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

Card 1 of 32

0

Didn't Know

0

Didn't Know

Knew It

0

1 of 2019 left

Question

Find the vertical asymptote of the equation.

Tap to reveal answer

Answer

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

← Didn't Know|Knew It →

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.

Tap to reveal answer

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$

← Didn't Know|Knew It →

Question

Which of the choices represents asymptote(s), if any?

Tap to reveal answer

Answer

Factor the numerator and denominator.

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.

$x-9\neq0$

$x\neq9$

The asymptote is located at .

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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}$$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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}$+\pi k$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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.

← Didn't Know|Knew It →

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}$)$ .

Tap to reveal answer

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}$=k$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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}$+\pi k$

← Didn't Know|Knew It →

Question

Find the vertical asymptote of the equation.

Tap to reveal answer

Answer

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

← Didn't Know|Knew It →

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.

Tap to reveal answer

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$

← Didn't Know|Knew It →

Question

Which of the choices represents asymptote(s), if any?

Tap to reveal answer

Answer

Factor the numerator and denominator.

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.

$x-9\neq0$

$x\neq9$

The asymptote is located at .

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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}$$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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.

← Didn't Know|Knew It →

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}$)$ .

Tap to reveal answer

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}$=k$

← Didn't Know|Knew It →

Question

Find the vertical asymptote of the equation.

Tap to reveal answer

Answer

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

← Didn't Know|Knew It →

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.

Tap to reveal answer

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$

← Didn't Know|Knew It →

Question

Which of the choices represents asymptote(s), if any?

Tap to reveal answer

Answer

Factor the numerator and denominator.

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.

$x-9\neq0$

$x\neq9$

The asymptote is located at .

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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}$$

← Didn't Know|Knew It →

0

Knew It