For what values of x, if any, does #f(x) = 1/((x-4)(x-8)) # have vertical asymptotes?

1 Answer
Feb 27, 2016

Answer:

#x=4# and #x=8#

Explanation:

Vertical asymptotes occur whenever the function is undefined at a particular #x#-value. When we're dealing with rational functions like the one here, this simply means that we get asymptotes whenever a zero is in the denominator (because #1/0# is undefined).

Our denominator is #(x-4)(x-8)#. Luckily for us, it's already factored, allowing us to solve it immediately. We are finding when this is equal to zero, so we write:
#(x-4)(x-8) = 0#

Using the zero-product property, we can say that #x-4 = 0# and #x-8 = 0#. Solving these two equations tells us that #x=4# and #x=8#. Therefore, our vertical asymptotes are at #x=4# and #x=8# because at these #x#-values, #f(x)# is undefined.