Find the Holes in the Graph f(x)=(x^2-81)/(x^3+7x^2-18x)

$f (x) = \frac{x^{2} - 81}{x^{3} + 7 x^{2} - 18 x}$

Factor $x^{2} - 81$ .

Rewrite $81$ as $9^{2}$ .

$f (x) = \frac{x^{2} - 9^{2}}{x^{3} + 7 x^{2} - 18 x}$

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 $b = 9$ .

$f (x) = \frac{(x + 9) (x - 9)}{x^{3} + 7 x^{2} - 18 x}$

Factor $x^{3} + 7 x^{2} - 18 x$ .

Factor $x$ out of $x^{3} + 7 x^{2} - 18 x$ .

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

$f (x) = \frac{(x + 9) (x - 9)}{x \cdot x^{2} + 7 x^{2} - 18 x}$

Factor $x$ out of $7 x^{2}$ .

$f (x) = \frac{(x + 9) (x - 9)}{x \cdot x^{2} + x (7 x) - 18 x}$

Factor $x$ out of $- 18 x$ .

$f (x) = \frac{(x + 9) (x - 9)}{x \cdot x^{2} + x (7 x) + x \cdot - 18}$

Factor $x$ out of $x \cdot x^{2} + x (7 x)$ .

$f (x) = \frac{(x + 9) (x - 9)}{x (x^{2} + 7 x) + x \cdot - 18}$

Factor $x$ out of $x (x^{2} + 7 x) + x \cdot - 18$ .

$f (x) = \frac{(x + 9) (x - 9)}{x (x^{2} + 7 x - 18)}$

Factor.

Factor $x^{2} + 7 x - 18$ 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 $- 18$ and whose sum is $7$ .

$- 2, 9$

Write the factored form using these integers.

$f (x) = \frac{(x + 9) (x - 9)}{x ((x - 2) (x + 9))}$

Remove unnecessary parentheses.

$f (x) = \frac{(x + 9) (x - 9)}{x (x - 2) (x + 9)}$

Cancel the common factor of $x + 9$ .

Cancel the common factor.

$f (x) = \frac{(x + 9) (x - 9)}{x (x - 2) (x + 9)}$

Rewrite the expression.

$f (x) = \frac{x - 9}{x (x - 2)}$

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

$x + 9$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{x - 9}{x (x - 2)}$ .

Set $x + 9$ equal to $0$ .

$x + 9 = 0$

Subtract $9$ from both sides of the equation.

$x = - 9$

Substitute $- 9$ for $x$ in $\frac{x - 9}{x (x - 2)}$ and simplify.

Substitute $- 9$ for $x$ to find the $y$ coordinate of the hole.

$\frac{- 9 - 9}{- 9 (- 9 - 2)}$

Simplify.

Cancel the common factor of $- 9 - 9$ and $- 9$ .

Factor $9$ out of $- 9$ .

$\frac{9 (- 1) - 9}{- 9 (- 9 - 2)}$

Factor $9$ out of $- 9$ .

$\frac{9 \cdot - 1 + 9 \cdot - 1}{- 9 (- 9 - 2)}$

Factor $9$ out of $9 \cdot - 1 + 9 \cdot - 1$ .

$\frac{9 \cdot (- 1 - 1)}{- 9 (- 9 - 2)}$

Cancel the common factors.

Factor $9$ out of $- 9 (- 9 - 2)$ .

$\frac{9 \cdot (- 1 - 1)}{9 (- (- 9 - 2))}$

Cancel the common factor.

$\frac{9 \cdot (- 1 - 1)}{9 (- (- 9 - 2))}$

Rewrite the expression.

$\frac{- 1 - 1}{- (- 9 - 2)}$

Subtract $1$ from $- 1$ .

$\frac{- 2}{- (- 9 - 2)}$

Subtract $2$ from $- 9$ .

$\frac{- 2}{- - 11}$

Multiply $- 1$ by $- 11$ .

$\frac{- 2}{11}$

Move the negative in front of the fraction.

$- \frac{2}{11}$

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

$(- 9, - \frac{2}{11})$

Find the Holes in the Graph f(x)=(x^2-81)/(x^3+7x^2-18x)

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