Chapter 9.6, Problem 13E

To determine

Write the function for sinusoid.

Answer to Problem 13E

  $\boxed{y=3\sin 2x}$

Explanation of Solution

Given information:

  

Formula used:

Sinusoidal function $y=a\sin \left(bx-h\right)+k$

Here a=amplitude, h=Horizontal Shift, k=vertical shift, b=period

Vertical shift $=k=\frac{\left(\text{maximum value}\right)+\left(\text{minimum value}\right)}{2}$

Amplitude $=\left|a\right|=\frac{\left(\text{maximum value}\right)-\left(\text{minimum value}\right)}{2}$

Step 1:

Find the maximum and minimum values. From the graph the maximum value is 3 and the minimum value is -3.

Step 2:

Identify the vertical shift, $k$ .The value of $k$ is the mean of the maximum and minimum values:

  $\begin{array}{l}k=\frac{\left(\text{maximum value}\right)+\left(\text{minimum value}\right)}{2}=\frac{3+\left(-3\right)}{2}=\frac{3-3}{2}=\frac{0}{2}\\ k=0\end{array}$

Step 3:

Decide whether the graph should be modelled by sin or cosine function. Because the line crosses the midline $y=0$ on the y-axis, the graph is a sine curve with no horizontal shift. So, $h=0$ .

Step 4:

Find the amplitude and period. The period is:

  $\begin{array}{l}\pi =\frac{2\pi }{b}\\ b=2\end{array}$

The amplitude is:

  $\begin{array}{l}\left|a\right|=\frac{\left(\text{maximum value}\right)-\left(\text{minimum value}\right)}{2}\\ \left|a\right|=\frac{3-\left(-3\right)}{2}=\frac{3+3}{2}=\frac{6}{2}\\ \left|a\right|=3\end{array}$

Now putting the value of $a,b,h,$ and $k$ in equation $y=a\sin \left(bx-h\right)+k$

  $\boxed{y=3\sin 2x}$