#### Answer

The amplitude is $\frac{1}{2}$, the period is $4\pi$, there is no vertical translation and the phase shift is $2\pi$ 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, $y=-\frac{1}{2}\sin (\frac{1}{2}x+\pi)$ becomes $y=0-\frac{1}{2}\sin [\frac{1}{2}(x+2\pi)]$.
Comparing the two equations, $a=-\frac{1}{2},b=\frac{1}{2},c=0$ and $d=-2\pi$.
The amplitude is $|a|=|-\frac{1}{2}|=\frac{1}{2}.$
The period is $\frac{2\pi}{b}=\frac{2\pi}{0.5}=4\pi$.
The vertical translation is $c=0$.
The phase shift is $|d|=|-2\pi|=2\pi$
Therefore, the amplitude is $\frac{1}{2}$, the period is $4\pi$, there is no vertical translation and the phase shift is $2\pi$ to the left since $d$ is less than zero.