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

Question

1250 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-07T16:10:50

#x = 2, x=-2, x=ln3#

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.