amplitude is 1, the graph of (2 ) = − sin will lie between −1 and 1 on the y x π Because the period is , one cycle will y-axis. (3x ) = sin y π begin at = 0 and end at = . We divide the 48. Comparing (ω ) x x = sin , y A x to [0, ] π interval into four subintervals, each of we π length by finding the following values: Because the 4 find ω = 3 . Therefore, the amplitude π π π 0, , , 3π , and = 1 and A is 1 = 1 and the period is 2π . Because the 4 2 4 (3x ) determine the of x-coordinates = sin will y the five key points on the graph. To obtain the amplitude is 1, the graph of y- lie 3 These values of x −1 between and 1 on the Because the y-axis. coordinates of the five key points for (2 ) = − sin , we multiply the of y x y-coordinates = period is 2π , one cycle will begin at 0 and x 3 = sin by = −1 .The y x A the five key points for end at = 2π . We divide the interval ⎡0, 2π ⎤ x five key points are 3 ⎢ 3 ⎥ π π ⎛ ⎞ ⎛ ⎞ ⎛ 3π ⎞ (0, ⎣ ⎦ 0) , ⎜ ⎟ ⎜ ,1⎟ , ,0 , (π , −1⎟ , ⎜ ,0) ⎝ 4 ⎠ ⎝ 2 ⎠ ⎝ ⎠ 4 π into four subintervals, each of length 2π / 3 = 4 6 by finding the following values: We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
48. Comparing y = sin (3x ) to y = A sin (ω x ) , we Because the amplitude is 1, the graph of y = − sin (2 x ) will lie between −1 and 1 on the y-axis. Because the period is π , one cycle will begin at x = 0 and end at x = π . We divide the interval [0, π ] into four subintervals, each of length π by finding the following values: 4 find A = 1 and ω = 3 . Therefore, the amplitude 0, π , π , 3π , and π is 1 = 1 and the period is 2π . Because the 4 2 4 amplitude is 1, the graph of 3 y = sin (3x ) will lie These values of x determine the x-coordinates of the five key points on the graph. To obtain the y- between −1 and 1 on the y-axis. Because the coordinates of the five key points for y = − sin (2 x ) , we multiply the y-coordinates of period is 2π , one cycle will begin at x = 0 and 3 end at x = 2π . We divide the interval ⎡0, 2π ⎤ the five key points for five key points are y = sin x by A = −1 .The 3 ⎢ 3 ⎥ ⎛ π ⎞ ⎛ π ⎞ ⎛ 3π ⎞ ⎣ ⎦ (0, 0) , ⎜ , −1⎟ , ⎜ ,0 ⎟ , ⎜ ,1⎟ , (π ,0) ⎝ 4 ⎠ ⎝ 2 ⎠ ⎝ 4 ⎠ into four subintervals, each of length 2π / 3 = π 4 6 by finding the following values: We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
62514-0132318806_ISM_02-1.docx