Definition: Translation to the Right or Left
If h7 0, then the graph of y= ƒ 1x - h2 is a translation of h units to the right of the graph of y =ƒ 1x2. If h 60, then the graph of y = ƒ 1x- h2 is a translation of 0h0 units to the left of the graph of y = ƒ 1x2.
䉴TRY THIS. Graph ƒ 1x2 = 1x, g1x2 = 1x -2, and h1x2 = 1x + 1 on the
same coordinate plane. ■
Notice that y = 1x - 3 lies 3 units to the right and y= 1x + 5 lies 5 units to the left of y = 1x. The next example shows two more horizontal translations.
EXAMPLE 1 Translations to the right or left
Graph ƒ1x2 = 1x, g1x2= 1x - 3, and h1x2 = 1x + 5 on the same coordinate plane.
Solution
First sketch ƒ 1x2 = 1x through 10, 02, 11, 12, and 14, 22 as shown in Fig. 2.32.
Since the first operation of g is to subtract 3, we get the corresponding points by adding 3 to each x-coordinate. So g goes through 13, 02, 14, 12, and 17, 22. Since the first operation of h is to add 5, we get corresponding points by subtracting 5 from the x-coordinates. So h goes through 1-5, 02, 1-4, 12, and 1-1, 22.
The calculator graphs of ƒ, g, and h are shown in Fig. 2.33. On many calcula- tors the radical does not contain the radicand. Be sure to note the difference between
y= 1 1x2 -3 and y = 1 1x- 32 on a calculator.
Figure 2.32 y
– 1 x – 3 –4 – 5
–1 – 2 1
2
2 3 4 5 6
– 5 + 3
h(x) = √x + 5 f(x) = √x g(x) = √x – 3
Figure 2.33
⫺5 12
⫺1 3
EXAMPLE 2 Horizontal translations
Sketch the graph of each function.
a. ƒ 1x2 = 0x - 10 b. ƒ 1x2= 1x + 322 Solution
a. The function ƒ 1x2 = 0x - 10 is in the absolute value family, and its graph is a translation one unit to the right of g1x2= 0x0. Calculate a few ordered pairs to get an accurate graph. The points 10, 12, 11, 02, and 12, 12 are on the graph of ƒ 1x2 = 0x - 10 shown in Fig. 2.34.
b. The function ƒ 1x2= 1x + 322 is in the square family, and its graph is a transla- tion three units to the left of the graph of g1x2 =x2. Calculate a few ordered pairs to get an accurate graph. The points 1-3, 02, 1-2, 12, and 1-4, 12 are on the graph shown in Fig. 2.35.
Figure 2.35 (– 2, 1)
(– 4, 1) (– 3, 0)
y
–1 x – 4
– 1
1 2
2 1 4 3
–5
f(x) = (x + 3)2 g(x) = x2
Figure 2.34 (2, 1) (0, 1)
(1, 0) y
– 1 x –3
– 1
2 3
– 2 2 1
4 3
g(x) = ⏐x⏐ f(x) = ⏐x – 1⏐
䉴TRY THIS. Graph ƒ 1x2 = 1x - 222. ■
Reflection
In the formula y= aƒ 1x- h2 +k, we multiply ƒ 1x - h2 by a before adding k. So next we study the effect that a has on the graph. We first consider the case where a = -1. Multiplying by -1 is the same as taking the opposite, so we usually just write y = -ƒ 1x2. All points of the graph of y = -ƒ 1x2 can be obtained by simply changing the signs of all of the y-coordinates of the points on the graph of y= ƒ 1x2.
This causes the graphs to be mirror images of each other with respect to the x-axis.
Definition: Reflection The graph of y= -ƒ1x2 is a reflection in the x-axis of the graph of y =ƒ1x2.
