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

$\frac{x^{2} - 9}{x - 3}$

Factor $x^{2} - 9$ .

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

$\frac{x^{2} - 3^{2}}{x - 3}$

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 $b = 3$ .

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

Cancel the common factor of $x - 3$ .

Cancel the common factor.

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

Divide $x + 3$ by $1$ .

$x + 3$

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 $x + 3$ .

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Substitute $3$ for $x$ in $x + 3$ and simplify.

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

$3 + 3$

Add $3$ and $3$ .

$6$

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

$(3, 6)$