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

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

Step 1

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.

$f (x) = \frac{(x - 3) (x + 1)}{x^{2} - 9}$

Step 2

Factor $x^{2} - 9$ .

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

$f (x) = \frac{(x - 3) (x + 1)}{x^{2} - 3^{2}}$

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 $b = 3$ .

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

Step 3

Cancel the common factor of $x - 3$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

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

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

Simplify.

Add $3$ and $1$ .

$\frac{4}{3 + 3}$

Add $3$ and $3$ .

$\frac{4}{6}$

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.

$\frac{2}{3}$

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

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

Step 6

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

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