636 CHAPTER 6 The Circular Functions and Their Graphs y 0 y = –5 + f(x) y = f(x) y = 3 + f(x) 6 3 –4 Vertical translations of y = f(x) x Figure 35 y2 = −2 cos 3x −3 7 11p 4 11p 4 − Vertical Translations The graph of a function of the form y =c +ƒ1x2 1or y =ƒ1x2 +c2 is translated vertically compared to the graph of y = ƒ1x2. See Figure 35. The translation is up c units if c 70 and down c units if c 60. EXAMPLE 4 Graphing y =c +a cos bx Graph y = 3 - 2 cos 3x over two periods. SOLUTION We use Method 1. The values of y will be 3 greater than the corresponding values of y in y = -2 cos 3x. This means that the graph of y = 3 - 2 cos 3x is the same as the graph of y = -2 cos 3x, translated vertically up 3 units. The period of y = -2 cos 3x is 2p 3 , so the key points have the following x-values. We use these x-values to make a table of points. 0, p 6 , p 3 , p 2 , 2p 3 Key x-values x 0 p 6 p 3 p 2 2p 3 3x 0 p 2 p 3p 2 2p cos 3x 1 0 -1 0 1 2 cos 3x 2 0 -2 0 2 3 −2 cos 3x 1 3 5 3 1 The key points and two periods of the graph are shown in Figure 36. x 0 5 y = 3 – 2 cos 3x 1 4 y = 3 y – 2p 3 2p 3 –p 3 p 6 p 3 p 2 –p 6 p 2 – Figure 36 S Now Try Exercise 53. CAUTION If we use Method 2 to graph the function y = 3 - 2 cos 3x in Example 4, we must first graph y = -2 cos 3x and then apply the vertical translation up 3 units. To begin, use the fact that a = -2 and b = 3 to determine that the amplitude is 2, the period is 2p 3 , and the graph is the reflection of the graph of y = 2 cos 3x across the x-axis. Then, because c = 3, translate the graph of y = -2 cos 3x up 3 units. See Figure 36. If the vertical translation is applied first, then the reflection must be across the line y =3, not across the x-axis.
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