Find the Holes in the Graph f(x)=(x^2-9)/(x+3)

$f (x) = \frac{x^{2} - 9}{x + 3}$

Factor $x^{2} - 9$ .

Rewrite $9$ as $3^{2}$ .

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

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

Cancel the common factor of $x + 3$ .

Cancel the common factor.

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

Divide $x - 3$ by $1$ .

$f (x) = x - 3$

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 $x - 3$ .

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Substitute $- 3$ for $x$ in $x - 3$ and simplify.

Substitute $- 3$ for $x$ to find the $y$ coordinate of the hole.

$- 3 - 3$

Subtract $3$ from $- 3$ .

$- 6$

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

$(- 3, - 6)$