For what values of x, if any, does #f(x) = 1/(x^2-3x+4) # have vertical asymptotes?

1 Answer
Dec 15, 2016

Answer:

There are no vertical asymptotes/

Explanation:

#f(x) = 1/(x^2-3x+4) #

We can complete the square of the denominator:
# x^2-3x+4 = (x-3/2)^2-(3/2)^2+4 #
# " " = (x-3/2)^2-9/4+4 #
# " " = (x-3/2)^2+7/4 #

So the denominator has its smallest values of #7/4# when #x=3/2# and as the denominator can ever be less than this value it has no zero's and hence there are no values for which the function #f(x) = 1/(x^2-3x+4) # is undefined.

graph{1/(x^2-3x+4) [-3.328, 3.602, -0.978, 2.486]}