Chapter 4.6, Problem 35E

To determine

To sketch: the graph of the given function.

Explanation of Solution

Given information:

Given function is

  $y=2\sec \left(x+\pi \right)$ .

Calculation:

The vertical asymptotes for $y=\sec x$ are at $x=-\frac{\pi }{2}$ and $x=\frac{\pi }{2}$ and its period is $2p$ .

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

On comparing we get,

  $\begin{array}{c}x+\pi =-\frac{\pi }{2}\\ x=-\frac{\pi }{2}-\pi \\ x=-\frac{3\pi }{2}\end{array}$

Also,

  $\begin{array}{c}x+\pi =\frac{\pi }{2}\\ x=\frac{\pi }{2}-\pi \\ x=-\frac{\pi }{2}\end{array}$

Therefore, the two consecutive asymptotes occur at $x=-\frac{3\pi }{2}$ and $x=-\frac{\pi }{2}$ .

Next, calculate the values of $y$ foe a few values of $x$ between these two asymptotes.

For example,

Substituting $x=-\frac{5\pi }{4}$

We get,

  $\begin{array}{c}y=2\sec \left(-\frac{5\pi }{4}+\pi \right)\\ y=2\sec \left(-\frac{\pi }{4}\right)\\ y=2\left(\sqrt{2}\right)\\ y=2\sqrt{2}\end{array}$

Similarly, calculate other values of $y$ and write them down in tabular form.

$x$	$-\frac{3\pi }{2}$	$-\frac{5\pi }{4}$	$-\pi$	$-\frac{7\pi }{4}$	$-\frac{\pi }{2}$
$y$	Not defined	$2\sqrt{2}$	$-2$	$-2\sqrt{2}$	Not defined

Now using these values, plot the graph including two full periods which means from $-2\pi$ to $2\pi$ .

The above graph represents the equation $y=2\sec \left(x+\pi \right)$ .