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

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

Step 1

Cancel the common factor of $x - 2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 2

Move the negative in front of the fraction.

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

Step 3

Multiply $3$ by $x$ .

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

Step 4

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

$x - 2$

Step 5

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $- \frac{3 x}{x + 2}$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

Substitute $2$ for $x$ to find the $y$ coordinate of the hole.

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

Simplify.

Cancel the common factor of $2$ and $2 + 2$ .

Factor $2$ out of $3 \cdot 2$ .

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

Cancel the common factors.

Factor $2$ out of $2$ .

$- \frac{2 \cdot 3}{2 \cdot 1 + 2}$

Factor $2$ out of $2$ .

$- \frac{2 \cdot 3}{2 \cdot 1 + 2 (1)}$

Factor $2$ out of $2 \cdot 1 + 2 (1)$ .

$- \frac{2 \cdot 3}{2 \cdot (1 + 1)}$

Cancel the common factor.

$- \frac{2 \cdot 3}{2 \cdot (1 + 1)}$

Rewrite the expression.

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

Add $1$ and $1$ .

$- \frac{3}{2}$

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

$(2, - \frac{3}{2})$

Step 6