Chapter 9.4, Problem 33E

To determine

To find: The graph for the given function.

Answer to Problem 33E

The required graph is shown in Figure 1.

Explanation of Solution

Given:

The given function is $g\left(x\right)=\sin \frac{1}{2}\left(x+2\pi \right)+3$ .

Calculation:

Consider the given function is,

  $g\left(x\right)=\sin \frac{1}{2}\left(x+2\pi \right)+3$

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,

  $\begin{array}{c}\frac{2\pi }{b}=\frac{2\pi }{\frac{1}{2}}\\ =4\pi \end{array}$

Thus, the vertical shift is $k=3$ and the horizontal shift is $h=2\pi$ .

Then, determine the five key points as,

First,

  $\left(h,k\right)=\left(2\pi ,3\right)$

Second,

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

Third, the maximums

  $\begin{array}{c}\left(\frac{2\pi }{b}+h,0+k\right)=\left(\frac{2\pi }{\frac{1}{2}}+2\pi ,0+3\right)\\ =\left(6\pi ,3\right)\end{array}$

Fourth point the maximum.

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

Fifth point, the minimum.

  $\begin{array}{c}\left(\frac{3}{4}\cdot \frac{2\pi }{b}+h,-a+k\right)=\left(\frac{3}{4}\cdot \frac{2\pi }{\frac{1}{2}}+2\pi ,-1+3\right)\\ =\left(5\pi ,2\right)\end{array}$

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

  

Figure 1