Find the Holes in the Graph f(x)=(-2x^2+6x)/(x^2-4x+3)

$f (x) = \frac{- 2 x^{2} + 6 x}{x^{2} - 4 x + 3}$

Factor $2 x$ out of $- 2 x^{2} + 6 x$ .

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

$f (x) = \frac{2 x (- x) + 6 x}{x^{2} - 4 x + 3}$

Factor $2 x$ out of $6 x$ .

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

Factor $2 x$ out of $2 x (- x) + 2 x (3)$ .

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

Factor $x^{2} - 4 x + 3$ 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 $3$ and whose sum is $- 4$ .

$- 3, - 1$

Write the factored form using these integers.

$f (x) = \frac{2 x (- x + 3)}{(x - 3) (x - 1)}$

Cancel the common factor of $- x + 3$ and $x - 3$ .

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

$f (x) = \frac{2 x (- (x) + 3)}{(x - 3) (x - 1)}$

Rewrite $3$ as $- 1 (- 3)$ .

$f (x) = \frac{2 x (- (x) - 1 \cdot - 3)}{(x - 3) (x - 1)}$

Factor $- 1$ out of $- (x) - 1 (- 3)$ .

$f (x) = \frac{2 x (- (x - 3))}{(x - 3) (x - 1)}$

Rewrite $- (x - 3)$ as $- 1 (x - 3)$ .

$f (x) = \frac{2 x (- 1 (x - 3))}{(x - 3) (x - 1)}$

Cancel the common factor.

$f (x) = \frac{2 x (- 1 (x - 3))}{(x - 3) (x - 1)}$

Rewrite the expression.

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

Multiply $- 1$ by $2$ .

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

Move the negative in front of the fraction.

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

Multiply $2$ by $x$ .

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

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

$x - 3$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $- \frac{2 x}{x - 1}$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Substitute $3$ for $x$ in $- \frac{2 x}{x - 1}$ and simplify.

Substitute $3$ for $x$ to find the $y$ coordinate of the hole.

$- \frac{2 \cdot 3}{3 - 1}$

Simplify.

Multiply $2$ by $3$ .

$- \frac{6}{3 - 1}$

Subtract $1$ from $3$ .

$- \frac{6}{2}$

Divide $6$ by $2$ .

$- 1 \cdot 3$

Multiply $- 1$ by $3$ .

$- 3$

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

$(3, - 3)$

Find the Holes in the Graph f(x)=(-2x^2+6x)/(x^2-4x+3)

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