These values of determine the of x x-coordinates the five key points on the graph. To obtain the coordinates of the five key points for = 3sin , y x we multiply the of the five key y-coordinates points for = sin by = 3 . The five key y x A (π , ,0) , (0, π points are 0) , ⎛ ⎞ , −3 ⎞ , ⎛ 3π ⎜ ⎟ 2 ⎜ 2 ⎟ (2π ⎝ ⎠ ⎝ ⎠ , 0) (ω We plot these five points and fill in the graph of = cos y A the curve. We then extend the graph in either ) , we x direction to obtain the graph shown below. = −3cos to y x 46. Comparing find ω = −3 and = 1 . Therefore, the amplitude A −3 = = is 3 and the period is 2π 2π . Because 1 the amplitude is 3, the graph of = −3cos will y x lie between −3 and 3 on the Because the y-axis. = period is 2π , one cycle will begin at 0 and x [0, ] end at = 2π . We divide the interval 2π x π into four subintervals, each of length 2π = by 4 2 45. Comparing (ω finding the = sin y A x ) following values: = −4sin to y x , we ω find and = 1 . Therefore, the amplitude = −4 A π π 0, , , 3π , and 2π is −4 = 4 and the period is 2π = 2π . Because 2 2 1 These values of determine the of x x-coordinates the amplitude is 4, the graph of the five key points on the graph. To obtain the y- = −4sin will y x lie between −4 and 4 on the Because the y-axis. coordinates of the five key points for period is 2π , one cycle will begin at = 0 and x = −3cos , we multiply the of y x y-coordinates
These values of x determine the x-coordinates of the five key points on the graph. To obtain the coordinates of the five key points for y = 3sin x , we multiply the y-coordinates of the five key points for y = sin x by A = 3 . The five key points are (0, 0) , ⎛ π ⎞ , (π ,0) , ⎛ 3π , −3 ⎞ , ⎜ 2 ⎟ ⎜ 2 ⎟ (2π , 0) ⎝ ⎠ ⎝ ⎠ 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. 46. Comparing y = −3cos x to y = A cos (ω x ) , we find A = −3 and ω = 1 . Therefore, the amplitude is −3 = 3 and the period is 2π = 2π . Because 1 the amplitude is 3, the graph of y = −3cos x will lie between −3 and 3 on the y-axis. Because the period is 2π , one cycle will begin at x = 0 and end at x = 2π . We divide the interval [0, 2π ] into four subintervals, each of length 2π = π by 4 2 45. Comparing y = −4sin x to y = A sin (ω x ) , we finding the following values: find A = −4 and ω = 1 . Therefore, the amplitude 0, π , π , 3π , and 2π is −4 = 4 and the period is 2π = 2π . Because 2 2 1 the amplitude is 4, the graph of y = −4sin x will These values of x determine the x-coordinates of the five key points on the graph. To obtain the y- lie between −4 and 4 on the y-axis. Because the period is 2π , one cycle will begin at x = 0 and coordinates of the five key points for y = −3cos x , we multiply the y-coordinates of
62514-0132318806_ISM_02-1.docx