Find points of discontinuity y=x^2+5x+6/x^2-16

  x=4 and x=-4

Step-by-step explanation:

We assume you intend ...

  y = (x^2 +5x +6)/(x^2 -16)

There will generally be points of discontinuity where the denominator is zero. Sometimes a denominator factor will be cancelled by a numerator factor, but the "hole" remains and there is still a discontinuity there.

Here, the denominator factors as the difference of squares:

  x^2 -16 = (x -4)(x +4)

There will be discontinuities where these factors are zero, at x=4 and x=-4.