EXAMPLE 3 Graphing using reflection
Graph each pair of functions on the same coordinate plane.
a. ƒ 1x2 =x2, g1x2= -x2 b. ƒ 1x2 =x3, g1x2= -x3 c. ƒ 1x2 = 0x0, g1x2= -0x0 Solution
a. The graph of ƒ 1x2= x2 goes through 10, 02, 1{1, 12, and 1{2, 42. The graph of g1x2= -x2 goes through 10, 02, 1{1, -12, and 1{2, -42 as shown in Fig. 2.36.
b. Make a table of ordered pairs for ƒ as follows:
x -2 -1 0 1 2
ƒ 1x2= x3 -8 -1 0 1 8
Sketch the graph of ƒ through these ordered pairs as shown in Fig. 2.37. Since g1x2 = -ƒ 1x2, the graph of g can be obtained by reflecting the graph of ƒ in the x-axis. Each point on the graph of ƒ corresponds to a point on the graph of g with the opposite y-coordinate. For example, the point 12, 82 on the graph of ƒ corre- sponds to the point 12, -82 on the graph of g. Both graphs are shown in Fig. 2.37.
c. The graph of ƒ is the familiar V-shaped graph of the absolute value function as shown in Fig. 2.38. Since g1x2= -ƒ 1x2, the graph of g can be obtained by re- flecting the graph of ƒ in the x-axis. Each point on the graph of ƒ corresponds to a point on the graph of g with the opposite y-coordinate. For example, 12, 22 on ƒ corresponds to 12, -22 on g. Both graphs are shown in Fig. 2.38.
Figure 2.36 (– 2, 4)
(– 2, – 4) y
–1 x – 3
–1 –2 –3 –4
1 2 3
2 1 4 3
g(x) = –x2 f(x) = x2
Figure 2.37 (2, 8)
(2, – 8) g(x) = –x3
f(x) = x3 y
–2 x –6
– 2 – 4
4 6
4 2 8 6
– 6 – 8 –4
(2, 2)
(2, – 2) f(x) = ⏐x⏐ y
– 1 x – 3
–1 –2 –3
1 3
– 2 2 1 3
g (x) = –⏐x⏐ Figure 2.38
䉴TRY THIS. Graph ƒ1x2=1x and g1x2= -1x on the same coordinate plane. ■ We do not include reflections in the y-axis in our function families, but it is interesting to note that you get a reflection in the y-axis simply by replacing x with -x in the formula for the function. For example, graph y1= 1x and y2= 1-x on your graphing calculator.
Stretching and Shrinking
If a7 1 then all of the y-coordinates on y = aƒ 1x2 are obtained by multiplying the y-coordinates on the graph of y= ƒ 1x2 by a. So the graph of y = ƒ 1x2 is stretched by a factor of a. If 0 6 a6 1 then the graph of y= ƒ 1x2 is shrunk by a factor of a.
If a is negative, then reflection occurs along with the stretching or shrinking.
Definitions:
Stretching and Shrinking
The graph of y = aƒ1x2 is obtained from the graph of y = ƒ1x2 by 1. stretching the graph of y = ƒ1x2 by a when a7 1, or 2. shrinking the graph of y = ƒ1x2 by a when 06 a6 1.
EXAMPLE 4 Graphing using stretching and shrinking
In each case graph the three functions on the same coordinate plane.
a. ƒ 1x2 = 1x, g1x2 = 21x, h1x2= 1 21x b. ƒ 1x2 =x2, g1x2= 2x2, h1x2= 1
2 x2 Solution
a. The graph of ƒ 1x2 = 1x goes through 10, 02, 11, 12, and 14, 22 as shown in Fig. 2.39. The graph of g is obtained by stretching the graph of ƒ by a factor of 2.
So g goes through 10, 02, 11, 22, and 14, 42. The graph of h is obtained by shrink- ing the graph of ƒ by a factor of 12. So h goes through 10, 02, 11, 122, and 14, 12.
The functions ƒ, g, and h are shown on a graphing calculator in Fig. 2.40.
