Answer
$|A|=3$, $p=\frac{2\pi}{2}=\pi $, $\phi=-\frac{\pi}{4}$.
See graph.
Work Step by Step
Step 1. Given $y=-3sin(2x+\frac{\pi}{2})=-3sin[2(x+\frac{\pi}{4})]$, we can find the amplitude $|A|=3$, period $p=\frac{2\pi}{2}=\pi $, and phase shift $\phi=-\frac{\pi}{4}$.
Step 2. See graph.
