Answer
The amplitude is $4$, the period is $\pi$, thevertical translation is $3$ units up since $c$ is more than zero and the phase shift is $\frac{\pi}{4}$ units to the left since $d$ is less than zero.
Work Step by Step
We first write the equation in the form $y=c+a \sin [b(x-d)]$. Therefore, $y=3-4\sin(2x+\frac{\pi}{2})$ becomes $y=3-4\sin [2(x+\frac{\pi}{4})]$.
Comparing the two equations, $a=-4,b=2,c=3$ and $d=-\frac{\pi}{4}$.
The amplitude is $|a|=|-4|=4.$
The period is $\frac{2\pi}{b}=\frac{2\pi}{2}=\pi$.
The vertical translation is $c=3$.
The phase shift is $|d|=|-\frac{\pi}{4}|=\frac{\pi}{4}$
Therefore, the amplitude is $4$, the period is $\pi$, thevertical translation is $3$ units up since $c$ is more than zero and the phase shift is $\frac{\pi}{4}$ units to the left since $d$ is less than zero.
