Find the Holes in the Graph f(x)=(x^2+6x+9)/(x^2-9)

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

Factor using the perfect square rule.

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

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

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 x = 2 \cdot x \cdot 3$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 3$ .

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

Factor $x^{2} - 9$ .

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

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

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 $b = 3$ .

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

Cancel the common factor of ${(x + 3)}^{2}$ and $x + 3$ .

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

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

Cancel the common factors.

Cancel the common factor.

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

Rewrite the expression.

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

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

$x + 3$

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

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

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

Simplify.

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

Factor $3$ out of $- 3$ .

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

Factor $3$ out of $3$ .

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

Factor $3$ out of $3 \cdot - 1 + 3 \cdot 1$ .

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

Cancel the common factors.

Factor $3$ out of $- 3$ .

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

Factor $3$ out of $- 3$ .

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

Factor $3$ out of $3 \cdot - 1 + 3 \cdot - 1$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Add $- 1$ and $1$ .

$\frac{0}{- 1 - 1}$

Subtract $1$ from $- 1$ .

$\frac{0}{- 2}$

Divide $0$ by $- 2$ .

$0$

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

$(- 3, 0)$

Find the Holes in the Graph f(x)=(x^2+6x+9)/(x^2-9)

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