Answer
The amplitude is $1$, the period is $2$, there is no vertical translation and the phase shift is $\frac{1}{3}$ units to the right since $d$ is more than zero.
Work Step by Step
We first write the equation in the form $y=c+a \cos [b(x-d)]$. Therefore, $y=-\cos [\pi(x-\frac{1}{3})]$ becomes $y=0-\cos [\pi(x-\frac{1}{3})]$.
Comparing the two equations, $a=-1,b=\pi,c=0$ and $d=\frac{1}{3}$.
The amplitude is $|a|=|-1|=1.$
The period is $\frac{2\pi}{b}=\frac{2\pi}{\pi}=2$.
The vertical translation is $c=0$.
The phase shift is $|d|=|\frac{1}{3}|=\frac{1}{3}$
Therefore, the amplitude is $1$, the period is $2$, there is no vertical translation and the phase shift is $\frac{1}{3}$ units to the right since $d$ is more than zero.