Find the Holes in the Graph f(x)=(-x-3)/(4x+12)

$f (x) = \frac{- x - 3}{4 x + 12}$

Factor $4$ out of $4 x + 12$ .

Factor $4$ out of $4 x$ .

$f (x) = \frac{- x - 3}{4 (x) + 12}$

Factor $4$ out of $12$ .

$f (x) = \frac{- x - 3}{4 x + 4 \cdot 3}$

Factor $4$ out of $4 x + 4 \cdot 3$ .

$f (x) = \frac{- x - 3}{4 (x + 3)}$

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

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

$f (x) = \frac{- (x) - 3}{4 (x + 3)}$

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

$f (x) = \frac{- (x) - 1 \cdot 3}{4 (x + 3)}$

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

$f (x) = \frac{- (x + 3)}{4 (x + 3)}$

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

$f (x) = \frac{- 1 (x + 3)}{4 (x + 3)}$

Cancel the common factor.

$f (x) = \frac{- 1 (x + 3)}{4 (x + 3)}$

Rewrite the expression.

$f (x) = \frac{- 1}{4}$

Move the negative in front of the fraction.

$f (x) = - \frac{1}{4}$

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

$x + 3$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $- \frac{1}{4}$ .

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.

$- \frac{1}{4}$

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

$(- 3, - \frac{1}{4})$