What is the discontinuity of the function f(x) = (x^2-3x-28)/(x+4) ?

1 Answer
Aug 30, 2014

If 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^2-3x-28)/(x+4), the only possible discontinuity would be when

x+4=0

or

x=-4

Now there are two types of discontinuity possible here.

If x+4 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+4 is not a factor of the numerator, then you will have a vertical asymptote at x=-4.

Let's see:

f(x)=(x^2-3x-28)/(x+4)

f(x)=((x+4)(x-7))/(x+4)

f(x)=x-7, as long as x doesn't equal -4.

The discontinuity is removable. The graph would look like the line f(x)=x-7 except at x=-4, there is a hole.

enter image source here

Note that most graphers will not properly show a hole (as in this case); this is known as a graphing error.