Answer
The amplitude is $1$, the period is $\frac{2\pi}{3}$, the vertical translation is $2$ units up as $c$ is greater than zero and the phase shift is $\frac{\pi}{15}$ units to the right since $d$ is more than zero.
Work Step by Step
We first write the equation in the form $y=c+a \sin[b(x-d)]$. Therefore, $y=2-\sin [3x-\frac{\pi}{5})]$ becomes $y=2-\sin [3(x-\frac{\pi}{15})]$.
Comparing the two equations, $a=-1,b=3,c=2$ and $d=\frac{\pi}{15}$.
The amplitude is $|a|=|-1|=1.$
The period is $\frac{2\pi}{b}=\frac{2\pi}{3}$.
The vertical translation is $c=2$.
The phase shift is $|d|=|\frac{\pi}{15}|=\frac{\pi}{15}$
Therefore, the amplitude is $1$, the period is $\frac{2\pi}{3}$, the vertical translation is $2$ units up as $c$ is greater than zero and the phase shift is $\frac{\pi}{15}$ units to the right since $d$ is more than zero.