How do you find the Vertical, Horizontal, and Oblique Asymptote given #(x^2-4)/(x^3+4x^2)#?

1 Answer

Answer:

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

Explanation:

Let rewrite the function
#(x^2-4)/(x^3+4x^2)=(x^2-4)/(x^2(x+4))#

As we cannot divide by #0#, the function is not defined for #x=0# and #x=-4#
So the vertical asymptotes are #x=0# and #x=-4#

The degree of the denominator is greater than the degree of the numeratos, so we don't have oblique asymptotes

Let's find the limit of the function as #x->oo#

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

limit #(x^2-4)/(x^3+4x^2)=x^2/x^3=1/x=0^+#
#x->+oo#
So #y=0# is a horizontal asymptote
graph{(x^2-4)/(x^3+4x^2) [-10, 10, -5, 5]}