Find the Holes in the Graph f(x)=(x^2+x-2)/(3x+6)

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

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

$- 1, 2$

Write the factored form using these integers.

$f (x) = \frac{(x - 1) (x + 2)}{3 x + 6}$

Factor $3$ out of $3 x + 6$ .

Factor $3$ out of $3 x$ .

$f (x) = \frac{(x - 1) (x + 2)}{3 (x) + 6}$

Factor $3$ out of $6$ .

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

Factor $3$ out of $3 x + 3 \cdot 2$ .

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

Cancel the common factor of $x + 2$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

$x + 2$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{x - 1}{3}$ .

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Substitute $- 2$ for $x$ in $\frac{x - 1}{3}$ and simplify.

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

$\frac{- 2 - 1}{3}$

Simplify.

Subtract $1$ from $- 2$ .

$\frac{- 3}{3}$

Divide $- 3$ by $3$ .

$- 1$

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

$(- 2, - 1)$

Find the Holes in the Graph f(x)=(x^2+x-2)/(3x+6)

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