Chapter 4.6, Problem 47E

To determine

To graph: the given function.

Explanation of Solution

Given information:

Given function is

  $y=0.1\tan \left(\frac{\pi x}{4}+\frac{\pi }{4}\right)$ .

Calculation:

The vertical asymptotes for $y=\tan x$ are at $x=-\left(\frac{\pi }{2}\right)$ and $x=\frac{\pi }{2}$ and its period is $\pi$ .

So for the given equation, let us find out the values of the two consecutive asymptotes.

On comparing,

  $\begin{array}{c}\frac{\pi x}{4}+\frac{\pi }{4}=-\frac{\pi }{2}\\ \frac{\pi x}{4}=-\frac{\pi }{2}-\frac{\pi }{4}\\ \frac{\pi x}{4}=\frac{-3\pi }{4}\\ x=\left(\frac{-3\pi }{4}\right)\left(\frac{4}{\pi }\right)\\ x=3\end{array}$

Also,

  $\begin{array}{c}\frac{\pi x}{4}+\frac{\pi }{4}=\frac{\pi }{2}\\ \frac{\pi x}{4}=\frac{\pi }{2}-\frac{\pi }{4}\\ \frac{\pi x}{4}=\frac{\pi }{4}\\ x=\left(\frac{\pi }{4}\right)\left(\frac{4}{\pi }\right)\\ x=1\end{array}$

Therefore, the two consecutive asymptotes occur at $x=3$ and $x=1$ .

Now, let us use the software to plot the graph including two full periods which means from $-6$ to $6$ using the command:

  $\text{plot}\left(0.1\cdot \tan \left(\frac{\pi \cdot x}{4}+\frac{\pi }{4}\right),x=-\mathrm{6...6}\right)$

The above graph represents the equation $y=0.1\tan \left(\frac{\pi x}{4}+\frac{\pi }{4}\right)$ .