Find the Holes in the Graph f(x)=(-2x-8)/(x^2+6x+8)

$f (x) = \frac{- 2 x - 8}{x^{2} + 6 x + 8}$

Factor $2$ out of $- 2 x - 8$ .

Factor $2$ out of $- 2 x$ .

$f (x) = \frac{2 (- x) - 8}{x^{2} + 6 x + 8}$

Factor $2$ out of $- 8$ .

$f (x) = \frac{2 (- x) + 2 (- 4)}{x^{2} + 6 x + 8}$

Factor $2$ out of $2 (- x) + 2 (- 4)$ .

$f (x) = \frac{2 (- x - 4)}{x^{2} + 6 x + 8}$

Factor $x^{2} + 6 x + 8$ 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 $8$ and whose sum is $6$ .

$2, 4$

Write the factored form using these integers.

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

Cancel the common factor of $- x - 4$ and $x + 4$ .

Factor $- 1$ out of $- x$ .

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

Rewrite $- 4$ as $- 1 (4)$ .

$f (x) = \frac{2 (- (x) - 1 \cdot 4)}{(x + 2) (x + 4)}$

Factor $- 1$ out of $- (x) - 1 (4)$ .

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

Rewrite $- (x + 4)$ as $- 1 (x + 4)$ .

$f (x) = \frac{2 (- 1 (x + 4))}{(x + 2) (x + 4)}$

Cancel the common factor.

$f (x) = \frac{2 (- 1 (x + 4))}{(x + 2) (x + 4)}$

Rewrite the expression.

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

Simplify the numerator.

Multiply $2$ by $- 1$ .

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

Multiply $2$ by $- 1$ .

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

Move the negative in front of the fraction.

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

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

$x + 4$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $- \frac{2}{x + 2}$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

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

Substitute $- 4$ for $x$ to find the $y$ coordinate of the hole.

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

Simplify.

Cancel the common factor of $2$ and $- 4 + 2$ .

Factor $2$ out of $2$ .

$- \frac{2 \cdot 1}{- 4 + 2}$

Cancel the common factors.

Factor $2$ out of $- 4$ .

$- \frac{2 \cdot 1}{2 (- 2) + 2}$

Factor $2$ out of $2$ .

$- \frac{2 \cdot 1}{2 \cdot - 2 + 2 \cdot 1}$

Factor $2$ out of $2 \cdot - 2 + 2 \cdot 1$ .

$- \frac{2 \cdot 1}{2 \cdot (- 2 + 1)}$

Cancel the common factor.

$- \frac{2 \cdot 1}{2 \cdot (- 2 + 1)}$

Rewrite the expression.

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

Add $- 2$ and $1$ .

$- \frac{1}{- 1}$

Divide $1$ by $- 1$ .

$- - 1$

Multiply $- 1$ by $- 1$ .

$1$

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

$(- 4, 1)$

Find the Holes in the Graph f(x)=(-2x-8)/(x^2+6x+8)

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