Find the Holes in the Graph (-x+3)/(x-3)

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

Cancel the common factor of $- x + 3$ and $x - 3$ .

Factor $- 1$ out of $- x$ .

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

Rewrite $3$ as $- 1 (- 3)$ .

$\frac{- (x) - 1 (- 3)}{x - 3}$

Factor $- 1$ out of $- (x) - 1 (- 3)$ .

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

Rewrite $- (x - 3)$ as $- 1 (x - 3)$ .

$\frac{- 1 (x - 3)}{x - 3}$

Cancel the common factor.

$\frac{- 1 (x - 3)}{x - 3}$

Divide $- 1$ by $1$ .

$- 1$

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 $- 1$ .

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.

$- 1$

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

$(3, - 1)$