A transformation of y = cos (x)
Determine amplitude, period, and phase shift, and sketch one cycle of y = -3 cos(2x – π) – 1.
Rewrite the function in the general form as
y = -3 cos[2(x – π/2)] – 1.
The amplitude is 3. Since the period is 2π/B, the period is 2π/2 or π. The fundamental cycle is shrunk to the interval [0, π]. Sketch one cycle of y = cos 2x on [0, π], as shown in Fig. 5.55. This cycle is reflected in the x-axis and stretched by a factor of 3. A shift of π/2 to the right means that one cycle of the original function occurs for π/2 ≤ x ≤ 3π/2. Evaluate the function at the endpoints and midpoint of the interval [π/2, 3π/2] to get (π/2, -4), (π, 2), and (3π/2, -4) for the endpoints and midpoint of this cycle. Since the graph is translated downward one unit, midway between these maximum and minimum points we get points where the graph intersects the line y = -1. These points are (3π/4, -1) and (5π/4, -1). Draw one cycle of the graph through these five points, as shown in Fig. 5.55.
The calculator graph in Fig. 5.56 supports our conclusions.