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)#

1 Answer
Nov 2, 2016

Answer:

#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!