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

Jan 16, 2016

x = ± 2 and x = - 3

#### Explanation:

Require to factorise the denominator completely.

Note that $\left({x}^{2} - 4\right)$ is a difference of 2 squares.

and factorises to $\left({x}^{2} - 4\right) = \left(x - 2\right) \left(x + 2\right)$

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

Vertical asymptotes occur when denominator = 0

(RW - because as the denominator gets closer and closer to zero, the value of the function gets bigger and bigger and the graph of the function will effectively become vertical)

when (x - 2 )( x + 2 )(x + 3 ) = 0

rArr x = ± 2 and x = - 3