Question

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

Answer 1

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]}