Question

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

Answer 1

`x = 2, x=-2, x=ln3`

Explanation:

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

`(x-2)(x+2)(e^x-3)=0`

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

`x-2=0 -> x=2`
`x+2=0 -> x=-2`
`e^x-3 =0 -> x = ln(3)`

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.