Find the Holes in the Graph f(x)=(2x+8)/(3x+12)

$f (x) = \frac{2 x + 8}{3 x + 12}$

Factor $2$ out of $2 x + 8$ .

Factor $2$ out of $2 x$ .

$f (x) = \frac{2 (x) + 8}{3 x + 12}$

Factor $2$ out of $8$ .

$f (x) = \frac{2 x + 2 \cdot 4}{3 x + 12}$

Factor $2$ out of $2 x + 2 \cdot 4$ .

$f (x) = \frac{2 (x + 4)}{3 x + 12}$

Factor $3$ out of $3 x + 12$ .

Factor $3$ out of $3 x$ .

$f (x) = \frac{2 (x + 4)}{3 (x) + 12}$

Factor $3$ out of $12$ .

$f (x) = \frac{2 (x + 4)}{3 x + 3 \cdot 4}$

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

$f (x) = \frac{2 (x + 4)}{3 (x + 4)}$

Cancel the common factor of $x + 4$ .

Cancel the common factor.

$f (x) = \frac{2 (x + 4)}{3 (x + 4)}$

Rewrite the expression.

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

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

$x + 4$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{2}{3}$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

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

$\frac{2}{3}$

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

$(- 4, \frac{2}{3})$