Find the Holes in the Graph f(x)=(x^2-36)/(x+6)

$f (x) = \frac{x^{2} - 36}{x + 6}$

Step 1

Factor $x^{2} - 36$ .

Rewrite $36$ as $6^{2}$ .

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

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 $b = 6$ .

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

Step 2

Cancel the common factor of $x + 6$ .

Cancel the common factor.

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

Divide $x - 6$ by $1$ .

$f (x) = x - 6$

Step 3

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

$x + 6$

Step 4

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $x - 6$ .

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

Substitute $- 6$ for $x$ in $x - 6$ and simplify.

Substitute $- 6$ for $x$ to find the $y$ coordinate of the hole.

$- 6 - 6$

Subtract $6$ from $- 6$ .

$- 12$

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

$(- 6, - 12)$

Step 5