Question

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

Answer 1

Horizontal asymptote: `y= 1`

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

Discontinuity at `(0, 1/4)`

Explanation:

Given `f(x) = (x^3 -x)/(x^3 -4x)`

Let's start by factoring the function `f(x) = (x(x^2-1))/(x(x^2-4) ) => (x(x-1)(x+1))/((x)(x-2)(x+2))`

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

`f(x) =( (x-1)(x+1))/((x-2)(x+2)) , x!=0`

Discontinuity aka hole on the graph at
`f(0) = [(0-1)(0+1)]/[(0-2)(0+2)] = 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]}