Find the Holes in the Graph f(x)=(-3x+12)/(x^2-4x)

$f (x) = \frac{- 3 x + 12}{x^{2} - 4 x}$

Step 1

Factor $3$ out of $- 3 x + 12$ .

Factor $3$ out of $- 3 x$ .

$f (x) = \frac{3 (- x) + 12}{x^{2} - 4 x}$

Factor $3$ out of $12$ .

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

Factor $3$ out of $3 (- x) + 3 (4)$ .

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

Step 2

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

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

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

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

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

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

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

Step 3

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

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

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

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

$f (x) = \frac{3 (- (x) - 1 \cdot - 4)}{x (x - 4)}$

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

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

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

$f (x) = \frac{3 (- 1 (x - 4))}{x (x - 4)}$

Cancel the common factor.

$f (x) = \frac{3 (- 1 (x - 4))}{x (x - 4)}$

Rewrite the expression.

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

Step 4

Simplify the numerator.

Multiply $3$ by $- 1$ .

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

Multiply $3$ by $- 1$ .

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

Step 5

Move the negative in front of the fraction.

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

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 $- \frac{3}{x}$ .

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

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

$- \frac{3}{4}$

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

$(4, - \frac{3}{4})$

Step 8

Find the Holes in the Graph f(x)=(-3x+12)/(x^2-4x)

Related Questions

Support LCMGCF.com by disabling your adblocker.