Question

i dont know how to solve this questions without a graphing on a calculator but i can't use calculator for this question pls help

Answer 1

Answer:

D

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.

A and B can be eliminated

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

Answer 2

Answer:

We have 1 removable discontinuity and 2 non-removable discontinuities

Step-by-step explanation:

Note that

$x^2+8x+12 = (x+2)(x+6)$

Initially, we have discontinuities at

$x = -2; x = -6; x = 3$

Considering

$f(x)=\dfrac{(x+2)(x-6)}{(x^2+8x+12)(x-3)}$

But

$\dfrac{(x+2)(x-6)}{(x^2+8x+12)(x-3)}= \dfrac{(x+2)(x-6)}{ (x+2)(x+6)(x-3)}= \dfrac{(x-6)}{ (x+6)(x-3)}$

We have now two discontinuities at

$x = -6; x = 3$