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High School Math : Algebra II

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Example Questions

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 : Parabolic Functions

What are the -intercepts of the equation?

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

Possible Answers:

x=3

x=3,-3

There are no -intercepts.

x=4,-4

x=2

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 And Graphing Exponential Equations

Find the vertical asymptote of the equation.

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

Possible Answers:

x=4

x=1

x=0

There are no vertical asymptotes.

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 #11 : Understanding Asymptotes

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

What is the horizontal asymptote of this equation?

Possible Answers:

y=-2

y=1

There is no horizontal asymptote.

y=32

y=0

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 #1 : Solving Exponential Equations

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

Possible Answers:

$\frac{2}{3}$

$\frac{3}{2}$

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 #2 : Solving Exponential Equations

Solve for :

2x^2 - 32 =0

Possible Answers:

x = 16

x = -4

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

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

x = 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 #1 : Solving Exponential Equations

Solve for (nearest hundredth):

$8^{y} = 100$

Possible Answers:

1.23

2.21

2.53

1.10

1.78

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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Example Question #1 : Solving Exponential Equations

Solve for (nearest hundredth):

$4 ^{2x - 7} = \frac{1}{256}$

Possible Answers:

x=-4

x=2

x = - 5.5

x = 1.5 x = 1 .5

x = 5.5

Correct answer:

x = 1.5 x = 1 .5

Explanation:

$\frac{1}{256} = \frac{1}{4 ^{4}} = 4 ^{-4}$ , so $4 ^{2x - 7} = \frac{1}{256}$ can be rewritten as

$4 ^{2x - 7} = 4 ^{-4}$

$\log_{4} 4 ^{2x - 7} =\log_{4} 4 ^{-4}$

2x - 7 = -4

2x - 7 + 7 = -4+ 7

2x = 3

x = 1 .5

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Example Question #5 : Solving Exponential Equations

Solve for (nearest hundredth):

$80 ^{t} = 39$

Possible Answers:

t=1.20

t=0.62

t=0.72

t=0.31

t =0.84

Correct answer:

t =0.84

Explanation:

One method: Take the natural logarithm of both sides and solve for :

$80 ^{t} = 39$

$\ln 80 ^{t} = \ln 39$

$t \ln 80 = \ln 39$

$t =\frac{ \ln 39}{ \ln 80} \approx \frac{ 3.66}{ 4.38} \approx 0.84$

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Example Question #1 : Solving Exponential Equations

Solve for :

$3 ^{4x + 7} = 9 ^{2x +3}$

Possible Answers:

The equation has no solution.

x = 1

x=0

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

x = 2

Correct answer:

The equation has no solution.

Explanation:

Since $9 = 3^{2}$ , we can rewrite this equation by subsituting and applying the power rule:

$3 ^{4x + 7} = 9 ^{2x +3}$

$3 ^{4x + 7} = \left (3 ^{2} \right ) ^{2x +3}$

$3 ^{4x + 7} =3 ^{2 \left ( 2x +3 \right )}$

$3 ^{4x + 7} =3 ^{4x +6}$

$\log_{3} 3 ^{4x + 7} = \log_{3} 3 ^{4x +6}$

4x + 7 = 4x + 6

4x + 7 -4x = 4x + 6-4x

7 = 6

This statement is identically false, which means that the original equation is identically false. There is no solution.

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All High School Math Resources

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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 #21 : Exponents

Example Question #1 : Parabolic Functions

Example Question #1 : Solving And Graphing Exponential Equations

Example Question #11 : Understanding Asymptotes

Example Question #1 : Solving Exponential Equations

Example Question #2 : Solving Exponential Equations

Example Question #1 : Solving Exponential Equations

Example Question #1 : Solving Exponential Equations

Example Question #5 : Solving Exponential Equations

Example Question #1 : Solving Exponential Equations

All High School Math Resources

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