Find the Holes in the Graph (-4x-12)/(x^2-9)

$\frac{- 4 x - 12}{x^{2} - 9}$

Step 1

Factor $4$ out of $- 4 x - 12$ .

Factor $4$ out of $- 4 x$ .

$\frac{4 (- x) - 12}{x^{2} - 9}$

Factor $4$ out of $- 12$ .

$\frac{4 (- x) + 4 (- 3)}{x^{2} - 9}$

Factor $4$ out of $4 (- x) + 4 (- 3)$ .

$\frac{4 (- x - 3)}{x^{2} - 9}$

Step 2

Factor $x^{2} - 9$ .

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

$\frac{4 (- x - 3)}{x^{2} - 3^{2}}$

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 $b = 3$ .

$\frac{4 (- x - 3)}{(x + 3) (x - 3)}$

Step 3

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

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

$\frac{4 (- (x) - 3)}{(x + 3) (x - 3)}$

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

$\frac{4 (- (x) - 1 (3))}{(x + 3) (x - 3)}$

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

$\frac{4 (- (x + 3))}{(x + 3) (x - 3)}$

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

$\frac{4 (- 1 (x + 3))}{(x + 3) (x - 3)}$

Cancel the common factor.

$\frac{4 (- 1 (x + 3))}{(x + 3) (x - 3)}$

Rewrite the expression.

$\frac{4 \cdot (- 1)}{x - 3}$

Step 4

Simplify the numerator.

Multiply $4$ by $- 1$ .

$\frac{4 (- 1)}{x - 3}$

Multiply $4$ by $- 1$ .

$\frac{- 4}{x - 3}$

Step 5

Move the negative in front of the fraction.

$- \frac{4}{x - 3}$

Step 6

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

$x + 3$

Step 7

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $- \frac{4}{x - 3}$ .

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Substitute $- 3$ for $x$ in $- \frac{4}{x - 3}$ and simplify.

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

$- \frac{4}{- 3 - 3}$

Simplify.

Subtract $3$ from $- 3$ .

$- \frac{4}{- 6}$

Cancel the common factor of $4$ and $- 6$ .

Factor $2$ out of $4$ .

$- \frac{2 (2)}{- 6}$

Cancel the common factors.

Factor $2$ out of $- 6$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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

Multiply $- - \frac{2}{3}$ .

Multiply $- 1$ by $- 1$ .

$1 (\frac{2}{3})$

Multiply $\frac{2}{3}$ by $1$ .

$\frac{2}{3}$

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

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

Step 8

Find the Holes in the Graph (-4x-12)/(x^2-9)

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