Problem 21 - Algebra and Trigonometry

Question

Determine the amplitude, period, and phase shift of each function. Then graph one period of the function. $$y=3 \sin (2 x-\pi)$$

Step-by-Step Solution

Answer

The amplitude of the function \(y = 3 \sin (2 x-\pi)\) is 3, its period is \(\pi\), and has a phase shift of -\(\frac{\pi}{2}\).

Step 1: Amplitude

The coefficient of the sine function gives the amplitude of the function. Here, the coefficient is 3, so the amplitude of the function \(y = 3 \sin (2 x-\pi)\) is 3.

Step 2: Period

The period can be calculated using the formula \(T = \frac{2\pi}{|B|}\) where B is the coefficient of x. Here B = 2, hence the period \(T = \frac{2\pi}{2} = \pi\).

Step 3: Phase Shift

Phase shift can be calculated by the formula \(-\frac{C}{B}\) where C is the constant term inside the function. Here, C is \(\pi\), B is 2, so the phase shift is \(-\frac{\pi}{2}\). It signifies the function graph to the right of origin.

Step 4: Graph the function

To plot the function, first mark the amplitude on the y-axis and period on the x-axis. Then, shift the sine wave graph to the right by the phase shift amount. Plot the graph for one complete period. Now, the graph starts at -\(\frac{\pi}{2}\) (due to phase shift) and completes one cycle at \(\frac{\pi}{2}\), the height of wave (amplitude) is 3, and the function repeats this wave pattern every \(\pi\) units (period).