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

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.

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The function f(x) has vertical asymptotes at x = 2, x = -2, and x = ln(3).

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