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

Nov 4, 2016

The vertical asymptotes are $x = 2$ and $x = - 2$
The horizontal asymptote is $y = 1$

Explanation:

Let's do some simplification
$f \left(x\right) = \frac{{x}^{3} - x}{{x}^{3} - 4 x} = \frac{x \left({x}^{2} - 1\right)}{x \left({x}^{2} - 4\right)} = \frac{\left(x + 1\right) \left(x - 1\right)}{\left(x + 2\right) \left(x - 2\right)}$

As we cannot divide by $0$,

The degree of the numerator is identical to the degree of the denominator, so there are no oblique asymptotes.
Limit $f \left(x\right) = {x}^{2} / {x}^{2} = 1$
$x \to \pm \infty$

So a horizontal asymptote is $y = 1$
graph{(x^2-1)/(x^2-4) [-10, 10, -5, 5]}