Question 1

For the rational function

f(x) = \frac{2x^2 + 3x - 1}{x^2 - 4}

, what are the vertical asymptotes?

Accepted Answer

The vertical asymptotes are at

f(x) = \frac{2x^2 + 3x - 1}{x^2 - 4}

and

x = -2

.

Answer

There are no vertical asymptotes.

Answer

The vertical asymptotes are at

f(x) = \frac{2x^2 + 3x - 1}{x^2 - 4}

and

x = -2

.

Answer

The vertical asymptote is at

f(x) = \frac{2x^2 + 3x - 1}{x^2 - 4}

.

Question 2

Which of the following rational functions has a vertical asymptote at

f(x) = \frac{2x^2 + 3x - 1}{x^2 - 4}

and a horizontal asymptote at

x = -2

?

Accepted Answer

f(x) = \frac{1}{x - 2}

Answer

f(x) = \frac{1}{x - 2}

Answer

f(x) = \frac{1}{x - 2}

Answer

f(x) = \frac{1}{x - 2}

Question 3

What is the horizontal asymptote of the rational function

f(x) = \frac{2x^2 + 3x - 1}{x^2 - 4}

?

Accepted Answer

f(x) = \frac{1}{x - 2}

Answer

f(x) = \frac{1}{x - 2}

Answer

f(x) = \frac{1}{x - 2}

Answer

f(x) = \frac{1}{x - 2}

Question 4

Which of the following rational functions has a vertical asymptote at

f(x) = \frac{2x^2 + 3x - 1}{x^2 - 4}

and a horizontal asymptote at

x = -2

?

Accepted Answer

f(x) = \frac{1}{x - 2}

Answer

f(x) = \frac{1}{x - 2}

Answer

f(x) = \frac{1}{x - 2}

Answer

f(x) = \frac{1}{x - 2}

Question 5

What are the horizontal and vertical asymptotes of the rational function

f(x) = \frac{2x^2 + 3x - 1}{x^2 - 4}

?

Accepted Answer

Horizontal asymptote:

f(x) = \frac{2x^2 + 3x - 1}{x^2 - 4}

. Vertical asymptote:

x = -2

.

Answer

Horizontal asymptote:

f(x) = \frac{2x^2 + 3x - 1}{x^2 - 4}

. Vertical asymptote:

x = -2

.

Answer

Horizontal asymptote:

f(x) = \frac{2x^2 + 3x - 1}{x^2 - 4}

. Vertical asymptote:

x = -2

.

Answer

Horizontal asymptote:

f(x) = \frac{2x^2 + 3x - 1}{x^2 - 4}

. Vertical asymptote:

x = -2

.

Question 6

Given the rational function

f(x) = \frac{2x^2 + 3x - 1}{x^2 - 4}

, what is its vertical asymptote?

Accepted Answer

The function has a hole at

f(x) = \frac{2x^2 + 3x - 1}{x^2 - 4}

, not a vertical asymptote.

Answer

The vertical asymptote is at

f(x) = \frac{2x^2 + 3x - 1}{x^2 - 4}

.

Answer

The function has no vertical asymptote or hole.

Answer

The vertical asymptote is at

f(x) = \frac{2x^2 + 3x - 1}{x^2 - 4}

.

Question 7

For the rational function

f(x) = \frac{2x^2 + 3x - 1}{x^2 - 4}

, what are the vertical asymptotes?

Accepted Answer

The vertical asymptotes are at

f(x) = \frac{2x^2 + 3x - 1}{x^2 - 4}

and

x = -2

.

Answer

The vertical asymptotes are at

f(x) = \frac{2x^2 + 3x - 1}{x^2 - 4}

and

x = -2

.

Answer

There are no vertical asymptotes.

Answer

There is a vertical asymptote at

f(x) = \frac{2x^2 + 3x - 1}{x^2 - 4}

only.

Question 8

What are the horizontal and vertical asymptotes of the rational function

f(x) = \frac{2x^2 + 3x - 1}{x^2 - 4}

?

Accepted Answer

No horizontal asymptote. Vertical asymptotes at

f(x) = \frac{2x^2 + 3x - 1}{x^2 - 4}

and

x = -2

.

Answer

Horizontal asymptote at

f(x) = \frac{2x^2 + 3x - 1}{x^2 - 4}

. No vertical asymptote.

Answer

Horizontal asymptote at

f(x) = \frac{2x^2 + 3x - 1}{x^2 - 4}

. Vertical asymptotes at

x = -2

and

x = 0

.

Answer

Horizontal asymptote at

f(x) = \frac{2x^2 + 3x - 1}{x^2 - 4}

. Vertical asymptote at

x = -2

.

Question 9

Determine the horizontal asymptote of the function

f(x) = \frac{2x^2 + 3x - 1}{x^2 - 4}

.

Accepted Answer

f(x) = \frac{1}{x - 2}

Answer

There is no horizontal asymptote.

Answer

f(x) = \frac{1}{x - 2}

Answer

f(x) = \frac{1}{x - 2}

Question 10

Given the rational function

f(x) = \frac{2x^2 + 3x - 1}{x^2 - 4}

, where are the vertical asymptotes located?

Accepted Answer

The vertical asymptotes are at

f(x) = \frac{2x^2 + 3x - 1}{x^2 - 4}

and

x = -2

.

Answer

There is a vertical asymptote at

f(x) = \frac{2x^2 + 3x - 1}{x^2 - 4}

only.

Answer

The function has no vertical asymptotes.

Answer

The vertical asymptotes are at

f(x) = \frac{2x^2 + 3x - 1}{x^2 - 4}

and

x = -2

.

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