Answered

Give the equations of any vertical, horizontal, or oblique asymptotes for the graph of the rational function

[tex] f(x)=\frac{x^2-8}{x+5} [/tex]

Vertical Asymptote:

Select the correct answer below and, if necessary, fill in the answer box to complete your choice.

A. The equation of the vertical asymptote is [tex] x = \square [/tex].
(Type an integer or a simplified fraction.)

B. There is no vertical asymptote.

Horizontal Asymptote:

Select the correct answer below and, if necessary, fill in the answer box to complete your choice.

A. The equation of the horizontal asymptote is [tex] y = \square [/tex].
(Type an integer or a simplified fraction.)

B. There is no horizontal asymptote.

Oblique Asymptote:

Select the correct answer below and, if necessary, fill in the answer box to complete your choice.

A. The equation of the oblique asymptote is [tex] y = \square [/tex].
(Type an integer or a simplified fraction.)

B. There is no oblique asymptote.



Answer :

To find the equations of the vertical, horizontal, or oblique asymptotes for the graph of the rational function [tex]\( f(x) = \frac{x^2 - 8}{x + 5} \)[/tex], we follow a step-by-step procedure:

### Vertical Asymptote

A vertical asymptote occurs where the denominator of the rational function is zero, provided that the numerator is not zero at the same point.

1. Set the denominator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ x + 5 = 0 \][/tex]
2. Solve for [tex]\( x \)[/tex]:
[tex]\[ x = -5 \][/tex]

Therefore, the equation of the vertical asymptote is:
[tex]\[ x = -5 \][/tex]

So, the correct answer for the vertical asymptote is:

A. The equation of the vertical asymptote is [tex]\( x= -5 \)[/tex].

### Horizontal Asymptote

A horizontal asymptote is determined by comparing the degrees of the numerator and the denominator.

1. The degree of the numerator [tex]\( x^2 - 8 \)[/tex] is 2.
2. The degree of the denominator [tex]\( x + 5 \)[/tex] is 1.

Since the degree of the numerator (2) is greater than the degree of the denominator (1), there is no horizontal asymptote for this function.

So, the correct answer for the horizontal asymptote is:

B. There is no horizontal asymptote.

### Oblique Asymptote

An oblique (or slant) asymptote occurs when the degree of the numerator is exactly one more than the degree of the denominator.

1. Since the degree of the numerator [tex]\( x^2 - 8 \)[/tex] is 2 and the degree of the denominator [tex]\( x + 5 \)[/tex] is 1, we expect an oblique asymptote.
2. To find the equation of the oblique asymptote, perform polynomial long division of [tex]\( x^2 - 8 \)[/tex] by [tex]\( x + 5 \)[/tex]:

Dividing [tex]\( x^2 - 8 \)[/tex] by [tex]\( x + 5 \)[/tex]:
- [tex]\( x^2 / x = x \)[/tex]
- Multiply [tex]\( x \)[/tex] by [tex]\( x + 5 \)[/tex] to get [tex]\( x^2 + 5x \)[/tex]
- Subtract [tex]\( x^2 + 5x \)[/tex] from [tex]\( x^2 - 8 \)[/tex] to get [tex]\( -5x - 8 \)[/tex]
- Now divide [tex]\( -5x - 8 \)[/tex] by [tex]\( x + 5 \)[/tex]:
- [tex]\( -5x / x = -5 \)[/tex]
- Multiply [tex]\( -5 \)[/tex] by [tex]\( x + 5 \)[/tex] to get [tex]\( -5x - 25 \)[/tex]
- Subtract [tex]\( -5x - 25 \)[/tex] from [tex]\( -5x - 8 \)[/tex] to get [tex]\( 17 \)[/tex]

3. The quotient is [tex]\( x - 5 \)[/tex] and the remainder is 17, which gets ignored for large values of [tex]\( x \)[/tex].

Therefore, the equation of the oblique asymptote is:
[tex]\[ y = x - 5 \][/tex]

So, the correct answer for the oblique asymptote is:

A. The equation of the oblique asymptote is [tex]\( y = x - 5 \)[/tex].

Summarizing all the results:
1. The vertical asymptote is [tex]\( x = -5 \)[/tex].
2. There is no horizontal asymptote.
3. The oblique asymptote is [tex]\( y = x - 5 \)[/tex].