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High School Math : Mathematical Relationships and Basic Graphs

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Example Questions

Example Question #101 : Algebra Ii

Order the following from least to greatest:

$25^{100}$

$2^{300}$

$3^{500}$

$4^{400}$

$2^{600}$

Possible Answers:

$2^{600}, 2^{300}, 25^{100}, 3^{500}, 4^{400}$

$3^{500}, 4^{400}, 2^{300}, 25^{100}, 2^{600}$

$2^{300}, 25^{100}, 2^{600}, 3^{500}, 4^{400}$

$25^{100}, 2^{300}, 2^{600}, 3^{500}, 4^{400}$

$2^{300}, 2^{600}, 3^{500}, 25^{100}, 4^{400}$

Correct answer:

$2^{300}, 25^{100}, 2^{600}, 3^{500}, 4^{400}$

Explanation:

In order to solve this problem, each of the answer choices needs to be simplified.

Instead of simplifying completely, make all terms into a form such that they have 100 as the exponent. Then they can be easily compared.

2^3^0^0 = (2^3)^1^0^0 = 8^1^0^0 , 3^5^0^0 = (3^5)^1^0^0 = 243^1^0^0 , 4^4^0^0 = (4^4)^1^0^0 = 256^1^0^0 , and 2^6^0^0 = (2^6)^1^0^0 = 64^1^0^0 .

Thus, ordering from least to greatest: 8^1^0^0, 25^1^0^0, 64^1^0^0, 243^1^0^0, 256^1^0^0 .

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Example Question #1 : Expressions & Equations

Simplify the expression:

$(3x^4y^2)(4xy^2)^{-3}$

Possible Answers:

$\frac{3x}{16y^3}$

Cannot be simplified

$\frac{3x}{64y^4}$

$\frac{3x}{64y^3}$

Correct answer:

$\frac{3x}{64y^4}$

Explanation:

Begin by distributing the exponent through the parentheses. The power rule dictates that an exponent raised to another exponent means that the two exponents are multiplied:

$(3x^{4}y^2)(4xy^2)^{-3}= (3xy^2)(4^{-3}x^{-3}y^{-6})$

Any negative exponents can be converted to positive exponents in the denominator of a fraction:

$\frac{3x^4y^2}{64x^3y^6}$

The like terms can be simplified by subtracting the power of the denominator from the power of the numerator:

$\frac{3x^4y^2}{64x^3y^6}=\frac{3}{64}x^{4-3}y^{2-6}=\frac{3}{64}xy^{-4}$

$\frac{3x}{64y^4}$

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Example Question #12 : Exponents

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

What are the y-intercepts of this equation?

Possible Answers:

$y=\frac{9}{16}$

y=3,-3

y=0

y=4,-4

$y=\frac{16}{9}$

Correct answer:

$y=\frac{9}{16}$

Explanation:

To find the y-intercepts, set the value equal to and solve.

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

$y=\frac{(0)^2-9}{(0)^2-16}$

$y=\frac{-9}{-16}$

$y=\frac{9}{16}$

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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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High School Math : Mathematical Relationships and Basic Graphs

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #101 : Algebra Ii

Example Question #1 : Expressions & Equations

Example Question #12 : Exponents

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

All High School Math Resources

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