Find the Holes in the Graph f(x)=(x^2-81)/(x-9)

$f (x) = \frac{x^{2} - 81}{x - 9}$

Factor $x^{2} - 81$ .

Rewrite $81$ as $9^{2}$ .

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

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 $b = 9$ .

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

Cancel the common factor of $x - 9$ .

Cancel the common factor.

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

Divide $x + 9$ by $1$ .

$f (x) = x + 9$

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

$x - 9$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $x + 9$ .

Set $x - 9$ equal to $0$ .

$x - 9 = 0$

Add $9$ to both sides of the equation.

$x = 9$

Substitute $9$ for $x$ in $x + 9$ and simplify.

Substitute $9$ for $x$ to find the $y$ coordinate of the hole.

$9 + 9$

Add $9$ and $9$ .

$18$

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

$(9, 18)$