Problem 49 - Precalculus Essentials

Question

Determine the amplitude, period, and phase shift of each function. Then graph one period of the function. $$y=-3 \cos \left(2 x-\frac{\pi}{2}\right)$$

Step-by-Step Solution

Answer

The amplitude of the function is 3, the period is \( \pi \), and the phase shift is \( \frac{\pi}{4} \) units to the right.

Step 1: Identify the Amplitude

The amplitude of a trigonometric function is the absolute value of the coefficient of the trigonometric term. Here the coefficient is -3, so the amplitude will be |-3| = 3.

Step 2: Identify the Period

The period of a trigonometric function is calculated as \( \frac{2\pi}{|b|} \), where b is the coefficient of x. In our function, b = 2. Hence, the period would be \( \frac{2\pi}{|2|} \) which equals \( \pi \).

Step 3: Identify the Phase Shift

The phase shift of a trigonometric function, which is the horizontal shift, can be determined by \( \frac{c}{|b|} \), where c is the term being subtracted or added to x. Here c equals \( \frac{\pi}{2} \) and b equals 2. Thus, the phase shift is \( \frac{\pi/2}{|2|} \) which is \( \frac{\pi}{4} \). Because c is positive and subtracted, we have phase shift of \( \frac{\pi}{4} \) units to the right.

Step 4: Graph One Period of the Function

Now that we have the amplitude, period, and phase shift, let's graph one period of the function. It will be a cosine function flipped upside down (because of -3), starts from \( \frac{\pi}{4} \) (due to phase shift), completes one full cycle at \( \frac{\pi}{4} + \pi \) and has maximum and minimum values of 3 and -3 respectively (due to amplitude).