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High School Math : Understanding Asymptotes

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Example Question #21 : Exponents

Find the vertical asymptote(s) of $y=\frac{x^3+2x+8}{x^2-36}$ .

Possible Answers:

x=6

x=36

There are no real vertical asymptotes.

x=-6 and x=6

$x=\frac{3}{2}$

Correct answer:

x=-6 and x=6

Explanation:

To find the vertical asymptotes, we set the denominator of the fraction equal to zero, as dividing anything by zero is "undefined." Since it's undefined, there's no way for us to graph that point!

Take our given equation, $y=\frac{x^3+2x+8}{x^2-36}$ , and now set the denominator equal to zero:

0=x^2-36 .

36=x^2

$\sqrt{36}=\sqrt{x^2}$

Don't forget, the root of a positive number can be both positive or negative ( -6^2=36 as does 6^2 ), so our answer will be $\pm6=x$ .

Therefore the vertical asymptotes are at x=-6 and x=6 .

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Example Question #121 : Algebra Ii

Find the horizontal asymptote(s) of $y=\frac{x^3+2x+8}{x^2-36}$ .

Possible Answers:

y=1

There are no real horizontal asymptotes.

y=0

y=6 and y=-6

$y=\frac{3}{2}$

Correct answer:

There are no real horizontal asymptotes.

Explanation:

To find the horizontal asymptote of the function, look at the variable with the highest exponent. In the case of our equation, $y=\frac{x^3+2x+8}{x^2-36}$ , the highest exponent is x^3 in the numerator.

When the variable with the highest exponent is in the numberator, there are NO horizontal asymptotes. Horizontal asymptotes only appear when the greatest exponent is in the denominator OR when the exponents have same power in both the denominator and numerator.

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Example Question #1 : Quadratic Functions

What are the -intercepts of the equation?

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

Possible Answers:

x=3

x=3,-3

x=2

x=4,-4

There are no -intercepts.

Correct answer:

x=3,-3

Explanation:

To find the x-intercepts of the equation, we set the numerator equal to zero.

0=x^2-9

9=x^2

$\sqrt{9}=\sqrt{x^2}$

3,-3=x

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Example Question #3 : 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:

x=4, -4

x=1

x=0

x=4

There are no vertical asymptotes.

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 #1311 : High School Math

$y=\frac{x^2-64}{x+2}$

What is the horizontal asymptote of this equation?

Possible Answers:

There is no horizontal asymptote.

y=0

y=1

y=-2

y=32

Correct answer:

There is no horizontal asymptote.

Explanation:

Since the exponent of the leading term in the numerator is greater than the exponent of the leading term in the denominator, there is no horizontal asymptote.

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High School Math : Understanding Asymptotes

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #21 : Exponents

Example Question #121 : Algebra Ii

Example Question #1 : Quadratic Functions

Example Question #3 : Find The Equations Of Vertical Asymptotes Of Tangent, Cosecant, Secant, And Cotangent Functions

Example Question #1311 : High School Math

All High School Math Resources

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