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

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Answer 1 · 2016-11-12T13:34:09

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

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]}

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