How do you find vertical, horizontal and oblique asymptotes for #(x^2-4)/(x^3+4x^2)#?

1 Answer
Nov 12, 2016

Answer:

The vertical asymptotes are #x=0# and #x=-4#
The horizontal asymptote is #y=0#

Explanation:

The denominator #=x^3+4x^2=(x^2)(x+4)#
as you cannot divide by #0#,
So, #x=0# and #x=-4# are vertical asymptotes.

The degree of the numerator is #<# than the degree of the denominator, there is no oblique asymptote.

#lim_(x->-oo)(x^2-4)/(x^3+4x^2)=lim_(x->-oo)x^2/x^3=lim_(x->-oo)1/x=0^(-)#

#lim_(x->+oo)(x^2-4)/(x^3+4x^2)=lim_(x->+oo)x^2/x^3=lim_(x->+oo)1/x=0^(+)#

Therefore #y=0# is a horizontal asymptote
graph{(x^2-4)/(x^3+4x^2) [-7.436, 8.364, -3.85, 4.05]}