How do you find the removable discontinuity(hole) for the graph of #y=(x^2 - 9x -10)/ (2x^2 - 2)#?

Question

Scarlett Allison · Accepted Answer

Factoring the numerator and denominator reveals a common factor of #(x+1)# which can be removed so the (modified) fraction is no longer discontinuous at #x=-1# #y=(x^2-9x-10)/(2x^2-2)# Factoring: #color(white)("XXXX")##y = ((x-10)(x+1))/(2(x^2-1))# Continue by factoring the difference of squares in the denominator #color(white)("XXXX")##y = ((x-10)(x+1))/(2(x+1)(x-1))# Divide the numerator and denominator by #(x+1)# #color(white)("XXXX")##=y = ((x-10)cancel((x+1)))/(2cancel((x+1))(x-1))# #color(white)("XXXX")##y = (x-10)/(2(x-1))# Note that the discontinuity at #x=1# (which would cause an attempt to divide by zero) can not be removed.

Ezra Adams · Answer

undefined To find the removable discontinuity (hole) for the graph of y=(x^2 - 9x -10)/ (2x^2 - 2), we need to factor both the numerator and denominator. Factoring the numerator, we get (x - 10)(x + 1). Factoring the denominator, we get 2(x - 1)(x + 1).

We can see that (x + 1) appears in both the numerator and denominator. Therefore, there is a removable discontinuity (hole) at x = -1. To find the y-coordinate of the hole, we substitute x = -1 into the simplified expression (x - 10)/(2(x - 1)). Simplifying this expression, we get y = -11/2.

Therefore, the removable discontinuity (hole) for the graph of y=(x^2 - 9x -10)/ (2x^2 - 2) is at the point (-1, -11/2).