For what values of x, if any, does #f(x) = 1/(x^2-4x+4) # have vertical asymptotes?

1 Answer
Feb 6, 2016

Answer:

#x=2#.

Explanation:

To find the vertical asymptotes we have to find for which values the denominator is equal to 0. Simply set the polynomial on the denominator to 0 then solve for #x#:

#x^2-4x+4=0#
#(x-2)(x-2)=0#
#therefore x = 2#

Thus there is one vertical asymptote at #x=2#.

Here is a graph of the function, the vertical asymptote is clearly indicated by a steep incline towards #x=2#.

From Mathematica