Find the Holes in the Graph f(x)=(-2x-6)/(x^2+4x+3)

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

Factor $2$ out of $- 2 x - 6$ .

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

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

Factor $2$ out of $- 6$ .

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

Factor $2$ out of $2 (- x) + 2 (- 3)$ .

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

Factor $x^{2} + 4 x + 3$ 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 $3$ and whose sum is $4$ .

$1, 3$

Write the factored form using these integers.

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

Cancel the common factor of $- x - 3$ and $x + 3$ .

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

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

Rewrite $- 3$ as $- 1 (3)$ .

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

Factor $- 1$ out of $- (x) - 1 (3)$ .

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

Rewrite $- (x + 3)$ as $- 1 (x + 3)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the numerator.

Multiply $2$ by $- 1$ .

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

Multiply $2$ by $- 1$ .

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

Move the negative in front of the fraction.

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

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

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

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

Simplify.

Add $- 3$ and $1$ .

$- \frac{2}{- 2}$

Divide $2$ by $- 2$ .

$- - 1$

Multiply $- 1$ by $- 1$ .

$1$

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

$(- 3, 1)$

Find the Holes in the Graph f(x)=(-2x-6)/(x^2+4x+3)

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