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

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Answer 1 · 2016-10-28T07:04:27

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

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]}

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