Find the Holes in the Graph f(x)=((x+3)(6-x))/((x-2)(x+3))

$f (x) = \frac{(x + 3) (6 - x)}{(x - 2) (x + 3)}$

Step 1

Cancel the common factor of $x + 3$ .

Cancel the common factor.

$f (x) = \frac{(x + 3) (6 - x)}{(x - 2) (x + 3)}$

Rewrite the expression.

$f (x) = \frac{6 - x}{x - 2}$

Step 2

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

$x + 3$

Step 3

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{6 - x}{x - 2}$ .

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Substitute $- 3$ for $x$ in $\frac{6 - x}{x - 2}$ and simplify.

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

$\frac{6 + 3}{- 3 - 2}$

Simplify.

Add $6$ and $3$ .

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

Subtract $2$ from $- 3$ .

$\frac{9}{- 5}$

Move the negative in front of the fraction.

$- \frac{9}{5}$

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

$(- 3, - \frac{9}{5})$

Step 4