# How do you determine the value of A such that f(x) has a point discontinuity without using trial and error?

## $f \left(x\right) = \frac{{x}^{2} + 2 x + A}{{x}^{2} - 4 x - 5}$

Nov 2, 2016

$A = - 3$

#### Explanation:

A point of discontinuity is also known as a hole. For example, the graph of $y = \frac{{x}^{2} + 2 x + 1}{{x}^{2} + 3 x + 2}$ would have a hole at $x = - 1$.

The trick about holes is that they occur when a common factor is cancelled out in the numerator and the denominator.

Start by factoring the expression that we know: the denominator.

${x}^{2} - 4 x - 5 = \left(x - 4\right) \left(x - 1\right)$

Now let's compare this to the numerator. There is no way that we can make ${x}^{2} + 2 x + A$ a factorable trinomial with $x - 4$ as a factor. So, $x - 1$ will have to be one of the factors.

To make $x - 1$ a factor, we have that the sum of the two factors is $2$ and their product is A. The two roots are therefore $- 1$ and $3$ (the only possible combination for a sum), and so the value of A is $- 1 \times 3 = - 3$.

Let's do a final check.

$f \left(x\right) = \frac{{x}^{2} + 2 x - 3}{{x}^{2} - 4 x - 5}$

$f \left(x\right) = \frac{\left(x + 3\right) \left(x - 1\right)}{\left(x - 4\right) \left(x - 1\right)}$

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

So, there will be a discontinuity at $x = 1$.

Hopefully you understand now!