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

1 Answer
Jan 28, 2016

Answer:

#x=2# and #x=ln3#

Explanation:

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

#(x-2)(x-2)(e^x-3)=0#

#(x-2)^2(e^x-3)=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#.

#mathbf((1))#

#(x-2)^2=0#

#x-2=0#

#x=2#

#mathbf((2))#

#e^x-3=0#

#e^x=3#

#x=ln3#

The vertical asymptotes occur at #x=2# and #x=ln3#.