For what values of x, if any, does #f(x) = 1/(x^2-3x+4) # have vertical asymptotes?
1 Answer
Dec 15, 2016
There are no vertical asymptotes/
Explanation:
We can complete the square of the denominator:
So the denominator has its smallest values of
graph{1/(x^2-3x+4) [-3.328, 3.602, -0.978, 2.486]}