resulting graph 1 unit upward.
(d) To obtain the graph of = ( − 2) − 2, we shift the graph of = () 2 units to the right (for the “−2” inside the
parentheses), and then shift the resulting graph 2 units downward.
(e) To obtain the graph of = −(), we reflect the graph of = () about the axis.
(f) To obtain the graph of = −1 (), we reflect the graph of = () about the line = (assuming is one–toone).
10. (a) To obtain the graph of = ( − 8), we shift the
graph of = () right 8 units.
(b) To obtain the graph of = −(), we reflect the
graph of = () about the axis.
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CHAPTER 1 FUNCTIONSAND MODELS
(c) To obtain the graph of = 2 − (), we reflect the
graph of = () about the axis, and then shift the
resulting graph 2 units upward.
(d) To obtain the graph of =
1
2 () − 1, we shrink the
graph of = () by a factor of 2, and then shift the
resulting graph 1 unit downward.
(e) To obtain the graph of = −1 (), we reflect the
graph of = () about the line = .
(f) To obtain the graph of = −1 ( + 3), we reflect the
graph of = () about the line = [see part (e)],
and then shift the resulting graph left 3 units.
11. () = 3 + 2. Start with the graph of = 3 and
shift 2 units upward.
12 . () = ( − 3) 2 . Start with the graph of = 2 and
shift 3 units to the right.
13. =
√ + 2. Start with the graph of = √ and shift
2 units to the left.
14 . = ln( + 5). Start with the graph of = ln and
shift 5 units to the left.
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CHAPTER 1 REVIEW
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63
15. () = 1 + cos2. Start with the graph of = cos, compress horizontally by a factor of 2, and then shift 1 unit upward.
16. () = − + 2. Start with the graph of = , reflect about the xaxis, and then shift 2 units upward.
17. () = 1 + 05 . Start with the graph of = 05 =
1
2
and shift 1 unit upward.
18. () =
− if 0
− 1 if ≥ 0
On (−∞0), graph = − (the line with slope −1 and intercept 0)
with open endpoint (00).
On [0∞), graph = − 1 (the graph of = shifted 1 unit downward)
with closed endpoint (00).
19. (a) () = 2 5 − 3 2 + 2 ⇒ (−) = 2(−) 5 − 3(−) 2 + 2 = −2 5 − 3 2 + 2. Since (−) 6= () and
(−) 6= −(), is neither even nor odd.
(b) () = 3 − 7 ⇒ (−) = (−) 3 − (−) 7 = − 3 + 7 = −( 3 − 7 ) = −(), so is odd.
(c) () = −
2
⇒ (−) = −(−)
2
= −
2
= (), so is even.
(d) () = 1 + sin ⇒ (−) = 1 + sin(−) = 1 − sin. Now (−) 6= () and (−) 6= −(), so is neither
even nor odd.
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64
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CHAPTER 1 FUNCTIONSAND MODELS
(e) () = 1 − cos2 ⇒ (−) = 1 − cos[2(−)] = 1 − cos(−2) = 1 − cos2 = (), so is even.
(f) () = ( + 1) 2 = 2 + 2 + 1. Now (−) = (−) 2 + 2(−) + 1 = 2 − 2 + 1. Since (−) 6= () and
(−) 6= −(), is neither even nor odd.
20. For the line segment from (−22) to (−10), the slope is
0 − 2
−1 + 2
= −2, and an equation is − 0 = −2( + 1) or,
equivalently, = −2 − 2. The circle has equation 2 + 2 = 1; the top half has equation =
√ 1 − 2
(we have solved for
positive ). Thus, () =