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Precalculus : Graphs and Inverses of Trigonometric Functions

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Example Questions

Example Question #52 : Graphing The Sine And Cosine Functions

Find the amplitude of the following trig function: $y=-2-3sin(2\theta-5)$

Possible Answers:

$\frac{3}{2}$

Correct answer:

Explanation:

Rewrite $y=-2-3sin(2\theta-5)$ so that it is in the form of:

y=Asin(Bx+C)+D

$y=-3sin(2\theta-5)-2$

The absolute value of is the value of the amplitude.

$\textup{Amplitude= }\left | A\right |= \left | -3\right |=3$

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Example Question #4 : Find The Amplitude Of A Sine Or Cosine Function

Find the amplitude of the function.

$y=5sin\, x$

Possible Answers:

Correct answer:

Explanation:

For the sine function

$y=Asin\, x$ where $A\epsilon \, \mathbb{R}$

the amplitude is given as $\left | A\right |$ .

As such the amplitude for the given function

$y=5sin\, x$ is

$\left | 5\right |=5$ .

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Example Question #3 : Period And Amplitude

Which of the given functions has the greatest amplitude?

Possible Answers:

$2\sin(x)$

$\sin\frac{2\pi }{3}(x+2)$

$\frac{1}{2}\sin(4x)$

$\sin(x+3)$

$\sin(4x)$

Correct answer:

$2\sin(x)$

Explanation:

The amplitude of a function is the amount by which the graph of the function travels above and below its midline. When graphing a sine function, the value of the amplitude is equivalent to the value of the coefficient of the sine. Similarly, the coefficient associated with the x-value is related to the function's period. The largest coefficient associated with the sine in the provided functions is 2; therefore the correct answer is $2\sin(x)$ .

The amplitude is dictated by the coefficient of the trigonometric function. In this case, all of the other functions have a coefficient of one or one-half.

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Example Question #1 : Find The Equations Of Vertical Asymptotes Of Tangent, Cosecant, Secant, And Cotangent Functions

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.

Possible Answers:

$y=\frac{3\pi}{4} \pm \pi k$

$y=\frac{\pi}{4} \pm 2\pi k$

$y=\frac{\pi}{2} \pm \pi k$

$y=\frac{\pi k}{4}$

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

Correct answer:

$y=\frac{3\pi}{4} \pm \pi k$

Explanation:

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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Example Question #1 : Find The Equations Of Vertical Asymptotes Of Tangent, Cosecant, Secant, And Cotangent Functions

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

Possible Answers:

$x=1, 9\textup{ and } y= -\frac{2}{5}$

x=1, 9

$\textup{There are no asymptotes.}$

x=9

$x=9\textup{ and }y=\frac{2}{5}$

Correct answer:

x=9

Explanation:

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 (x+1) terms will cancel. The hole will be located at x=-1 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 x=9 .

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Example Question #1 : Solving Exponential Functions

Find the vertical asymptote of the equation.

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

Possible Answers:

x=4, -4

x=4

There are no vertical asymptotes.

x=1

x=0

Correct answer:

x=4, -4

Explanation:

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

0=x^2-16

16=x^2

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

4, -4=x

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Example Question #4 : Find The Equations Of Vertical Asymptotes Of Tangent, Cosecant, Secant, And Cotangent Functions

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

Possible Answers:

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

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

$y=\pi k$

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

Correct answer:

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

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:

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

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Example Question #5 : Find The Equations Of Vertical Asymptotes Of Tangent, Cosecant, Secant, And Cotangent Functions

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

Possible Answers:

$y=\frac{k}{2}$

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

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

$y=\pi k$

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

Correct answer:

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

Explanation:

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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Example Question #6 : Find The Equations Of Vertical Asymptotes Of Tangent, Cosecant, Secant, And Cotangent Functions

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

Possible Answers:

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

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

$\theta = \pi k$

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

Correct answer:

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

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

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Example Question #1 : Find The Equations Of Vertical Asymptotes Of Tangent, Cosecant, Secant, And Cotangent Functions

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

Possible Answers:

k=2

k=5

k=1

k=10

Correct answer:

k=1

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.

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

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

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

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

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

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

1=k

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Precalculus : Graphs and Inverses of Trigonometric Functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #52 : Graphing The Sine And Cosine Functions

Example Question #4 : Find The Amplitude Of A Sine Or Cosine Function

Example Question #3 : Period And Amplitude

Example Question #1 : Find The Equations Of Vertical Asymptotes Of Tangent, Cosecant, Secant, And Cotangent Functions

Example Question #1 : Find The Equations Of Vertical Asymptotes Of Tangent, Cosecant, Secant, And Cotangent Functions

Example Question #1 : Solving Exponential Functions

Example Question #4 : Find The Equations Of Vertical Asymptotes Of Tangent, Cosecant, Secant, And Cotangent Functions

Example Question #5 : Find The Equations Of Vertical Asymptotes Of Tangent, Cosecant, Secant, And Cotangent Functions

Example Question #6 : Find The Equations Of Vertical Asymptotes Of Tangent, Cosecant, Secant, And Cotangent Functions

Example Question #1 : Find The Equations Of Vertical Asymptotes Of Tangent, Cosecant, Secant, And Cotangent Functions

All Precalculus Resources

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