Chapter 3 Transformations of Graphs and Data
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Chapter 3 Transformations of Graphs and Data 6. Convert grams to ounces. median: Self-Test
(pp. 215)
1. The graph of h is the image of the graph 1 of f : x under the translation x T(x, y) = (x + 7, y – 9). The asymptotes of the graph of f are the x-axis and the y-axis. By the Graph Translation Theorem, the asymptotes of the graph of h are 7 units to the right and 9 units below, that is, x = 7 and y = –9. 2. Under T, the graph of y = x2 is translated 4 units to the right and 2 units down. So, by the Graph Translation Theorem, an equation for the image is y + 2 = (x 4)2 . 3. Since the vertex of the graph of y = x2 is (0, 0), the vertex of the image is 4 units to the right and 2 units down, or (4, –2).
(153.8 g)( 0.035
oz g
) ≈ 5.383 oz;
range: (13.2 g)( 0.035
oz g
) ≈ 0.462 oz;
IQR: 1 g = 0.035 oz. 7. Let y = f(x) and switch x and y. x = 43 y + 2 8. Solve the Equation in Question 7 for y. x 4
3
3
= y + 2 , from which y = x 2 . 4
This is an equation for a function because each value of x yields exactly one value of y. 9. f and f -1are reflection images of each other over the line y = x.
4. Under S, the image of (4, 0) is (8, 0). The image of (–1, 2) is (2, –2), and the image of (–4, –4) is (–8, 4). Connecting these points in order as in the preimage yields the graph here. 10. n(m(x)) = n(16 – 5x) = 16 5x + 16 5x
5. a. The median is 150 larger; 3.8 + 150 = 153.8 g. b. The range is the same; 13.6 - 0.4 = 13.2 g. c. The IQR is the same; 4.6 - 3.6 = 1 g.
11. The domain of n • m is the set of all values x in the domain of m for which m(x) is in the domain of n. The domain of m is the set of all real numbers, and so is the range of m. Thus m(x) can be any real number. But if m(x) is negative, then m(x) is not real. So the domain of n • m is the set of all x with m(x) ≥ 0, that is, with 16 – 5x ≥ 0, the set
{x x } . 16 5
12. Use the Graph Scale Change Theorem with a = 3 and b = 0.5. An equation for 89
Functions, Statistics and Trigonometry Solution Manual
Chapter 3, Self-Test
Back to Self-Test
the image is and b, 2y =
y b
= 1
2 x 3
1 x a
2
Substituting for a
or y =
4.5 x2
.
13. Use the Graph Standardization Theorem with μ = 83 and = 7. The transformation is x
x83 . 7
14. Since j(-x) = (-x)2 + 5 = x2 + 5 = j(x) an even function. 15. a. The parent function is the function whose graph was transformed to get the graph of g. That is the absolute value function A with A(x) = y = x . b. The graph of g can be viewed as the image of the graph of A under the scale change S2,0.5. So, by the Graph Scale Change Theorem, an equation for g is
y 0.5
=
x 2
.
c. The graph is symmetric to the y-axis. d. {y| y ≥ 0} 16. Let T be the transformation. When a point (x, y) is reflected over the line y = x, its image is (y, x). So a rule is T(x, y) = (y, x). 17. Standardize these scores to compare them. 37 30 mouse: z = 2.06 ; 3.4 1260 910 moose: z = ; Since the 185 1.89 mouse has a higher z-score, it is heavier relative to its population than the moose.
90
Functions, Statistics and Trigonometry Solution Manual
Chapter 3, Self-Test
Chapter 3 Transformations of Graphs and Data 6. Convert grams to ounces. median: Self-Test
(pp. 215)
1. The graph of h is the image of the graph 1 of f : x under the translation x T(x, y) = (x + 7, y – 9). The asymptotes of the graph of f are the x-axis and the y-axis. By the Graph Translation Theorem, the asymptotes of the graph of h are 7 units to the right and 9 units below, that is, x = 7 and y = –9. 2. Under T, the graph of y = x2 is translated 4 units to the right and 2 units down. So, by the Graph Translation Theorem, an equation for the image is y + 2 = (x 4)2 . 3. Since the vertex of the graph of y = x2 is (0, 0), the vertex of the image is 4 units to the right and 2 units down, or (4, –2).
(153.8 g)( 0.035
oz g
) ≈ 5.383 oz;
range: (13.2 g)( 0.035
oz g
) ≈ 0.462 oz;
IQR: 1 g = 0.035 oz. 7. Let y = f(x) and switch x and y. x = 43 y + 2 8. Solve the Equation in Question 7 for y. x 4
3
3
= y + 2 , from which y = x 2 . 4
This is an equation for a function because each value of x yields exactly one value of y. 9. f and f -1are reflection images of each other over the line y = x.
4. Under S, the image of (4, 0) is (8, 0). The image of (–1, 2) is (2, –2), and the image of (–4, –4) is (–8, 4). Connecting these points in order as in the preimage yields the graph here. 10. n(m(x)) = n(16 – 5x) = 16 5x + 16 5x
5. a. The median is 150 larger; 3.8 + 150 = 153.8 g. b. The range is the same; 13.6 - 0.4 = 13.2 g. c. The IQR is the same; 4.6 - 3.6 = 1 g.
11. The domain of n • m is the set of all values x in the domain of m for which m(x) is in the domain of n. The domain of m is the set of all real numbers, and so is the range of m. Thus m(x) can be any real number. But if m(x) is negative, then m(x) is not real. So the domain of n • m is the set of all x with m(x) ≥ 0, that is, with 16 – 5x ≥ 0, the set
{x x } . 16 5
12. Use the Graph Scale Change Theorem with a = 3 and b = 0.5. An equation for 89
Functions, Statistics and Trigonometry Solution Manual
Chapter 3, Self-Test
Back to Self-Test
the image is and b, 2y =
y b
= 1
2 x 3
1 x a
2
Substituting for a
or y =
4.5 x2
.
13. Use the Graph Standardization Theorem with μ = 83 and = 7. The transformation is x
x83 . 7
14. Since j(-x) = (-x)2 + 5 = x2 + 5 = j(x) an even function. 15. a. The parent function is the function whose graph was transformed to get the graph of g. That is the absolute value function A with A(x) = y = x . b. The graph of g can be viewed as the image of the graph of A under the scale change S2,0.5. So, by the Graph Scale Change Theorem, an equation for g is
y 0.5
=
x 2
.
c. The graph is symmetric to the y-axis. d. {y| y ≥ 0} 16. Let T be the transformation. When a point (x, y) is reflected over the line y = x, its image is (y, x). So a rule is T(x, y) = (y, x). 17. Standardize these scores to compare them. 37 30 mouse: z = 2.06 ; 3.4 1260 910 moose: z = ; Since the 185 1.89 mouse has a higher z-score, it is heavier relative to its population than the moose.
90
Functions, Statistics and Trigonometry Solution Manual
Chapter 3, Self-Test
