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High School Math : Mathematical Relationships and Basic Graphs

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Example Questions

Example Question #1 : Solving And Graphing Exponential Equations

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

What is the horizontal asymptote of this equation?

Possible Answers:

y=3

y=2

y=0

There is no horizontal asymptote.

y=1

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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Example Question #21 : Exponents

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

Possible Answers:

There are no real horizontal asymptotes.

y=0

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

y=1

y=6 and y=-6

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,-3

x=4,-4

x=2

x=3

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 #1 : Solving Exponential Functions

Find the vertical asymptote of the equation.

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

Possible Answers:

x=4

There are no vertical asymptotes.

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 #111 : Mathematical Relationships And Basic Graphs

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

What is the horizontal asymptote of this equation?

Possible Answers:

y=-2

y=0

y=32

y=1

There is no horizontal asymptote.

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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Example Question #122 : Algebra Ii

Which value for satisfies the equation $4^x = \frac{192}{x}$ ?

Possible Answers:

$\frac{3}{2}$

$\frac{2}{3}$

Correct answer:

Explanation:

is the only choice from those given that satisfies the equation. Substition of for gives:

$4^3 = \frac{192}{3} \rightarrow 64 = 64$

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Example Question #123 : Algebra Ii

Solve for :

2x^2 - 32 =0

Possible Answers:

x = 4

$x = \{-4, 4\}$

x = -4

x = 16

$x = \{-4, 0, 4\}$

Correct answer:

$x = \{-4, 4\}$

Explanation:

To solve for in the equation 2x^2 - 32 =0

Factor out of the expression on the left of the equation:

$\rightarrow 2(x^2 - 16) =0$

Use the "difference of squares" technique to factor the parenthetical term on the left side of the equation.

$\rightarrow 2(x+4)(x-4)=0$

Any variable that causes any one of the parenthetical terms to become will be a valid solution for the equation. x+4 becomes when is , and x-4 becomes when is , so the solutions are and .

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Example Question #2 : Solving Exponential Equations

Solve for (nearest hundredth):

$8^{y} = 100$

Possible Answers:

2.21

1.78

1.10

1.23

2.53

Correct answer:

2.21

Explanation:

Take the common logarithm of both sides and solve for :

$8^{y} = 100$

$\log 8^{y} = \log 100$

$y \log 8 =2$

$y =\frac{2}{\log 8} \approx \frac{2}{0.9031} \approx 2.21$

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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 #1 : Solving And Graphing Exponential Equations

Example Question #1 : Understanding Asymptotes

Example Question #115 : Algebra Ii

Example Question #21 : Exponents

Example Question #1 : Quadratic Functions

Example Question #1 : Solving Exponential Functions

Example Question #111 : Mathematical Relationships And Basic Graphs

Example Question #122 : Algebra Ii

Example Question #123 : Algebra Ii

Example Question #2 : Solving Exponential Equations

All High School Math Resources

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