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

Jan 28, 2016

$x = 2$ and $x = \ln 3$

#### Explanation:

Vertical asymptotes occur when the denominator of a function equals $0$.

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

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

Like when factoring and solving a quadratic equation, when we have multiple terms being multiplied by one another and equalling $0$, any of the terms can be equal to $0$ for the expression. Thus, we can set both of the multiplied equal to $0$.

$m a t h b f \left(\left(1\right)\right)$

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

$x - 2 = 0$

$x = 2$

$m a t h b f \left(\left(2\right)\right)$

${e}^{x} - 3 = 0$

${e}^{x} = 3$

$x = \ln 3$

The vertical asymptotes occur at $x = 2$ and $x = \ln 3$.