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

Dec 15, 2016

There are no vertical asymptotes/

Explanation:

$f \left(x\right) = \frac{1}{{x}^{2} - 3 x + 4}$

We can complete the square of the denominator:
${x}^{2} - 3 x + 4 = {\left(x - \frac{3}{2}\right)}^{2} - {\left(\frac{3}{2}\right)}^{2} + 4$
$\text{ } = {\left(x - \frac{3}{2}\right)}^{2} - \frac{9}{4} + 4$
$\text{ } = {\left(x - \frac{3}{2}\right)}^{2} + \frac{7}{4}$

So the denominator has its smallest values of $\frac{7}{4}$ when $x = \frac{3}{2}$ 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 $f \left(x\right) = \frac{1}{{x}^{2} - 3 x + 4}$ is undefined.

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