648 CHAPTER 6 The Circular Functions and Their Graphs EXAMPLE 2 Graphing y =a tan bx Graph y = -3 tan 1 2 x. SOLUTION The period is p 1 2 = p , 1 2 = p # 2 1 = 2p. Adjacent asymptotes are at x = -p and x = p. Dividing the interval 1-p, p2 into four equal parts gives key x-values of - p 2 , 0, and p 2 . Evaluating the function at these x-values gives the following key points. a- p 2 , 3b, 10, 02, a p 2 , -3b Key points By plotting these points and joining them with a smooth curve, we obtain the graph shown in Figure 48. Because the coefficient -3 is negative, the graph is reflected across the x-axis compared to the graph of y = 3 tan 1 2 x. S Now Try Exercise 21. x y 0 3 –3 y = –3 tan x Period: 2p 1 2 –p 2 p p –p 2 Figure 48 EXAMPLE 3 Graphing y =a cot bx Graph y = 1 2 cot 2x. SOLUTION Because this function involves the cotangent, we can locate two adjacent asymptotes by solving the equations 2x = 0 and 2x = p. The lines x = 0 (the y-axis) and x = p 2 are two such asymptotes. We divide the interval A0, p 2 B into four equal parts, obtaining key x-values of p 8 , p 4 , and 3p 8 . Evaluating the function at these x-values gives the key points Ap 8 , 1 2B , A p 4 , 0B , A 3p 8 , - 1 2B . We plot these points and join them with a smooth curve approaching the asymptotes to obtain the graph shown in Figure 49. x y 0 1 –1 Period: y = cot 2x 1 2 p 2 p 8 p 4 p 2 3p 8 Figure 49 S Now Try Exercise 23. NOTE The function y = -3 tan 1 2 x in Example 2, graphed in Figure 48, has a graph that compares to the graph of y = tan x as follows. 1. The period is larger because b = 1 2 , and 1 2 61. 2. The graph is stretched vertically because a = -3, and -3 71. 3. Each branch of the graph falls from left to right (that is, the function decreases) between each pair of adjacent asymptotes because a = -3, and -3 60. When a 60, the graph is reflected across the x-axis compared to the graph of y = a tan bx.
RkJQdWJsaXNoZXIy NjM5ODQ=