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

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

Factor $x^{2} - 9$ .

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

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

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.

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

Rewrite the expression.

$f (x) = \frac{1}{x + 3}$

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 $\frac{1}{x + 3}$ .

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Substitute $3$ for $x$ in $\frac{1}{x + 3}$ and simplify.

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

$\frac{1}{3 + 3}$

Add $3$ and $3$ .

$\frac{1}{6}$

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

$(3, \frac{1}{6})$