# How do you find the vertical, horizontal and slant asymptotes of: #(x^2-25)/(x^2+5x)#?

##### 1 Answer

vertical asymptote:

horizontal asymptote:

slant asymptote: does not exist

#### Explanation:

**Finding the Vertical Asymptote**

Given,

#f(x)=(x^2-25)/(x^2+5x)#

Factor the numerator.

#f(x)=((x+5)(x-5))/(x(x+5)#

Cancel out any factors that appear in the numerator and denominator.

#f(x)=(x-5)/x#

Set the denominator equal to

#color(green)(|bar(ul(color(white)(a/a)color(black)(x=0)color(white)(a/a)|)))#

**Finding the Horizontal Asymptote**

Given,

#f(x)=(color(darkorange)1x^2-25)/(color(purple)1x^2+5x)#

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^2-25)/(x^2+5x)#

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

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