Note how the viewing window affects the shape of the graph. The curves do not appear as separated on the calculator as they do in Fig. 2.39. 䊐
Figure 2.39 (4, 4)
(4, 2) (4, 1) y
1 2 1 4 3
x
2 3 4
× 2
×1 2–
f(x) = √x h(x) = 1– x
2√ g(x) = 2√x
Figure 2.40
⫺2 4
⫺2 4
Figure 2.41 y
–1 x – 3
–1
1 2 3
–2 2 1 4
4 3
5
h(x) = x1– 2 2 g(x) = 2x2
f(x) = x2
b. The graph of ƒ1x2 = x2 is the familiar parabola shown in Fig. 2.41. We stretch it by a factor of 2 to get the graph of g and shrink it by a factor of 12 to get the graph of h.
䉴TRY THIS. Graph ƒ1x2 = 0x0 , g1x2 = 30x0 , and h1x2 =130x0 on the same coor-
dinate plane. ■
Note that the graph of y = 2x2 has exactly the same shape as the graph of y = x2 if we simply change the scale on the y-axis as in Fig. 2.42.
Figure 2.42 y
–1 x
–2 1 2
2 1 4
3
y = x2
y
–1 x
–2 1 2
4 2 8
6
y = 2x2
Vertical Translation
In the formula y = aƒ 1x- h2 +k, the last operation is addition of k. If we add the constant k to all y-coordinates on the graph of y =ƒ 1x2, the graph will be translated up or down depending on whether k is positive or negative.
Definition: Translation Upward or Downward
If k 7 0, then the graph of y = ƒ 1x2 + k is a translation of k units upward of the graph of y = ƒ 1x2. If k6 0, then the graph of y =ƒ 1x2 + k is a translation of 0k0 units downward of the graph of y = ƒ 1x2.
EXAMPLE 5 Translations upward or downward
Graph the three given functions on the same coordinate plane.
a. ƒ 1x2= 1x, g1x2= 1x + 3, h1x2= 1x - 5 b. ƒ 1x2= x2, g1x2 = x2+ 2, h1x2 = x2 - 3 Solution
a. First sketch ƒ 1x2 = 1x through 10, 02, 11, 12, and 14, 22 as shown in Fig. 2.43.
Since g1x2 = 1x+ 3 we can add 3 to the y-coordinate of each point to get 10, 32, 11, 42, and 14, 52. Sketch g through these points. Every point on ƒ can be moved up 3 units to obtain a corresponding point on g. For h1x2 = 1x- 5, we subtract 5 from the y-coordinates on ƒ to obtain points on h. So h goes through 10, -52, 11, -42, and 14, -32.
Figure 2.43 y
–1 x –3 – 4
–5 – 1
– 2 – 3 – 4
1 2 3
–2 2 1 4 5
5 ( 0, 3)
( 0, 0)
( 0, – 5)
– 5 + 3
h(x) = f(x) = g(x) =
√x x + 3
√
x – 5
√
Figure 2.44
⫺3 5
⫺5 6
Figure 2.45 (–1, – 2)
y
– 3 x
–1
1 2 3
–2 1 4
4 3
5
–2 (–1, 3)
(–1, 1)
h(x) = x2 – 3 f(x) = x2 g(x) = x2 + 2
The relationship between ƒ, g, and h can be seen with a graphing calculator in Fig. 2.44. You should experiment with your graphing calculator to see how a change in the formula changes the graph. 䊐
b. First sketch the familiar graph of ƒ 1x2 = x2 through 1{2, 42, 1{1, 12, and 10, 02 as shown in Fig. 2.45. Since g1x2 =ƒ 1x2 + 2, the graph of g can be obtained by translating the graph of ƒ upward two units. Since h1x2 = ƒ 1x2 -3, the graph of h can be obtained by translating the graph of ƒ downward three units. For ex- ample, the point 1-1, 12 on the graph of ƒ moves up to 1-1, 32 on the graph of g and down to 1-1, -22 on the graph of h as shown in Fig. 2.45.
䉴TRY THIS. Graph ƒ1x2= 1x, g1x2= 1x+1, and h1x2=1x-2 on the same
coordinate plane. ■
Multiple Transformation
Any combination of stretching, shrinking, reflecting, horizontal translation, and vertical translation transforms one function into a new function. If a transformation does not change the shape of a graph, then it is a rigid transformation. If the shape changes, then it is nonrigid. Stretching and shrinking are nonrigid transformations.
