Answer
a.
amplitude=$\displaystyle \frac{3}{2}$,
period=$\displaystyle \frac{2\pi}{3}$,
horizontal shift=0
b.
$y=\displaystyle \frac{3}{2}\cos 3x$
Work Step by Step
For
$y=a\sin k(x-b),\ ,\quad y=a\cos k(x-b)$,
$(k>0)$,
Amplitude: $|a|$,
Period: $\displaystyle \frac{2\pi}{k}$,
Horizontal shift: $b$
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f(0)$\neq$0,
its maximum is at x=0, and after x=0, the graph descends.
These are characteristics of a cosine curve
with positive a,
and no horizontal shift (b=0).
The amplitude is $\displaystyle \frac{3}{2}$, so a can be $\displaystyle \pm\frac{3}{2}$,
but since it is positive, a=$\displaystyle \frac{3}{2}$.
The period is $\displaystyle \frac{4\pi}{6}=\frac{2\pi}{3}$ (the next unmarked value on the x-axis after $\displaystyle \frac{3\pi}{6}=\frac{\pi}{2}$)
So, from
$\displaystyle \frac{2\pi}{k} =\frac{2\pi}{3}$, it follows that k=$3$.
With these parameters,
$f(x)=y=a\cos k(x-b)$
$y=\displaystyle \frac{3}{2}\cos 3x$
a.
amplitude=$\displaystyle \frac{3}{2}$,
period=$\displaystyle \frac{2\pi}{3}$,
horizontal shift=0
b.
$y=\displaystyle \frac{3}{2}\cos 3x$