Find the Holes in the Graph f(x)=(-2x-6)/(x+3)

$f (x) = \frac{- 2 x - 6}{x + 3}$

Step 1

Factor $2$ out of $- 2 x - 6$ .

Factor $2$ out of $- 2 x$ .

$f (x) = \frac{2 (- x) - 6}{x + 3}$

Factor $2$ out of $- 6$ .

$f (x) = \frac{2 (- x) + 2 (- 3)}{x + 3}$

Factor $2$ out of $2 (- x) + 2 (- 3)$ .

$f (x) = \frac{2 (- x - 3)}{x + 3}$

Step 2

Cancel the common factor of $- x - 3$ and $x + 3$ .

Factor $- 1$ out of $- x$ .

$f (x) = \frac{2 (- (x) - 3)}{x + 3}$

Rewrite $- 3$ as $- 1 (3)$ .

$f (x) = \frac{2 (- (x) - 1 \cdot 3)}{x + 3}$

Factor $- 1$ out of $- (x) - 1 (3)$ .

$f (x) = \frac{2 (- (x + 3))}{x + 3}$

Rewrite $- (x + 3)$ as $- 1 (x + 3)$ .

$f (x) = \frac{2 (- 1 (x + 3))}{x + 3}$

Cancel the common factor.

$f (x) = \frac{2 (- 1 (x + 3))}{x + 3}$

Divide $2 \cdot (- 1)$ by $1$ .

$f (x) = 2 \cdot (- 1)$

$f (x) = 2 (- 1)$

Step 3

Multiply $2$ by $- 1$ .

$f (x) = - 2$

Step 4

To find the holes in the graph, look at the denominator factors that were cancelled.

$x + 3$

Step 5

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $- 2$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Because the cancelled expression is constant with respect to $x$ , it is the $y$ coordinate for every hole.

$- 2$

The holes in the graph are the points where any of the cancelled factors are equal to $0$ .

$(- 3, - 2)$

Step 6