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

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

Step 1

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

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

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

Factor $2$ out of $8$ .

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

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

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

Step 2

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

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

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

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

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

Factor $x$ out of $x \cdot x + x \cdot - 4$ .

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

Step 3

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

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

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

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

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

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

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

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 4

Simplify the numerator.

Multiply $2$ by $- 1$ .

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

Multiply $2$ by $- 1$ .

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

Step 5

Move the negative in front of the fraction.

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

Step 6

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

$x - 4$

Step 7

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

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

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

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

$- \frac{2}{4}$

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 $0$ .

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

Step 8

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

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