Find the Holes in the Graph f(x)=(x^2-16)/(x+4)

$f (x) = \frac{x^{2} - 16}{x + 4}$

Factor $x^{2} - 16$ .

Rewrite $16$ as $4^{2}$ .

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

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 $b = 4$ .

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

Cancel the common factor of $x + 4$ .

Cancel the common factor.

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

Divide $x - 4$ by $1$ .

$f (x) = x - 4$

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 $x - 4$ .

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

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Substitute $- 4$ for $x$ in $x - 4$ and simplify.

Substitute $- 4$ for $x$ to find the $y$ coordinate of the hole.

$- 4 - 4$

Subtract $4$ from $- 4$ .

$- 8$

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

$(- 4, - 8)$