#### Answer

Refer to the graph below.

#### Work Step by Step

The parent function of $y=-2+\frac{1}{2}\sin{(3x)}$ is $y=\sin{x}$.
The parent function $y=\sin{x}$ has a period of $2\pi$ so one period of its graph is in the interval $[0,2\pi]$
Note that the period of $y=d+\sin{(bx)}$ is $\frac{2\pi}{b}$.
The given function has $b=3$ so its period is $\frac{2\pi}{3}=\frac{2\pi}{3}$.
This means that one period of the given function is in the interval $[0, \frac{2\pi}{3}]$.
Dividing this interval into four equal parts yield the key x-values $0, \frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{2},$ and $\frac{2\pi}{3}$.
To graph the given function, perform the following steps:
(1) Create a table of values for $y=-2+\frac{1}{2}\sin{(3x)}$ using the key x-values listed above.
(Refer to the table below.)
(2) Plot each point from the table of values and connect them using a sinusoidal curve to complete one period in the interval $[0,\frac{2\pi}{3}]$.
(3) Extend the graph one more period by repeating the cycle in the interval $[\frac{2\pi}{3}, \frac{4\pi}{3}]$.
Refer to the graph in the answer part above.