Answer
a.
amplitude=4,
period=$ 2\pi$,
horizontal shift=0
b.
$f(x)=4\sin x$
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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a.
The shape infers the sine function as parent,
with no horizontal shift $(b=0)$,
f(0)=0, after which f(x) rises,
so a is positive.
The amplitude is 4, so $a=\pm 4.$
Since it is positive, a=4.
The period is $ 2\pi$
so from $\displaystyle \frac{2\pi}{k}$=2$\pi$, it follows that $k=1.$
b.
$y=a\sin k(x-b)$
$y=4\sin[1\cdot(x-0)]$
$y=4\sin x$
