Question

What are the asymptotes for `(x-4)/(16x-x^3)`?

Answer 1

The vertical asymptotes are `x=0 and x=4 and x=-4`
The horizontal asymptote is the line `y=0`
.

Explanation:

Set the denominator `16x-x^3` equal to zero, and solve for `x`.

`16x-x^3=0`

Factor out `x`.

`x(16-x^2)=0`

`x=0`

`x=0`

`16-x^2=0`

Factor to get

`(4+x)(4-x) =0`

`4+x =0` so `x = -4`

`4-x = 0` so `x=4`

The horizontal asymptote is `y = 0` (The `x`-axis)

because as `x` increases without bound, the value of

`(x-4)/(16x-x^3) = (cancel(x^3) (1/x^2 -4/x^3))/(cancel(x^3)(16/x^2 - 1))` gets closer and closer to `0/1`.