Answer
a.
amplitude=2,
period=$\pi$,
horizontal shift=0
b.
$y=2\cos 2x$
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 2, a can be $\pm$2,
but since it is positive, a=2.
The period is $\pi$ (the next unmarked value on the x-axis after $\displaystyle \frac{3\pi}{4}$)
So from $\displaystyle \frac{2\pi}{k} =\pi$, it follows that k=2.
With these parameters,
$f(x)=y=a\cos k(x-b)$
$y=2\cos 2x$
a.
amplitude=2,
period=$\pi$,
horizontal shift=0
b.
$y=2\cos 2x$