Answer
$Amplitude = \frac{1}{2}$
$Period = \frac{2\pi}{3}$
$Horizontal\ shift = (–\frac{\pi}{3})$ i.e left shift $\frac{\pi}{3}$
$Vertical\ shift = \frac{3}{2}$ ($\frac{3}{2} $units shift upward)
$Phase = \pi$
Work Step by Step
If C is any real number and $B> 0$, then the graphs of $y = k + A\sin(Bx+C)$ and $y = k + A\cos (Bx+C)$ will have
$Amplitude = |A|$
$Period = \frac{2\pi}{B}$
$Horizontal\ shift = –\frac{C}{B}$
$Vertical\ shift = k$
$Phase = C$
so for $y = \frac{3}{2} - \frac{1}{2}\sin (3x + \pi )$
$Amplitude = |-\frac{1}{2}| = \frac{1}{2}$
$Period = \frac{2\pi}{3}$
$Horizontal\ shift = (–\frac{\pi}{3})$ i.e left shift $\frac{\pi}{3}$
$Vertical\ shift = \frac{3}{2}$ ($\frac{3}{2}$ units shift upward)
$Phase = \pi$
