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High School Math : Algebra II

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Example Questions

Example Question #1 : Solving And Graphing Exponential Equations

$y=\frac{4x^2-2}{2x^3-27}$

What are the horizontal asymptotes of this equation?

Possible Answers:

y=2

$y=\frac{1}{2}$

y=0

There are no horizontal asymptotes.

y=1

Correct answer:

y=0

Explanation:

When looking for the horizontal asymptotes, examine the exponents of the variables. Because the variable in the denominator has a higher exponent than the variable in the numerator, the horizontal asymptote will be at y=0 .

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Example Question #2 : Solving And Graphing Exponential Equations

$y=\frac{4x^2-2}{2x^3-27}$

What are the vertical asymptotes of the equation?

Possible Answers:

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

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

$x=\frac{3\sqrt[3]{4}}{2}$

$x=\frac{3\sqrt[3]{4}}{2}, -\frac{3\sqrt[3]{4}}{2}$

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

Correct answer:

$x=\frac{3\sqrt[3]{4}}{2}$

Explanation:

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

0=2x^3-27

27=2x^3

$\frac{27}{2}=x^3$

$\sqrt[3]{\frac{27}{2}}=\sqrt[3]{x^3}$

${\frac{\sqrt[3]{27}}{\sqrt[3]2}}=\sqrt[3]{x^3}$

$\frac{3}{\sqrt[3]{2}}=x$

However, we need to rationalize from here. We need to get rid of the cubed root in the denominator.

$(\sqrt[3]{2})^3=\sqrt[3]{2}*\sqrt[3]{2}*\sqrt[3]{2}$ .

Therefore:

$\frac{3}{\sqrt[3]{2}}*\frac{\sqrt[3]{2}}{\sqrt[3]{2}}*\frac{\sqrt[3]{2}}{\sqrt[3]{2}}=x$

$\frac{3*(\sqrt[3]{2})^2}{(\sqrt[3]{2})^3}=x$

Bring the exponent from the numerator under the radical:

$\frac{3\sqrt[3]{2^2}}{2}=x$

Simplify:

$\frac{3\sqrt[3]{4}}{2}=x$

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Example Question #1 : Solving And Graphing Exponential Equations

$y=\frac{x-3}{x^2-12}$

What is the horizontal asymptote of this equation?

Possible Answers:

y=1

There is no horizontal asymptote.

$y=\sqrt{12}$

y=3

y=0

Correct answer:

y=0

Explanation:

To find the horizontal asymptotes, we compare the exponents of in our fraction. Because the denominator variable's exponent is greater than the numerator variable's exponent, our horizontal asymptote is at y=0 .

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Example Question #2 : Solving And Graphing Exponential Equations

$y=\frac{x-3}{x^2-12}$

What are the vertical asymptotes of the equation?

Possible Answers:

$x=2\sqrt{3},-2\sqrt{3}$

$x=\sqrt{12}$

There are no vertical asymptotes.

x=3

x=0

Correct answer:

$x=2\sqrt{3},-2\sqrt{3}$

Explanation:

To find the vertical asymptotes, we set the denominator equal to zero.

0=x^2-12

12=x^2

$\sqrt{12}=\sqrt{x^2}$

$\sqrt{4*3}=x$

Because the square root only gives us the absolute value, our answer will be:

$x=2\sqrt{3}, -2\sqrt{3}$

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Example Question #4 : Understanding Asymptotes

$y=\frac{3x^2-5}{2x^2+7}$

What are the horizontal asymptotes of this equation?

Possible Answers:

There are no horizontal asymptotes.

y=0

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

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

y=1

Correct answer:

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

Explanation:

Since the exponents of the variables in both the numerator and denominator are equal, the horizontal asymptote will be the coefficient of the numerator's variable divided by the coefficient of the denominator's variable.

For this problem, since we have $\frac{3x^2}{2x^2}$ , our asymptote will be $y=\frac{3}{2}$ .

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Example Question #3 : Solving And Graphing Exponential Equations

$y=\frac{3x^2-5}{2x^2+7}$

What are the vertical asymptotes of the equation?

Possible Answers:

There are no vertical asymptotes.

$x=-\sqrt{\frac{7}{2}}, \sqrt{\frac{7}{2}}$

$x=-\sqrt{\frac{7}{2}}$

$x=\frac{7}{2}$

There are no real vertical asymptotes.

Correct answer:

There are no real vertical asymptotes.

Explanation:

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

$0={2x^2+7}$

-7=2x^2

$\frac{-7}{2}=x^2$

$\sqrt{-\frac{7}{2}}=\sqrt{x^2}$

$\sqrt{-\frac{7}{2}}=x$

Since we'd be trying to find a negative number, we have no real solution. Therefore, there are no real vertical asymptotes.

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Example Question #4 : Solving And Graphing Exponential Equations

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

What are the vertical asymptotes of this equation?

Possible Answers:

x=8,-8

x=-2

There are no real vertical asymptotes for this function.

x=2,-2

x=0

Correct answer:

x=-2

Explanation:

To find the vertical asymptotes, we set the denominator equal to zero.

0=x+2

-2=x

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Example Question #1 : Understanding Asymptotes

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

What is the horizontal asymptote of this equation?

Possible Answers:

y=1

y=0

y=3

y=2

There is no horizontal asymptote.

Correct answer:

y=1

Explanation:

Look at the exponents of the variables. Both our numerator and denominator are 1x^2 . Therefore the horizontal asymptote is calculated by dividing the coefficient of the numerator by the coefficient of the denomenator.

$y=\frac{1}{1}=1$

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Example Question #1 : Understanding Asymptotes

Find the vertical asymptote(s) of $y=\frac{x^2+9x+8}{x^2-44}$ .

Possible Answers:

$x=2\sqrt{11}$

There are no real vertical asymptotes for this function.

$x=2\sqrt{11}$ and $x=-2\sqrt{11}$

x=1

x=-1 and x=-8

Correct answer:

$x=2\sqrt{11}$ and $x=-2\sqrt{11}$

Explanation:

To find the vertical asymptotes, we set the denominator of the fraction equal to zero, as dividing anything by zero is undefined.

Take our given equation, $y=\frac{x^2+9x+8}{x^2-44}$ , and now set the denominator equal to zero:

0=x^2-44

44=x^2

$\sqrt{44}=\sqrt{x^2}$

is not a perfect square, but let's see if we can pull anything out.

$\sqrt{4*11}=\sqrt{x^2}$

$\sqrt{4}*\sqrt{11}=\sqrt{x^2}$

$2\sqrt{11}=x$

Don't forget that there is a negative result as well:

$x=-2\sqrt{11}$ .

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Example Question #115 : Algebra Ii

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

Possible Answers:

x=-6 and x=6

x=36

x=6

There are no real vertical asymptotes.

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

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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High School Math : Algebra II

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 : Solving And Graphing Exponential Equations

Example Question #2 : Solving And Graphing Exponential Equations

Example Question #1 : Solving And Graphing Exponential Equations

Example Question #2 : Solving And Graphing Exponential Equations

Example Question #4 : Understanding Asymptotes

Example Question #3 : Solving And Graphing Exponential Equations

Example Question #4 : Solving And Graphing Exponential Equations

Example Question #1 : Understanding Asymptotes

Example Question #1 : Understanding Asymptotes

Example Question #115 : Algebra Ii

All High School Math Resources

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