# How do you find the vertical, horizontal and slant asymptotes of: #(x^2-1)/(x^2+4)#?

##### 1 Answer

#### Answer:

vertical asymptote: does not exist

horizontal asymptote:

slant asymptote: does not exist

#### Explanation:

**Finding the Vertical Asymptote**

Given,

#f(x)=(x^2-1)/(x^2+4)#

Factor the numerator.

#f(x)=((x+1)(x-1))/(x^2+4)#

Cancel out any factors that appear in the numerator and denominator. Since there aren't any, set the denominator equal to

#x^2+4=0#

#x^2=-4#

#x=+-sqrt(-4)#

Since you can't take the square root of a negative number in the domain of real numbers, there is no vertical asymptote.

#:.# , the vertical asymptote does not exist.

**Finding the Horizontal Asymptote**

Given,

#f(x)=(color(darkorange)1x^2-1)/(color(purple)1x^2+4)#

Divide the **numerator** by the **denominator**.

#f(x)=color(darkorange)1/color(purple)1#

#color(green)(|bar(ul(color(white)(a/a)color(black)(f(x)=1)color(white)(a/a)|)))#

**Finding the Slant Asymptote**

Given,

#f(x)=(x^2-1)/(x^2+4)#

There would be a slant asymptote if the degree of the leading term in the numerator was

#:.# , the slant asymptote does not exist.