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

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

Step 1

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 $- 1$ .

$- 3, 2$

Write the factored form using these integers.

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

Step 2

Cancel the common factor of $x + 2$ .

Cancel the common factor.

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

Divide $x - 3$ by $1$ .

$f (x) = x - 3$

Step 3

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

$x + 2$

Step 4

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

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Substitute $- 2$ for $x$ in $x - 3$ and simplify.

Substitute $- 2$ for $x$ to find the $y$ coordinate of the hole.

$- 2 - 3$

Subtract $3$ from $- 2$ .

$- 5$

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

$(- 2, - 5)$

Step 5

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

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