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

Feb 7, 2016

$x = 2 , x = - 2 , x = \ln 3$

#### Explanation:

To find the vertical asymptotes you simply have to set the denominator of the fraction to $0$ and solve the resulting equation:

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

Hence, we are presented with 3 possible values for $x$ hence 3 asymptotes.

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

Hence we have the equations of our vertical asymptotes

graph{1/((x-2)(x+2)(e^x-3)) [-5, 5, -2.5, 2.5]}

As you can see by the graph the asymptotes at our worked out locations are very distinct.