Translating (horizontally or vertically) and reflection are rigid transformations.
When graphing a function involving more than one transformation, y = aƒ 1x - h2 + k, apply the transformations in the order that we discussed them:
h@a@k. Remember that this order is simply the order of operations.
P R O C E D U R E
Multiple Transformations
To graph y=aƒ1x-h2+k apply the transformations in the following order:
1. Horizontal translation (h)
2. Reflecting/stretching/shrinking (a) 3. Vertical translation (k).
Note that the order in which you reflect and stretch or reflect and shrink does not matter. It does matter that you do vertical translation last. For example, if y= x2 is reflected in the x-axis and then translated up one unit, the equation for the graph is y = -x2 +1. If y= x2 is translated up one unit and then reflected in the x-axis, the equation for the graph in the final position is y = -1x2 +12 or y = -x2- 1.
Changing the order has resulted in different functions.
EXAMPLE 6 Graphing using several transformations
Use transformations to graph each function.
a. y = -21x- 322+ 4 b. y = 4- 21x +1 Solution
a. This function is in the square family. So the graph of y = x2 is translated 3 units to the right, reflected and stretched by a factor of 2, and finally translated 4 units upward. See Fig. 2.46.
Figure 2.46 x y
1 –1 –2
–3 2 3 4 5
1 –1 –2 –3 –4 –5 –6 2 4 3 y = x2
y = –2(x – 3)2 + 4
y = –2(x – 3)2 y = (x – 3)2
b. Rewrite the function as y = -21x+ 1 +4 and recognize that it is in the square root family. So we start with the graph of y = 1x. The graph of y = 1x +1 is a horizontal translation one unit to the left of y = 1x. Stretch by a factor of 2 to get y =21x + 1. Reflect in the x-axis to get y = -21x+ 1.
Finally, translate 4 units upward to get the graph of y = -21x +1 + 4. All of these graphs are shown in Fig. 2.47.
Figure 2.47 y
– 1 x –2 –3 –4
1 2 3
2 1 4 3
y =
–1
Translate left y
–1 x –2 –3 –4
1 2 3
2 1 4 3
y =
–1
y
–1 x – 2 – 3 – 4
1 2 3
2 1 4 3
– 1 Stretch
y = 2
Reflect y
x – 2
– 3 – 4
1 2 3
2 1 4 3
y = – 2
Translate up y
–1 x – 2 – 3 – 4
1 2 3
2 1 4 3
– 1
y = 4 – 2 x + 1
√
x + 1
√ √x + 1
x + 1
√
√x
䉴TRY THIS. Graph y =4 - 2 0x +10. ■
The Linear Family of Functions
The function ƒ 1x2= x is called the identity function because the coordinates in each ordered pair are identical. Its graph is a line through 10, 02 with slope 1. A member of the linear family, a linear function is a transformation of the identity function: ƒ1x2 = a1x - h2 + k where a ⬆0. Since a, h, and k are real numbers, we can rewrite this form as a multiple of x plus a constant. So a linear function has the form ƒ1x2 = mx + b, with m⬆ 0 (the slope-intercept form). If m = 0, then the function has the form ƒ1x2 = b and it is a constant function.
EXAMPLE 7 Graphing linear functions using transformations
Sketch the graphs of y = x, y= 2x, y= -2x, and y = -2x -3.
Solution
The graph of y= x is a line through 10, 02, 11, 12, and 12, 22. Stretch the graph of y = x by a factor of 2 to get the graph of y = 2x. Reflect in the x-axis to get the graph of y = -2x. Translate downward three units to get the graph of
y= -2x - 3. See Fig. 2.48.
Figure 2.48 y
–1 x –3
– 2 – 3 – 4
1 2 3
–2 2 1 4 3
y = x
Stretch y
– 1 x – 3
–2 –3 –4
1 2 3
– 2 2 1 4 3
y = 2x
Reflect y
–1 x
–3 –1
–2 –3 –4
1 2 3
–2 2 4 3
y = – 2x
Translate down y
–3 x – 1 – 3 – 4
1 2 3
–2 2 1 4 3
y = – 2x – 3
䉴TRY THIS. Graph y = -2x +5. ■
Symmetry
The graph of g1x2 = -x2 is a reflection in the x-axis of the graph of ƒ 1x2 = x2. If the paper were folded along the x-axis, the graphs would coincide. See Fig. 2.49.
