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

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

Factor $2 x$ out of $2 x^{2} + 2 x$ .

Factor $2 x$ out of $2 x^{2}$ .

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

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

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

Factor $2 x$ out of $2 x (x) + 2 x (1)$ .

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

Factor $x^{2} - 2 x - 3$ 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 $- 2$ .

$- 3, 1$

Write the factored form using these integers.

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

Cancel the common factor of $x + 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

$x + 1$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{2 x}{x - 3}$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

Substitute $- 1$ for $x$ to find the $y$ coordinate of the hole.

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

Simplify.

Multiply $2$ by $- 1$ .

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

Subtract $3$ from $- 1$ .

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

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

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

$\frac{- 2 (1)}{- 4}$

Cancel the common factors.

Factor $- 2$ out of $- 4$ .

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

Cancel the common factor.

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

Rewrite the expression.

$\frac{1}{2}$

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

$(- 1, \frac{1}{2})$

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

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