# How do you find the asymptotes for (x^3-x)/(x^3-4x)?

Dec 11, 2015

Horizontal asymptote: $y = 1$

Vertical asymptote: $x = 2$ and $x = - 2$

Discontinuity at $\left(0 , \frac{1}{4}\right)$

#### Explanation:

Given $f \left(x\right) = \frac{{x}^{3} - x}{{x}^{3} - 4 x}$

Let's start by factoring the function $f \left(x\right) = \frac{x \left({x}^{2} - 1\right)}{x \left({x}^{2} - 4\right)} \implies \frac{x \left(x - 1\right) \left(x + 1\right)}{\left(x\right) \left(x - 2\right) \left(x + 2\right)}$

If you notice, there is a same factor of $x$ on numerator and denominator, which can reduce the function to

$f \left(x\right) = \frac{\left(x - 1\right) \left(x + 1\right)}{\left(x - 2\right) \left(x + 2\right)} , x \ne 0$

Discontinuity aka hole on the graph at
$f \left(0\right) = \frac{\left(0 - 1\right) \left(0 + 1\right)}{\left(0 - 2\right) \left(0 + 2\right)} = \frac{1}{4}$

Discontinuity at (0, 1/4)

Horizontal asymptote : $y = 1$

Since degree and coefficient of the numerator and denominator are the same, hence the horizontal asymptote is $y = 1$

Vertical asymptote

Set the denominator of the reduce function equal to zero

x-2 = 0 ; x + 2 = 0 
x = 2 ; and x= -2#
graph{(x^3-x)/(x^3-4x) [-7.024, 7.024, -3.51, 3.51]}