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

1 Answer

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.