# For what values of x, if any, does f(x) = 1/((x^2-4)(x+1)(x-5))  have vertical asymptotes?

Vertical Asymptotes are $x = - 2$, $x = - 1$, $x = 2$, $x = 5$
graph{1/((x^2-4)(x+1)(x-5)) [-4.083, 5.917, -2.5, 2.5]}

#### Explanation:

The given function $f \left(x\right) = \frac{1}{\left({x}^{2} - 4\right) \left(x + 1\right) \left(x - 5\right)}$

Notice the denominator $\left(\left({x}^{2} - 4\right) \left(x + 1\right) \left(x - 5\right)\right)$

will be zero when x=2, x=-2, x=-1, x=5 and the value of the function at these values is not defined. These are the asymptotes (the vertical lines).

So, to determine these vertical lines, we simply, equate each factor of the denominator to zero.
${x}^{2} - 4 = 0$

${x}^{2} = 4$

$x = + 2$ vertical asymptote
$x = - 2$ vertical asymptote

also $x + 1 = 0$

$x = - 1$ vertical asymptote

also $x - 5 = 0$

$x = 5$ vertical asymptote