Q. 34 - Precalculus Enhanced with Graphing Utilities

Question

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

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

Step-by-Step Solution

Answer

The graph of the given function is graph (D) 

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)