Find the Holes in the Graph (x^3-9x)/(3x^2-6x-9)

$\frac{x^{3} - 9 x}{3 x^{2} - 6 x - 9}$

Step 1

Factor $x^{3} - 9 x$ .

Factor $x$ out of $x^{3} - 9 x$ .

Factor $x$ out of $x^{3}$ .

$\frac{x \cdot x^{2} - 9 x}{3 x^{2} - 6 x - 9}$

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

$\frac{x \cdot x^{2} + x \cdot - 9}{3 x^{2} - 6 x - 9}$

Factor $x$ out of $x \cdot x^{2} + x \cdot - 9$ .

$\frac{x (x^{2} - 9)}{3 x^{2} - 6 x - 9}$

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

$\frac{x (x^{2} - 3^{2})}{3 x^{2} - 6 x - 9}$

Factor.

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 $b = 3$ .

$\frac{x ((x + 3) (x - 3))}{3 x^{2} - 6 x - 9}$

Remove unnecessary parentheses.

$\frac{x (x + 3) (x - 3)}{3 x^{2} - 6 x - 9}$

Step 2

Factor $3 x^{2} - 6 x - 9$ .

Factor $3$ out of $3 x^{2} - 6 x - 9$ .

Factor $3$ out of $3 x^{2}$ .

$\frac{x (x + 3) (x - 3)}{3 (x^{2}) - 6 x - 9}$

Factor $3$ out of $- 6 x$ .

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

Factor $3$ out of $- 9$ .

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

Factor $3$ out of $3 x^{2} + 3 (- 2 x)$ .

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

Factor $3$ out of $3 (x^{2} - 2 x) + 3 \cdot - 3$ .

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

Factor.

Factor $x^{2} - 2 x - 3$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 3$ and whose sum is $- 2$ .

$- 3, 1$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

Step 3

Cancel the common factor of $x - 3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 4

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

$x - 3$

Step 5

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{x (x + 3)}{3 (x + 1)}$ .

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Substitute $3$ for $x$ in $\frac{x (x + 3)}{3 (x + 1)}$ and simplify.

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

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

Simplify.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Add $3$ and $3$ .

$\frac{6}{3 + 1}$

Add $3$ and $1$ .

$\frac{6}{4}$

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

Factor $2$ out of $6$ .

$\frac{2 (3)}{4}$

Cancel the common factors.

Factor $2$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

$\frac{3}{2}$

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

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

Step 6

Find the Holes in the Graph (x^3-9x)/(3x^2-6x-9)

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