Question 430404
{{{y = (x^2-3x-28)/(16-x^2) = ((x-7)(x+4))/((4-x)(4+x)) = (x-7)/(4-x)}}}.  There is a hole on the graph of y = f(x) at x = -4, and hence that is where the removable discontinuity is.   (There is an essential discontinuity at x = 4, or a vertical asymptote there.)