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

1 Answer
Nov 23, 2016

Answer:

The vertical asymptotes are #x=0# and #x=-4#
No slant asymptotes.
The horizontal asymptote is #y=0#

Explanation:

Let #f(x)=(x^2-4)/(x^3+4x^2)#

Let's factorise the denominator
#x^3+4x^2=x^2(x+4)#

The domain of #f(x)# is #D_f(x)=RR-{0,-4}#

As we cannot divide by #0#

So #x!=0# and #x!=-4#

The vertical asymptotes are #x=0# and #x=-4#

As the degree of the numerator is #<# the degree of the denominator, there is no slant asymptotes.

For calculating the limits, we take the terms of highest degree in the numerator and the denominator

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

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

So, #y=0# is a horizontal asymptote

graph{(x^2-4)/(x^3+4x^2) [-10, 10, -5, 5]}