What are the points of discontinuity y=x-8/x^2+6x-7

The points of discontinuity of the function y = [x - 8] / [x^2 + 6x - 7] are the values of x for which the denominator is 0, because the function is not defined in those points.

Then, you need to solve x^2 + 6x - 7 =0.

To do that you can complete squares:

x^2 + 6x - 7 = (x^2 + 6x + 9) - 9 - 7 = (x + 3)^2 - 16 = 0

=> (x+3)^2 = 16

=> x + 3 = (+/-) 4

=> x = 4 - 3   and x = -4 - 3 = - 7

=> x = 1 and x = -7

The points of discontinuity area x = 1 and x = -7