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

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

Step 1

Factor $x^{2} - 9$ .

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

$f (x) = \frac{x^{2} - 3^{2}}{x^{2} - x - 12}$

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 $b = 3$ .

$f (x) = \frac{(x + 3) (x - 3)}{x^{2} - x - 12}$

Step 2

Factor $x^{2} - x - 12$ 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 $- 12$ and whose sum is $- 1$ .

$- 4, 3$

Write the factored form using these integers.

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

Step 3

Cancel the common factor of $x + 3$ .

Cancel the common factor.

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

Rewrite the expression.

$f (x) = \frac{x - 3}{x - 4}$

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

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

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

Simplify.

Subtract $3$ from $- 3$ .

$\frac{- 6}{- 3 - 4}$

Subtract $4$ from $- 3$ .

$\frac{- 6}{- 7}$

Dividing two negative values results in a positive value.

$\frac{6}{7}$

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

$(- 3, \frac{6}{7})$

Step 6

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

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