Since f(x) = (x^2 + 5x + 6) / (2x + 16)
For discontinuities, they can be found at where the slope does NOT exist
Take the derivative of f(x):
f'(x) = (x^2 + 16x + 34) / 2(x+8)^2
Apparently, when x= -8, f'(x) is NOT defined
Therefore, the discontinuity is uniquely located at x = -8
