Chapter 9.4, Problem 31E

To determine

To find: The graph for the given function.

Answer to Problem 31E

The required graph is shown in Figure 1.

Explanation of Solution

Given:

The given function is $g\left(x\right)=\sin 2\left(x+\pi \right)$ .

Calculation:

Consider the given function is,

  $g\left(x\right)=\sin 2\left(x+\pi \right)$

Consider the general form is,

  $y=a\sin \left(x-h\right)+k$

From above and from the given function. The amplitude is 1, the period is obtained as,

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

Thus, the vertical shift is $k=0$ and the horizontal shift is $h=-\pi$ .

Then, determine the five key points as,

First,

  $\left(h,k\right)=\left(-\pi ,0\right)$

Second,

  $\begin{array}{c}\left(\frac{1}{2}\cdot \frac{2\pi }{b}+h,0+k\right)=\left(\frac{1}{2}\cdot \frac{2\pi }{2}-\pi ,0\right)\\ =\left(\frac{\pi }{2},0\right)\end{array}$

Third, the maximums

  $\begin{array}{c}\left(\frac{2\pi }{b}+h,0+k\right)=\left(\frac{2\pi }{2}-\pi ,0+0\right)\\ =\left(0,0\right)\end{array}$

Fourth point the maximum.

  $\begin{array}{c}\left(\frac{1}{4}\cdot \frac{2\pi }{b}+h,a+k\right)=\left(\frac{1}{4}\cdot \frac{2\pi }{2}-\pi ,1+0\right)\\ =\left(\frac{-3\pi }{4},-1\right)\end{array}$

Fifth point, the minimum.

  $\begin{array}{c}\left(\frac{3}{4}\cdot \frac{2\pi }{b}+h,-a+k\right)=\left(\frac{3}{4}\cdot \frac{2\pi }{2}-\pi ,1+0\right)\\ =\left(\frac{-\pi }{4},-1\right)\end{array}$

From the points, the graph for the function is shown in Figure 1.

  

Figure 1