Answer
The amplitude of the function is $\frac{1}{2}$ and period is $6$.
Work Step by Step
The given equation is in the form of $y=A\sin Bx$.
Here, $A=\frac{1}{2}\text{ and }B=\frac{\pi }{3}$
So, the amplitude is:
$\begin{align}
& \text{Amplitude}=\left| A \right| \\
& =\left| \frac{1}{2} \right| \\
& =\frac{1}{2}
\end{align}$
The period is given below:
$\begin{align}
& \text{Period = }\frac{2\pi }{B} \\
& =\frac{2\pi }{\left( \frac{\pi }{3} \right)} \\
& =6
\end{align}$
$\text{Phase shift=}\frac{C}{B}=0$
And the quarter period is as follows:
$\begin{align}
& \text{Quarter-period}=\frac{6}{4} \\
& =\frac{3}{2}
\end{align}$
Now, add quarter periods starting from x = 0 to generate x-values for the key points. The x-value for the first key point is as follows:
$x=\text{0}$
And the x-value for the second key point is:
$\begin{align}
& x=0+\frac{3}{2} \\
& =\frac{3}{2}
\end{align}$
And the x-value for the third key point is:
$\begin{align}
& x=\frac{3}{2}+\frac{3}{2} \\
& =3
\end{align}$
And the x-value for the fourth key point is:
$\begin{align}
& x=3+\frac{3}{2} \\
& =\frac{9}{2}
\end{align}$
And the x-value for the fifth key point is:
$\begin{align}
& x=\frac{9}{2}+\frac{3}{2} \\
& =6
\end{align}$
