Answer
$Amplitude = 3$
$Period = 4\pi$
$Horizontal\ shift = \frac{2\pi}{3} $ (right shift)
$Vertical\ shift = -2$ (2 units shift downward)
$Phase = -\frac{\pi}{3}$
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 $y = -2 + 3\cos (\frac{1}{2}x - \frac{\pi}{3})$
$Amplitude = |3| = 3$
$Period = \frac{2\pi}{\frac{1}{2}} = 4\pi$
$Horizontal\ shift = (–\frac{-\frac{\pi}{3}}{\frac{1}{2}}) = \frac{2\pi}{3} $ (right shift)
$Vertical\ shift = -2$ (2 units shift downward)
$Phase = -\frac{\pi}{3}$
