# What is the discontinuity of the function f(x) = (x^2-9)/(x^2-6x+9) ?

Aug 30, 2014

f a rational expression (basically a fraction with a polynomial in the numerator and another polynomial in the denominator) has a discontinuity, it will be where the denominator equals zero.

So for $f \left(x\right) = \frac{{x}^{2} - 9}{{x}^{2} - 6 x + 9}$, the only possible discontinuity would be when

${x}^{2} - 6 x + 9 = 0$

The key to the next step is remembering how to factor.

${x}^{2} - 6 x + 9 = 0$
$\left(x - 3\right) \left(x - 3\right) = 0$

So there is some type of discontinuity at $x = 3$.

Now there are two types of discontinuity possible here.
If $x - 3$ is also a factor of the numerator, you will have what's called a removable discontinuity. That will show up as a hole in the graph.

If $x - 3$ is not a factor of the numerator, then you will have a vertical asymptote at $x = 3$.

Let's see:
$f \left(x\right) = \frac{{x}^{2} - 9}{{x}^{2} - 6 x + 9}$

$f \left(x\right) = \frac{\left(x + 3\right) \left(x - 3\right)}{\left(x - 3\right) \left(x - 3\right)}$

$f \left(x\right) = \frac{x + 3}{x - 3}$

This is an interesting case. We removed the discontinuity, but instead of a hole in the graph, we have a vertical asymptote at the exact same point.

Hope this helps.