be sold.

    3. Find the price at which no hats can be sold.

  8. Revenue. The weekly demand for cashmere sweaters produced by Wool Shop, Inc., is x(p)=−3p+600. The revenue is given by R(p)=−3p2+600p. Describe how to sketch the graph of y=R(p) by applying transformations to the graph of y=p2. [Hint: First, write R(p) in the form −3(p−h)2+k.]

  9. Daylight. At 60° north latitude, the graph of y=f(t) gives the number of hours of daylight. (On the t-axis, note that 1=January and 12=December.)

    Sketch the graph of y=f(t)−12.

  10. Use the graph of y=f(t) of Exercise 127 to sketch the graph of y=24−f(t). Interpret the result.

Beyond the Basics

In Exercises 129 and 130, the graphs of y=f(x) and y=g(x) are given. Find an equation for g(x) if the graph of g is obtained from the graph of f by a sequence of transformations.

In Exercises 131–138, by completing the square on each quadratic expression, use transformations on y=x2 to sketch the graph of y=f(x).

  1. f(x)=x2+4x

    [Hint: x2+4x=(x2+4x+4)−4=(x+2)2−4.]

  2. f(x)=x2−6x

  3. f(x)=−x2+2x

  4. f(x)=−x2−2x

  5. f(x)=2x2−4x

  6. f(x)=2x2+6x+3.5

  7. f(x)=−2x2−8x+3

  8. f(x)=−2x2+2x−1

In Exercises 139–144, sketch the graph of each function.

  1. y=|2x+3|

  2. y=|〚x〛|

  3. y=|4−x2|

  4. y=|1x|

  5. y=√|x|

  6. y=〚|x|〛

Critical Thinking / Discussion / Writing

  1. The x-intercepts of the function f are −1, 0, 2. Find the corresponding x-intercepts for the following functions.

    1. y=f(x+2)

    2. y=f(x−2)

    3. y=−f(x)

    4. y=f(−x)

    5. y=f(2x)

    6. y=f(12x)

  2. The y-intercept of the function f is 2. Find the corresponding y-intercepts for the following functions.

    1. y=f(x)+2

    2. y=f(x)−2

    3. y=−f(x)

    4. y=f(−x)

    5. y=2f(x)

    6. y=12f(x)

  3. Let a function f have domain [−1, 3] and range [−2, 1]. Find the corresponding domain and range for the following functions.

    1. y=f(x+2)

    2. y=f(x)−2

    3. y=−f(x)

    4. y=f(−x)

    5. y=2f(x)

    6. y=12 f(x)

  4. Let a function f have a relative maximum at x=1 and a relative minimum at x=2. Find the corresponding relative maxima and minima for the following functions.

    1. y=f(x+2)

    2. y=f(x)−2

    3. y=−f(x)

    4. y=f(−x)

    5. y=2f(x)

    6. y=12f(x)

Getting Ready for the Next Section

In Exercises 149–154, perform the indicated operations.

  1. (5x2+5x+7)+(x2+9x−4)

  2. (x2+2x)+(6x3−2x+5)

  3. (5x2+6x−2)−(3x2−9x+1)

  4. (x3+2)−(2x3+5x−3)

  5. (x−2)(x2+2x+4)

  6. (x2+x+1)(x2−x+1)

In Exercises 155–158, find the domain of each function.

  1. f(x)=2x−3x2−5x+6

  2. f(x)=x−2x2−4

  3. f(x)=√2x−3

  4. f(x)=1√5−2x

In Exercises 159–162, solve the given inequality.

  1. x−1x−10<0

  2. −3(1−x)2>0

  3. −2x+8x2+1≤0

  4. (x−3)(x−1)(x−5)(x+1)≥0

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