Question

In Problems 31–34, match the given function to one of the graphs (A)– (D).

$y=-3\sin \left(\frac{1}{2}x\right)$

Short answer

The graph of the given function is graph (D)

Step-by-step solution

Step 1. Given information   

The trigonometric function is:

$y=-3\sin \left(\frac{1}{2}x\right)$

Step 2. To determine the amplitude and period of the sinusoidal function.     

By comparing given function $-3\sin \left(\frac{1}{2}x\right)$with standard sinusoidal function$A\sin \left(\omega x\right)$

We get amplitude$A=3$

Time period is$$\begin{gathered} T=\frac{2\mathrm{\pi }}{\omega } \\ T=\frac{2\mathrm{\pi }}{{\displaystyle \frac{1}{2}}} \\ T=4\mathrm{\pi } \end{gathered}$$

So the graph of the given function will have time period $4\mathrm{\pi }$and amplitude 3

Step 3. To find passing points of the graph  

When$x=0$

then $$\begin{gathered} y=-3\sin \left(\frac{1}{2}.0\right) \\ y=-3\sin 0 \\ y=-3.0 \\ y=0 \end{gathered}$$

Graph will pass through (0,0)

when $x=\mathrm{\pi }$

Then

$$\begin{gathered} y=-3\sin \left(\frac{1}{2}.\mathrm{\pi }\right) \\ y=-3\sin \left(\frac{\mathrm{\pi }}{2}\right) \\ y=-3.1 \\ y=-3 \end{gathered}$$

So the graph will pass through (0,0) and$\left(\mathrm{\pi },-3\right)$

Now look at the given graphs from A-D, we can see that

graph (D) has amplitude 3 , time period $4\mathrm{\pi }$and passing point (0,0) and $\left(\mathrm{\pi },-3\right)$

So the graph for given function is graph (D)