The symmetry that we call reflection occurs between two functions, but the graph of ƒ 1x2 =x2 has a symmetry within itself. Points such as 12, 42 and 1-2, 42 are on the graph and are equidistant from the y-axis. Folding the paper along the y-axis brings all such pairs of points together. See Fig. 2.50. The reason for this symmetry about the y-axis is the fact that ƒ 1-x2 = ƒ 1x2 for any real number x. We get the same y-coordinate whether we evaluate the function at a number or at its opposite.
Figure 2.49 y
–3
–2 –4 –6
3 2
6
–2 4
x
g(x) = –x2 f(x) = x2
(2, 4)
2
(2, – 4) (–2, – 4)
(–2, 4)
Figure 2.50 y
–1
–2 1 2
2 6
x f(x) = x2
(2, 4) (–2, 4) 4
(1, 1) (–1, 1)
Consider the graph of ƒ 1x2 = x3 shown in Fig. 2.51. On the graph of ƒ 1x2 = x3 we find pairs of points such as 12, 82 and 1-2, -82. The odd exponent in x3 causes the second coordinate to be negative when the sign of the first coordinate is changed from positive to negative. These points are equidistant from the origin and on oppo- site sides of the origin. So the symmetry of this graph is about the origin. In this case ƒ 1x2 and ƒ 1-x2 are not equal, but ƒ 1-x2 = -ƒ 1x2.
Definition: Symmetric about the y-Axis
If ƒ 1-x2 = ƒ 1x2 for every value of x in the domain of the function ƒ, then ƒ is called an even function and its graph is symmetric about the y-axis.
Figure 2.51 y
–2
–2 –4 –6
1 2
2 6 8
4
–8
x f(x) = x3
(2, 8)
(–2, –8) (0, 0)
Definition: Symmetric about the Origin
If ƒ 1-x2 = -ƒ 1x2 for every value of x in the domain of the function ƒ, then ƒ is called an odd function and its graph is symmetric about the origin.
A graph might look like it is symmetric about the y-axis or the origin, but the only way to be sure is to use the definitions of these terms as shown in the fol- lowing example. Note that an odd power of a negative number is negative and an even power of a negative number is positive. So for any real number x we have 1-x2n= -xn if n is odd and 1-x2n= xn if n is even.
EXAMPLE 8 Determining symmetry in a graph
Discuss the symmetry of the graph of each function.
a. ƒ 1x2= 5x3- x b. ƒ 1x2 = 0x0 + 3 c. ƒ 1x2= x2 - 3x + 6 Solution
a. Replace x by -x in the formula for ƒ 1x2 and simplify:
ƒ 1-x2 = 51-x23 - 1-x2= -5x3 + x
Is ƒ 1-x2 equal to ƒ 1x2 or the opposite of ƒ 1x2? Since -ƒ 1x2 = -5x3+ x, we have ƒ 1-x2 = -ƒ 1x2. So ƒ is an odd function and the graph is symmetric about the origin.
b. Since 0-x0 = 0x0 for any x, we have ƒ 1-x2 = 0-x0 + 3 = 0x0 +3. Because ƒ 1-x2 = ƒ 1x2, the function is even and the graph is symmetric about the y-axis.
c. In this case, ƒ 1-x2 = 1-x22- 31-x2 +6 = x2+ 3x+ 6. So ƒ 1-x2 ⬆ ƒ 1x2, and ƒ 1-x2 ⬆ -ƒ 1x2. This function is neither even nor odd and its graph has neither type of symmetry.
