Pre-Calculus - Graphing Tangent, Cosecant, Secant, and Cotangent Functions

Which of the following represents the asymptotes for the general parent function $\sec \left ( \theta \right )$ ?

$\theta =\frac{(2k+1)\pi }{2}$

Explanation

If you do not have these asymptotes memorized, they can be easily derived. Write $\sec$ in terms of $\cos$ .

$\sec \left ( \theta \right )= \frac{1}{\cos \left ( \theta \right )}$

Now we need to solve for $\cos \left ( \theta \right )= 0$ since it is the denominator of the function. When the denominator of a function is equal to zero, there is a vertical asymptote because that function is then undefined.

$\cos \left ( \theta \right )= 0$ when $\theta = \pm \frac{\pi }{2},\pm \frac{3\pi }{2},\pm \frac{5\pi }{2},..., \frac{(2k+1)\pi }{2}$ . So for any integer , we say that there is a vertical asymptote for $\sec \left ( \theta \right )$ when $\theta =\frac{(2k+1)\pi }{2}$ .

Which of the following represents the asymptotes for the general parent function $\sec \left ( \theta \right )$ ?

$\theta =\frac{(2k+1)\pi }{2}$

$\theta = \frac{\pi }{2}$

$\theta = \pi k$

$\theta = \left ( 2k+1 \right )\pi$

Explanation

If you do not have these asymptotes memorized, they can be easily derived. Write $\sec$ in terms of $\cos$ .

$\sec \left ( \theta \right )= \frac{1}{\cos \left ( \theta \right )}$

Find the vertical asymptote of the equation.

$y=\frac{x^2-9}{x^2-16}$

There are no vertical asymptotes.

Explanation

To find the vertical asymptotes, we set the denominator of the function equal to zero and solve.

$\sqrt{16}=\sqrt{x^2}$

True or false: There is a vertical asymptote for $\tan \left ( \theta \right )+10$ at $\frac{3\pi }{2}$

The statement is true.

The statement is false.

There is insufficient information to answer the question.

Explanation

We know that the parent function of $\tan \left ( \theta \right )$ has asymptotes at $\frac{\pi }{2}+\pi k$ where is any integer. Considering $\frac{3\pi }{2}$ we can set the asymptotic equation equal to this one and solve for to see if is an integer. If is an integer, then there is a vertical asymptote here.

$\frac{3\pi }{2}=\frac{\pi }2{}+\pi k$

$\pi = \pi k$

So when , then

$\frac{\pi }{2}+\pi \left ( 1 \right )=\frac{3\pi }{2}$ , so the given statement is true.

Given the function $\cot (\theta -\frac{\pi }{3})+5$ , determine the equation of all vertical asymptotes across the domain. Let be any integer.

$y=\frac{\pi }{3}+\pi$

$y=\frac{\pi }{3}$

$y=\pi k$

$y=\frac{\pi }{3}+\pi k$

Explanation

We know that the parent function $y=\cot (\theta )$ has vertical asymptotes at $\pi k$ where is any integer. We will set the quantity inside the $\cot$ function equal to zero to solve for the shift of the asymptote.

$\theta -\frac{\pi }{3}=0$

$\theta =\frac{\pi }{3}$

Now we must add this to the asymptotes of the parent function:

$\frac{\pi }{3}+\pi k$

Assume that there is a vertical asymptote for the function $tan(\theta -\frac{\pi }{6})$ at $\frac{5\pi }{3}$ , solve for from the equation of all vertical asymptotes at $tan(\theta -\frac{\pi }{6})$ .

Explanation

We know that the parent function $\tan \left ( \theta \right )$ has vertical asymptotes at $\frac{\pi }{2}+\pi k$ . So now we will set the inner quantity of the $\tan$ function equal to zero to find the shift of the asymptote.