Find the Holes in the Graph f(x)=(x^2-x-12)/(x^2+x-6)

$f (x) = \frac{x^{2} - x - 12}{x^{2} + x - 6}$

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.

$f (x) = \frac{(x - 4) (x + 3)}{x^{2} + x - 6}$

Factor $x^{2} + x - 6$ 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 $- 6$ and whose sum is $1$ .

$- 2, 3$

Write the factored form using these integers.

$f (x) = \frac{(x - 4) (x + 3)}{(x - 2) (x + 3)}$

Cancel the common factor of $x + 3$ .

Cancel the common factor.

$f (x) = \frac{(x - 4) (x + 3)}{(x - 2) (x + 3)}$

Rewrite the expression.

$f (x) = \frac{x - 4}{x - 2}$

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 $\frac{x - 4}{x - 2}$ .

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Substitute $- 3$ for $x$ in $\frac{x - 4}{x - 2}$ and simplify.

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

$\frac{- 3 - 4}{- 3 - 2}$

Simplify.

Subtract $4$ from $- 3$ .

$\frac{- 7}{- 3 - 2}$

Subtract $2$ from $- 3$ .

$\frac{- 7}{- 5}$

Dividing two negative values results in a positive value.

$\frac{7}{5}$

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

$(- 3, \frac{7}{5})$

Find the Holes in the Graph f(x)=(x^2-x-12)/(x^2+x-6)

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