䉴TRY THIS. Discuss the symmetry of the graph of ƒ 1x2 = -2x2+ 5. ■ Do you see why functions symmetric about the y-axis are called even and functions symmetric about the origin are called odd? In general, a function defined by a polyno- mial with even exponents only, such as ƒ 1x2 = x2 or ƒ 1x2 = x6- 5x4+ 2x2 + 3, is symmetric about the y-axis. (The constant term 3 has even degree because 3 =3x0.2 A function with only odd exponents such as ƒ 1x2 =x3 or ƒ 1x2= x5 -6x3 +4x is symmetric about the origin. A function containing both even- and odd-powered terms such as ƒ 1x2 =x2 + 3x has neither symmetry. For other types of functions (such as absolute value) you must examine the function more carefully to determine symmetry.
Note that functions cannot have x-axis symmetry. A graph that is symmetric about the x-axis fails the vertical line test. For example, the graph of x= y2 is symmetric about the x-axis, but the graph fails the vertical line test and y is not a function of x.
Reading Graphs to Solve Inequalities
In Chapter 1 we learned how to use a two-dimensional graph to help us solve an in- equality in one variable. Using transformations we can graph more quickly and use those graphs to solve a wide variety of inequalities in one variable.
EXAMPLE 9 Using a graph to solve an inequality
Solve the inequality 1x- 122- 26 0 by graphing.
Solution
The graph of y =1x - 122 - 2 is obtained by translating the graph of y = x2 one unit to the right and two units downward. See Fig. 2.52. To find the x-intercepts we solve 1x - 122 -2 = 0:
1x - 122 =2 x- 1 = {12
x =1 { 12
Figure 2.52
y > 0 for x > 1 +
y = (x – 1)2 – 2 y
–3 x
–4 –2
– 2 – 3
1 2 1 4 5
4 5 3
(1 – 2, 0)
(1 + 2, 0) y < 0
for 1 – 2 < x < 1 +
–1 6
y > 0 for x < 1 – 2√
√
√ √2
√2
√
The x-intercepts are 11 - 12, 02 and 11+ 12, 02. If the y-coordinate of a point on the graph is negative, then the x-coordinate satisfies 1x- 122- 26 0. So the solution set to 1x - 122 - 26 0 is the open interval 11 - 12, 1 + 122.
Although a graphing calculator will not find the exact solution to this inequal- ity, you can use TRACE to support the answer and see that y is negative between the x-intercepts. See Fig. 2.53.
䉴TRY THIS. Solve 2 - 0x - 10 … 0 by graphing. ■ Note that the solution set to 1x - 122 -2 Ú 0 can also be obtained from the graph in Fig. 2.52. If the y-coordinate of a point on the graph is positive or zero, then the x-coordinate satisfies 1x- 122- 2 Ú0. So the solution set to 1x -122- 2 Ú0 is 1- ⬁, 1 - 12 4傼31 + 12, ⬁2.
Figure 2.53
⫺4 5
⫺3 5
FUNCTION
gallery... Some Basic Functions and Their Properties
y
1 2 3 x –1
– 2 – 3
1 3 2
– 2 – 1
– 3
f(x) = x y
1 2 3 x –1
– 2 – 3
3 4 5
2
– 1 f(x) = 3x + 2 y
1 2 3 x –1
– 2 – 3
1 3 2
5 f(x) = 4 Constant Function
Domain (ⴚH, H) Range {4}
Constant on (ⴚH, H) Symmetric about y-axis
Identity Function
Domain (ⴚH, H) Range (ⴚH, H) Increasing on (ⴚH, H) Symmetric about origin
Linear Function
Domain (ⴚH, H) Range (ⴚH, H) Increasing on (ⴚH, H)
Absolute Value Function
Domain (ⴚH, H) Range [0, H) Increasing on [0, H) Decreasing on (ⴚH, 0]
Symmetric about y-axis
Cube Function
Domain (ⴚH, H) Range (ⴚH, H) Increasing on (ⴚH, H) Symmetric about origin
Square Function
Domain (ⴚH, H) Range [0, H) Increasing on [0, H) Decreasing on (ⴚH, 0]
Symmetric about y-axis
Cube-Root Function
Domain (ⴚH, H) Range (ⴚH, H) Increasing on (ⴚH, H) Symmetric about origin
Square-Root Function
Domain [0, H) Range [0, H) Increasing on [0, H)
