Question

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

`f(x)=(x^2+2x+A)/(x^2-4x-5)`

Answer 1

`A = -3`

Explanation:

A point of discontinuity is also known as a hole. For example, the graph of `y = (x^2 + 2x + 1)/(x^2 + 3x + 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 - 4x - 5 = (x - 4)(x - 1)`

Now let's compare this to the numerator. There is no way that we can make `x^2 + 2x + 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 xx 3= -3`.

Let's do a final check.

`f(x) = (x^2 + 2x - 3)/(x^2 - 4x - 5)`

`f(x) = ((x + 3)(x - 1))/((x - 4)(x - 1))`

`f(x) = (x + 3)/(x - 4)`

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

Hopefully you understand now!