# How to find the asymptotes of f(x) = (x^2+x-2)/( x^3-3x^2+2x)?

Dec 22, 2016

$f \left(x\right)$ has vertical asymptotes $x = 0$ and $x = 2$, a horizontal asymptote $y = 0$ and a hole at $\left(1 , - 3\right)$

#### Explanation:

Start by factoring both numerator and denominator and cancelling common factors:

$f \left(x\right) = \frac{{x}^{2} + x - 2}{{x}^{3} - 3 {x}^{2} + 2 x} = \frac{\left(x + 2\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 1\right)}}}}{x \left(x - 2\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 1\right)}}}} = \frac{\left(x + 2\right)}{x \left(x - 2\right)}$

with exclusion $x \ne 1$

Notice that when $x = 1$, both the numerator and denominator are zero in the given rational function. So $f \left(1\right)$ is undefined, but the simplified expression is well defined when $x = 1$:

$\frac{\textcolor{b l u e}{1} + 2}{\textcolor{b l u e}{1} \left(\textcolor{b l u e}{1} - 2\right)} = \frac{3}{- 1} = - 3$

So $f \left(x\right)$ has a hole (removable singularity) at $\left(1 , - 3\right)$

The remaining values of $x$ at which the denominator is zero are $x = 0$ and $x = 2$. In both of these cases the numerator is non-zero. So there are vertical asymptotes at each of these values of $x$.

Finally, since the degree of the numerator is greater than the denominator, $f \left(x\right) \to 0$ as $x \to \pm \infty$. That is: $f \left(x\right)$ has horizontal asymptote $y = 0$.

It has no slant (oblique) asymptotes. Such slant asymptotes can only occur if the degree of the numerator is $1$ greater than the denominator.

Here's a graph of $f \left(x\right)$ with the vertical asymptotes...

graph{(y-(x+2)/(x(x-2)))(0.9999x-2)(0.9999x+0.0001y) = 0 [-8.71, 11.29, -5.76, 4.24]}