Greatest Integer Function
Domain (ⴚH, H) Range {n ƒ n is an integer}
Constant on [n, n + 1) for every integer n y
1 2 3 x –1
– 2 – 3
1 2 3
f(x) = ⏐x⏐ y
1 2 3 x –1
– 2 – 3
1 2 3 4 5
f(x) = x2
y
1 2 3 4 5 x 1
2
3 f(x) = x√
y
1 2 x – 2
2 6 8
4
– 4 – 2
– 6 – 8 f(x) = x3
y
1 2 3 x –1
– 2 – 3
1 2
– 2 – 1
√ f(x) = x3
y
–1 x –3
– 2 – 3
1 2 3
–2 2 f(x) = [[ x]] 1
1. The graph of ƒ1x2 = 1-x24 is a reflection in the x-axis of the graph of g1x2= x4.
2. The graph of ƒ1x2 = x -4 lies four units to the right of the graph of ƒ 1x2 =x.
3. The graph of y= 0x+20 +3 is a translation two units to the right and three units upward of the graph of y= 0x0. 4. The graph of ƒ1x2 = -3 is a reflection in the x-axis of
the graph of g1x2 = 3.
5. The functions y = x2+ 4x+ 1 and y= 1x +222 -3 have the same graph.
6. The graph of y = -1x - 322 - 4 can be obtained by moving y= x2 three units to the right and down four units, and then reflecting in the x-axis.
7. If ƒ1x2 = -x3 +2x2- 3x +5, then ƒ1-x2 = x3+ 2x2 +3x +5.
8. The graphs of ƒ1x2= -1x and g1x2 = 1-x are identical.
9. If ƒ1x2 =x3 - x, then ƒ1-x2 = -ƒ1x2.
10. The solution set to 0x0 - 1 … 0 is 3-1, 14.
FOR thought… True or False? Explain.
Fill in the blank.
1. Translating and reflecting are ________ transformations.
2. Stretching and shrinking are ________ transformations.
3. The graph of a function of the form y =a1x -h22+ k where a ⬆0 is a ________.
4. The graph of y= ƒ 1x -h2 is a ________ of the graph of y =ƒ 1x2.
5. The graph of y= -ƒ 1x2 is a ________ of the graph of y =ƒ 1x2.
6. The function ƒ 1x2 =x is the ________ function.
7. The function ƒ 1x2 =mx +b with m⬆ 0 is a ________
function.
8. The function ƒ 1x2 =b is a ________ function.
9. If ƒ 1-x2= -ƒ 1x2 for every x in the domain of ƒ, then ƒ is an ________ function.
10. If ƒ 1-x2=ƒ 1x2 for every x in the domain of ƒ, then ƒ is an ________ function.
Sketch the graphs of each pair of functions on the same coordinate plane.
11. ƒ 1x2 = 0x0 , g1x2 = 0x0 -4 12. ƒ 1x2 = 1x, g1x2 = 1x+ 3
13. ƒ 1x2 =x, g1x2 =x +3 14. ƒ 1x2= x2, g1x2= x2 -5 15. y =x2, y= 1x -322 16. y= 0x0 , y = 0x+ 20 17. y = 1x, y = 1x +9 18. y= x2, y =1x- 122 19. ƒ 1x2 = 1x, g1x2 = -1x 20. ƒ 1x2= x, g1x2= -x 21. y = 1x, y =31x
22. y = 21- x2, y =421- x2 23. y =x2, y= 1
4 x2 24. y= 0x0, y= 1 30x0 25. y = 24- x2, y = -24 -x2
26. ƒ 1x2 =x2+1, g1x2 = -1x2+ 12
Match each function in Exercises 27–34 with its graph (a)–(h). See the procedure for multiple transformations on page 136.
27. y =x2 28. y= 1x -422 +2 29. y =1x +422- 2 30. y= -21x - 222
31. y = -21x +222 32. y = - 1
2 x2 -4 33. y = 1
2 1x +422 +2 34. y = -21x -422 -2