Unformatted text preview: 38 CHAPTER 1 FUNCTIONS AND MODELS 9. We start with the graph of y = 23” y
(Figure 2). reﬂect it about the
y—axis. and then about the ac—axis —i) x
(or just rotate 1800 to handle both
reﬂections) to obtain the graph of
y : —2_“”. In each graph. y = 0
is the horizontal asymptote. I
y = 2 y = —2‘“”
10. We start with the graph of y : ex (Figure 13). 11. We start with the graph ofy : 6“ (Figure 13),
vertically stretch by a factor of 2. and then shift reﬂect it about the m—axis, and then shift 3 units
1 unit upward. There is a horizontal asymptote upward. Note the horizontal asymptote of y : 3.
of y : 1.
_v
2
0 3
y : 26$ y : 1 + 261: 12. We start with the graph of y = e“E (Figure 13),
reﬂect it about the y—axis. and then about the asaxis
(or just rotate 180° to handle both reﬂections) to
obtain the graph of y 2 76’“. Now shift this graph
1 unit upward. vertically stretch by a factor of 5, and then shift 2 units upward. y:2+5(1—e—m) 13. (a) To ﬁnd the equation of the graph that results from shifting the graph of y : e“6 2 units downward, we subtract 2 from the original function to get y : e$ i 2. (b) To ﬁnd the equation of the graph that results from shifting the graph of y : em 2 units to the right, we replace m with m — 2 in the original function to get y : 6(172). (c) To ﬁnd the equation of the graph that results from reﬂecting the graph of y : 61 about the x—axis. we multiply
the original function by 71 to get y : —e“”. (d) To ﬁnd the equation of the graph that results from reﬂecting the graph of y : er” about the y—axis, we replace a: with 7:1: in the original function to get y : e". (e) To ﬁnd the equation of the graph that results from reﬂecting the graph of y : e“ about the ac—axis and then about
the y—axis. we ﬁrst multiply the original function by 71 (to get y = —e’) and then replace m with em in this equation to get y : #e ...
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 Spring '10
 Ban
 Calculus

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