Answer
a. amplitude = $3$,
period = $ 4\pi$,
horizontal shift = $0$
b. $y=3\displaystyle \sin\left(\frac{1}{2}x\right)$
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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$f(0)=0$,
and to the right of $x=0$, the graph rises.
These are characteristics of a sine curve with positive $a$,
and no horizontal shift ($b=0$).
The amplitude is $3$, so $a$ can be $\pm 3$, but since it is positive, $a=3$.
The period is $4\pi.$
So, from
$\displaystyle \frac{2\pi}{k} =4\pi$, it follows that $k=\displaystyle \frac{2\pi}{4\pi}=\frac{1}{2}$.
With these parameters,
$f(x)=y=a\sin k(x-b)$
$y=3\displaystyle \sin\left(\frac{1}{2}x\right)$
a.
amplitude=$3$,
period=$ 4\pi$,
horizontal shift=0
b.
$y=3\displaystyle \sin(\frac{1}{2}x)$
