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

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

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 $b = 6$ .

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

Cancel the common factor of $x - 6$ .

Cancel the common factor.

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

Divide $x + 6$ by $1$ .

$f (x) = x + 6$

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

$x - 6$

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 $0$ .

$x - 6 = 0$

Add $6$ to 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.

$6 + 6$

Add $6$ and $6$ .

$12$

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

$(6, 12)$