# 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) ?

Jun 12, 2017

$x = 0$ is an asymptote.

$x = 1$ is an asymptote.

$\left(3 , \frac{5}{18}\right)$ 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).

$f \left(x\right) = \frac{\left(x - 3\right) \left(x + 2\right) \left(x\right)}{\left({x}^{2} - x\right) \left({x}^{3} - 3 {x}^{2}\right)}$

$f \left(x\right) = \frac{\left(x - 3\right) \left(x + 2\right) \left(x\right)}{\left(x\right) \left(x - 1\right) \left({x}^{2}\right) \left(x - 3\right)}$

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 $f \left(x\right)$ as $x$ approaches each of these numbers.

${\lim}_{x \to 0} \frac{x \left(x - 3\right) \left(x + 2\right)}{{x}^{3} \left(x - 1\right) \left(x - 3\right)} = {\lim}_{x \to 0} \frac{\left(x - 3\right) \left(x + 2\right)}{{x}^{2} \left(x - 1\right) \left(x - 3\right)}$

$= \frac{- 3 \cdot 2}{0 \cdot \left(- 1\right) \cdot \left(- 3\right)} = \pm \infty$

So $x = 0$ is an asymptote.

${\lim}_{x \to 1} \frac{x \left(x - 3\right) \left(x + 2\right)}{{x}^{3} \left(x - 1\right) \left(x - 3\right)} = \frac{1 \cdot \left(- 2\right) \cdot 3}{1 \cdot 0 \cdot \left(- 2\right)} = \pm \infty$

So $x = 1$ is an asymptote.

${\lim}_{x \to 3} \frac{x \left(x - 3\right) \left(x + 2\right)}{{x}^{3} \left(x - 1\right) \left(x - 3\right)} = {\lim}_{x \to 3} \frac{\left(x + 2\right)}{{x}^{2} \left(x - 1\right)}$

$= \frac{5}{9 \cdot 2} = \frac{5}{18}$

So $\left(3 , \frac{5}{18}\right)$ is a hole in $f \left(x\right)$.