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

1 Answer
Apr 28, 2016

Answer:

vertical asymptote: does not exist
horizontal asymptote: #f(x)=1#
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 #0# and solve for #x#.

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

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.