First we rewrite the function in the form y = A sin[B(x – C)] + D by factoring 3 out of 3x + π:

From this equation we get A = 2, B = 3, and C = – π/3. So the amplitude is 2, the period is 2π/3, and the phase shift is -π/3. The period change causes the fundamental cycle of y = sin x on [0, 2π] to shrink to the interval [0, 2π/3]. Now draw one cycle of y = sin3x on [0, 2π/3] , as shown in Fig. 5.53. Stretch the

cycle vertically so that it has an amplitude of 2. The numbers π/3 and 1 shift the cycle a distance of π/3 to the left and up one unit. So one cycle of the function occurs on [-π/3, π/3]. Check by evaluating the function at the endpoints and midpoint of the interval [-π/3, π/3] to get (-π/3, 1), (0, 1), and (π/3, 1). Since the graph is shifted one unit upward, these points are the points where the curve intersects the line y = 1. Evaluate the function midway between these points to get (-π/6, 3) and (π/6, -1), the highest and lowest points of this cycle. One cycle is drawn through these five points and continued for another cycle, as shown in Fig. 5.53.

The calculator graph in Fig. 5.54 supports our conclusions about amplitude, period, and phase shift. Note that it is easier to obtain the amplitude, period, and phase shift from the equation than from the calculator graph.