Question

What are the asymptote(s) and hole(s), if any, of `f(x) =((x-3)(x+2)*x)/((x^2-x)(x^3-3x^2)`?

Answer 1

`x=0` is an asymptote.

`x=1` is an asymptote.

`(3, 5/18)` is a hole.

Explanation:

First, let's simplify our fraction without cancelling anything out (since we're going to be taking limits and cancelling stuff out might mess with that).

`f(x) = ((x-3)(x+2)(x))/((x^2-x)(x^3-3x^2))`

`f(x) = ((x-3)(x+2)(x))/((x)(x-1)(x^2)(x-3))`

`f(x) = (x(x-3)(x+2))/(x^3(x-1)(x-3)`

Now: holes and asymptotes are values which make a function undefined. Since we have a rational function, it will be undefined if and only if the denominator is equal to 0. We therefore only need to check the values of `x` which make the denominator `0`, which are:

`x=0`
`x=1`
`x=3`

To find out whether these are asymptotes or holes, let's take the limit of `f(x)` as `x` approaches each of these numbers.

`lim_(x->0)(x(x-3)(x+2))/(x^3(x-1)(x-3)) = lim_(x->0)((x-3)(x+2))/(x^2(x-1)(x-3))`

`= (-3*2)/(0*(-1)*(-3)) = +-oo`

So `x=0` is an asymptote.

`lim_(x->1)(x(x-3)(x+2))/(x^3(x-1)(x-3)) = (1*(-2)*3)/(1*0*(-2)) = +-oo`

So `x=1` is an asymptote.

`lim_(x->3)(x(x-3)(x+2))/(x^3(x-1)(x-3)) = lim_(x->3)((x+2))/(x^2(x-1))`

`= 5/(9*2) = 5/18`

So `(3, 5/18)` is a hole in `f(x)`.

Final Answer