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

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 $f \left(x\right) = \frac{{x}^{2} - 3 x - 28}{x + 4}$, 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:

$f \left(x\right) = \frac{{x}^{2} - 3 x - 28}{x + 4}$

$f \left(x\right) = \frac{\left(x + 4\right) \left(x - 7\right)}{x + 4}$

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

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

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