Find the Holes in the Graph f(x)=(-3x+12)/(x^2-5x+4)

$f (x) = \frac{- 3 x + 12}{x^{2} - 5 x + 4}$

Step 1

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

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

$f (x) = \frac{3 (- x) + 12}{x^{2} - 5 x + 4}$

Factor $3$ out of $12$ .

$f (x) = \frac{3 (- x) + 3 (4)}{x^{2} - 5 x + 4}$

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

$f (x) = \frac{3 (- x + 4)}{x^{2} - 5 x + 4}$

Step 2

Factor $x^{2} - 5 x + 4$ 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 $4$ and whose sum is $- 5$ .

$- 4, - 1$

Write the factored form using these integers.

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

Step 3

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

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

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

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

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

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

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

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 4

Simplify the numerator.

Multiply $3$ by $- 1$ .

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

Multiply $3$ by $- 1$ .

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

Step 5

Move the negative in front of the fraction.

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

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 $- \frac{3}{x - 1}$ .

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Substitute $4$ for $x$ in $- \frac{3}{x - 1}$ and simplify.

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

$- \frac{3}{4 - 1}$

Simplify.

Subtract $1$ from $4$ .

$- \frac{3}{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$- \frac{3}{3}$

Rewrite the expression.

$- 1 \cdot 1$

Multiply $- 1$ by $1$ .

$- 1$

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

$(4, - 1)$

Step 8

Find the Holes in the Graph f(x)=(-3x+12)/(x^2-5x+4)

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