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Algebra II : Asymptotes

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Example Questions

Example Question #1 : Solving Exponential Functions

What is the horizontal asymptote of the graph of the equation $y = -2e^{x-3} -5$ ?

Possible Answers:

y=5

y=-5

$y=-\frac{5}{2}$

y=3

y=-2

Correct answer:

y=-5

Explanation:

The asymptote of this equation can be found by observing that $e^{x-3} > 0$ regardless of . We are thus solving for the value of as $e^{x-3}$ approaches zero.

$-2e^{x-3} < -2\cdot 0$

$-2e^{x-3} < 0$

$-2e^{x-3} -5 < 0-5$

$-2e^{x-3} -5 < -5$

y<-5

So the value that cannot exceed is , and the line y = -5 is the asymptote.

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Example Question #2 : Solving Exponential Functions

What is/are the asymptote(s) of the graph of the function

$f(x) = 5e^{x-2}-2$ ?

Possible Answers:

x=2

y = -2, x=-2

y = -2, x=2

y = -2

x=-2

Correct answer:

y = -2

Explanation:

An exponential equation of the form $f(x) = ae^{x-h} +k$ has only one asymptote - a horizontal one at y = k . In the given function, k = -2 , so its one and only asymptote is y = -2 .

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

Find the vertical asymptote of the equation.

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

Possible Answers:

There are no vertical asymptotes.

x=4

x=1

x=0

x=4, -4

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 #1 : Asymptotes

Determine the asymptotes, if any: $y=\frac{x^2-2x+1}{x^2-1}$

Possible Answers:

x=-1

$x=-\frac{1}{2}$

$x=\pm1$

x=1

$\textup{There are no asymptotes.}$

Correct answer:

x=-1

Explanation:

Factorize both the numerator and denominator.

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

Notice that one of the binomials will cancel.

The domain of this equation cannot include $x=\pm1$ .

The simplified equation is:

$y=\frac{x-1}{x+1}$

Since the (x-1) term canceled, the (x-1) term will have a hole instead of an asymptote.

Set the denominator equal to zero.

x+1=0

Subtract one from both sides.

x=-1

There will be an asymptote at only: x=-1

The answer is: x=-1

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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 : Asymptotes

Where is an asymptote located, if any? $y=\frac{2x-4}{x^2-4}$

Possible Answers:

x=2

$\textup{There is no asymptote.}$

x=-2

y=2

y=-2

Correct answer:

x=-2

Explanation:

Factor the numerator and denominator.

2x-4 = 2(x-2)

x^2-4 = (x+2)(x-2)

Rewrite the equation.

$y=\frac{2x-4}{x^2-4}=\frac{ 2(x-2)}{(x+2)(x-2)}=\frac{2}{x+2}$

Notice that the (x-2) will cancel. This means that the root of (x-2) will be a hole instead of an asymptote.

Set the denominator equal to zero and solve for x.

x+2=0

x=-2

An asymptote is located at: x=-2

The answer is: x=-2

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Example Question #1 : Asymptotes

Consider the exponential function y=2^x+3 . Determine if there are any asymptotes and where they lie on the graph.

Possible Answers:

There are no asymptotes. goes to positive infinity in both the and directions.

There is one vertical asymptote at y=2 .

There is one horizontal asymptote at y=3 .

There is one vertical asymptote at y=3 .

Correct answer:

There is one horizontal asymptote at y=3 .

Explanation:

For positive values, increases exponentially in the direction and goes to positive infinity, so there is no asymptote on the positive -axis. For negative values, as decreases, the term 2^x becomes closer and closer to zero so approaches as we move along the negative axis. As the graph below shows, this is forms a horizontal asymptote.

Exp_asymp

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Algebra II : Asymptotes

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #1 : Solving Exponential Functions

Example Question #2 : Solving Exponential Functions

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

Example Question #1 : Asymptotes

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

Example Question #1 : Asymptotes

Example Question #1 : Asymptotes

All Algebra II Resources

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