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

1 Answer
Apr 30, 2016

Answer:

vertical asymptote: #x=0#
horizontal asymptote: #f(x)=1#
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 #0# and solve for #x#.

#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 #color(darkorange)("leading coefficient")# of the leading term in the numerator by the #color(purple)("leading coefficient")# of the leading term in 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 #1# value larger than the degree of the leading term in the denominator. In your case, we see that the degree in both the numerator and denominator are equal.

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