# For what values of x, if any, does f(x) = 1/((x+12)(x-3))  have vertical asymptotes?

Feb 10, 2016

In x=-12 and x=3

#### Explanation:

If we replace x by -12 or 3, the denominator will be zero, which means:

$f \left(x\right) = \infty$

But what kind of infinity you have when x has those values?

If you use ${3}^{-}$, which is a number a little smaller than 3, you will obtain
$f \left({3}^{-}\right) = \left(\frac{1}{0} ^ -\right) = - \infty$

By the other hand if you use ${3}^{+}$, which is a number a little higher than 3, you will obtain
$f \left({3}^{-}\right) = \left(\frac{1}{0} ^ +\right) = + \infty$.
If by on side you have $- \infty$ and by the other $+ \infty$, you've got a vertica lasymptote.