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

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

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

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

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

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

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

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

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

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 $- 6$ .

$- 4, - 2$

Write the factored form using these integers.

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

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

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

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

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

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

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

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

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 1$ by $2$ .

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

Move the negative in front of the fraction.

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

Multiply $2$ by $x$ .

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

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

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

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

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

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

Simplify.

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

Factor $2$ out of $2 \cdot 4$ .

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

Cancel the common factors.

Factor $2$ out of $4$ .

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

Factor $2$ out of $- 2$ .

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

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Subtract $1$ from $2$ .

$- \frac{4}{1}$

Divide $4$ by $1$ .

$- 1 \cdot 4$

Multiply $- 1$ by $4$ .

$- 4$

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

$(4, - 4)$

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

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