# For what values of x, if any, does f(x) = 1/((x^2-4)(x+3)(x-2))  have vertical asymptotes?

Jul 15, 2017

$\text{vertical asymptotes at "x=+-2" and } x = - 3$

#### Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

$\text{solve } \left({x}^{2} - 4\right) \left(x + 3\right) \left(x - 2\right) = 0$

$\Rightarrow {\left(x - 2\right)}^{2} \left(x + 2\right) \left(x + 3\right) = 0$

$\Rightarrow x = \pm 2 \text{ and "x=-3" are the asymptotes}$
graph{1/((x^2-4)(x+3)(x-2)) [-10, 10, -5, 5]}

Jul 15, 2017

$x = - 3 , - 2 , 2$

#### Explanation:

A vertical asymptote will occur when the denominator is zero.

We have:

$f \left(x\right) = \frac{1}{\left({x}^{2} - 4\right) \left(x + 3\right) \left(x - 2\right)}$
$\text{ } = \frac{1}{\left(x + 2\right) \left(x - 2\right) \left(x + 3\right) \left(x - 2\right)}$
$\text{ } = \frac{1}{\left(x + 2\right) {\left(x - 2\right)}^{2} \left(x + 3\right)}$

And so the denominator will be zero if:

$\left(x + 2\right) {\left(x - 2\right)}^{2} \left(x + 3\right) = 0 \implies x = - 3 , - 2 , 2$

Note that for $x = 2$ then as the factor in the denominator is ${\left(x - 2\right)}^{2}$ which is always positive then $f \left(x\right) \rightarrow \infty$ irrespective of the sign of $x$ as $x \rightarrow 2$

We can confirm this graphically:

graph{1/((x^2-4)(x+3)(x-2)) [-6, 6, -5, 5]}