Find the Holes in the Graph f(x)=(-4x+8)/(2x^2-8)

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

Factor $4$ out of $- 4 x + 8$ .

Factor $4$ out of $- 4 x$ .

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

Factor $4$ out of $8$ .

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

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

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

Factor $2 x^{2} - 8$ .

Factor $2$ out of $2 x^{2} - 8$ .

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

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

Factor $2$ out of $- 8$ .

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

Factor $2$ out of $2 x^{2} + 2 \cdot - 4$ .

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

Rewrite $4$ as $2^{2}$ .

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

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 2$ .

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

Remove unnecessary parentheses.

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

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

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

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

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

Rewrite $2$ as $- 1 (- 2)$ .

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

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

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

Rewrite $- (x - 2)$ as $- 1 (x - 2)$ .

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

Cancel the common factor.

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

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.

$x - 2$

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 $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

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

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

Simplify.

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

Factor $2$ out of $2$ .

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

Cancel the common factors.

Factor $2$ out of $2$ .

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

Factor $2$ out of $2$ .

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

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Add $1$ and $1$ .

$- \frac{1}{2}$

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

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

Find the Holes in the Graph f(x)=(-4x+8)/(2x^2-8)

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