Question

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

Answer 1

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(x)=(x^2-9)/(x^2-6x+9)`, the only possible discontinuity would be when

`x^2-6x+9=0`

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

`x^2-6x+9=0`
`(x-3)(x-3)=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(x)=(x^2-9)/(x^2-6x+9)`

`f(x)=((x+3)(x-3))/((x-3)(x-3))`

`f(x)=(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.