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

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

Factor $3$ out of $3 x - 9$ .

Factor $3$ out of $3 x$ .

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

Factor $3$ out of $- 9$ .

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

Factor $3$ out of $3 x + 3 \cdot - 3$ .

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

Cancel the common factor of $x - 3$ .

Cancel the common factor.

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

Divide $3$ by $1$ .

$f (x) = 3$

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 $3$ .

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Because the cancelled expression is constant with respect to $x$ , it is the $y$ coordinate for every hole.

$3$

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

$(3, 3)$