How do you find the Vertical, Horizontal, and Oblique Asymptote given #f(x)= -1/(x+1)^2#?

1 Answer
Apr 28, 2016

Answer:

vertical asymptote: #x=-1#
horizontal asymptote: #f(x)=0#
oblique asymptote: does not exist

Explanation:

Finding the Vertical Asymptote

Given,

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

Set the denominator equal to #0# and solve for #x#.

#(x+1)^2=0#

#x+1=0#

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

Finding the Horizontal Asymptote

Given,

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

If you foil the denominator, you will notice that the degree of the denominator is #2#.

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

Since the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is #f(x)=0#.

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

Finding the Oblique Asymptote

Given,

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

There would be a slant asymptote if the degree of the leading term in the numerator is #1# value larger than the degree of the leading term in the denominator. In your case, we see the opposite — ie. the degree in the denominator is greater than the degree in the numerator.

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