# How do you find the asymptote(s) or hole(s) of f(x) = (x(x^2-4))/((x^2-6)(x-2))?

Mar 2, 2017

Hole at $x = 2$, vertical asymptotes at $x = - \sqrt{6}$ and $x = \sqrt{6}$ and a horizontal asymptote at $y = 1$.

#### Explanation:

As $f \left(x\right) = \frac{x \left({x}^{2} - 4\right)}{\left({x}^{2} - 6\right) \left(x - 2\right)}$.

= $\frac{x \left(x + 2\right) \left(x - 2\right)}{\left(x + \sqrt{6}\right) \left(x - \sqrt{6}\right) \left(x - 2\right)}$

= $\frac{x \left(x + 2\right)}{\left(x + \sqrt{6}\right) \left(x - \sqrt{6}\right)}$

Observe that although $\left(x - 2\right)$ has cancelled, we cannot have $x - 2 = 0$ as then $f \left(x\right)$ becomes undefined.

Hence, we have a hole at $x = 2$.

As regards asymptotes, $f \left(x\right) = \frac{x \left(x + 2\right)}{\left(x + \sqrt{6}\right) \left(x - \sqrt{6}\right)}$

and as $x \to \pm \sqrt{6}$, $f \left(x\right) \to \pm \infty$ and as such we have vertical asymptotes at $x = - \sqrt{6}$ and $x = \sqrt{6}$.

Further, as $f \left(x\right) = \frac{x \left({x}^{2} - 4\right)}{\left({x}^{2} - 6\right) \left(x - 2\right)}$

= $\frac{x \left(x + 2\right)}{{x}^{2} - 6} = \frac{{x}^{2} + 2 x}{{x}^{2} - 6}$

= $\frac{{x}^{2} - 6 + 2 x + 6}{{x}^{2} - 6} = 1 + \frac{2 x + 6}{{x}^{2} - 6} = 1 + \frac{\frac{2}{x} + \frac{6}{x} ^ 2}{1 - \frac{6}{x} ^ 2}$

and as $x \to \infty$, $f \left(x\right) \to 1$

and hevce, we have a horizontal asymptote at $y = 1$.

graph{(x(x+2))/((x+sqrt6)(x-sqrt6)) [-10, 10, -5, 5]}