Answer:
<em><u>D</u></em>
Step-by-step explanation:
When the denominator is equal to 0, the function doesn't exist. So factor the bottom, luckily its already partially factored.
(x^2 + 8x + 12) = (x+2)(x+6)
So now the denominator is (x+2)(x+6)(x-3)
The function doesn't exist when x = -2, -6, 3
Now we know there are 3 total discontinuities.
<u>A and B can be eliminated</u>
Since there is an (x+2) on the top and the bottom, they don't affect the shape of the function, but only causes a removable discontinuity when x = -2.
However, (x+6) and (x-3) are only on the bottom, so they DO change the shape, so they are non-removable
The answer therefore is 1 removable and 2 non-removable; D
