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