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

1 Answer
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(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.