Find the Equations of Vertical Asymptotes of Tangent, Cosecant, Secant, and Cotangent Functions - Pre-Calculus

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Question

Find the vertical asymptote of the equation.

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

Answer

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

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

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Question

Given the function $y=tan(\theta-\frac{\pi}{4})+2$ , determine the equation of all the vertical asymptotes across the domain. Let be an integer.

Answer

For the function $y=tan(\theta-\frac{\pi}{4})+2$ , it is not necessary to graph the function. The y-intercept does not affect the location of the asymptotes.

Recall that the parent function $tan(\theta)$ has an asymptote at $\frac{\pi}{2} \pm \pi k$ for every period.

Set the inner quantity of $tan(\theta-\frac{\pi}{4})$ equal to zero to determine the shift of the asymptote.

$\theta-\frac{\pi}{4}=0$

$\theta=\frac{\pi}{4}$

This indicates that there is a zero at $\theta=\frac{\pi}{4}$ , and the tangent graph has shifted $\frac{\pi}{4}$ units to the right. As a result, the asymptotes must all shift $\frac{\pi}{4}$ units to the right as well. The period of the tangent graph is $\pi$ .

$\frac{\pi}{2} \pm \pi k + \frac{\pi}{4} = \frac{3\pi}{4} \pm \pi k$

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Question

Which of the choices represents asymptote(s), if any? $y=\frac{2x^2+2x}{x^2-8x-9}$

Answer

Factor the numerator and denominator.

$y=\frac{2x^2+2x}{x^2-8x-9} = \frac{2x(x+1)}{(x-9)(x+1)}$

Notice that the terms will cancel. The hole will be located at because this is a removable discontinuity.

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

The denominator cannot be equal to zero. Set the denominator to find the location where the x-variable cannot exist.

The asymptote is located at .

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Question

Given the function $y=\csc (\theta +\frac{\pi }{2})$ , determine the equation of all vertical asymptotes across the domain. Let be any integer.

Answer

We know that the parent function $y=\csc (\theta )$ has asymptotes at $\pi k$ where is any integer. We will set the quantity within the $\csc$ function equal to zero in order to find the shift of the asymptote.

$\theta +\frac{\pi }{2}=0$

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

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

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

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Question

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

Answer

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:

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

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Question

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

Answer

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

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Question

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

Answer

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.

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Question

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 $tan(\theta -\frac{\pi }{6})$ .

Answer

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.

$\theta -\frac{\pi }{6}=0$

$\theta =\frac{\pi }{6}$

Now we will add this to the parent function equation for vertical asymptotes

$\frac{\pi }{6}+\frac{\pi }{2}+\pi k$

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

Now we will set this equation for the given vertical asymptote at $\tan \left ( \theta \right )-\frac{\pi }{6}$

$\frac{5\pi }{3}=\frac{2\pi }{3}+\pi k$

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

$\frac{3}{3}=k$

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