Find the Holes in the Graph f(x)=(-3x^2+3x)/(2x^2-9x+7)

$f (x) = \frac{- 3 x^{2} + 3 x}{2 x^{2} - 9 x + 7}$

Step 1

Factor $3 x$ out of $- 3 x^{2} + 3 x$ .

Factor $3 x$ out of $- 3 x^{2}$ .

$f (x) = \frac{3 x (- x) + 3 x}{2 x^{2} - 9 x + 7}$

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

$f (x) = \frac{3 x (- x) + 3 x (1)}{2 x^{2} - 9 x + 7}$

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

$f (x) = \frac{3 x (- x + 1)}{2 x^{2} - 9 x + 7}$

Step 2

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot 7 = 14$ and whose sum is $b = - 9$ .

Factor $- 9$ out of $- 9 x$ .

$f (x) = \frac{3 x (- x + 1)}{2 x^{2} - 9 x + 7}$

Rewrite $- 9$ as $- 2$ plus $- 7$

$f (x) = \frac{3 x (- x + 1)}{2 x^{2} + (- 2 - 7) x + 7}$

Apply the distributive property.

$f (x) = \frac{3 x (- x + 1)}{2 x^{2} - 2 x - 7 x + 7}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$f (x) = \frac{3 x (- x + 1)}{(2 x^{2} - 2 x) - 7 x + 7}$

Factor out the greatest common factor (GCF) from each group.

$f (x) = \frac{3 x (- x + 1)}{2 x (x - 1) - 7 (x - 1)}$

Factor the polynomial by factoring out the greatest common factor, $x - 1$ .

$f (x) = \frac{3 x (- x + 1)}{(x - 1) (2 x - 7)}$

Step 3

Cancel the common factor of $- x + 1$ and $x - 1$ .

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

$f (x) = \frac{3 x (- (x) + 1)}{(x - 1) (2 x - 7)}$

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

$f (x) = \frac{3 x (- (x) - 1 \cdot - 1)}{(x - 1) (2 x - 7)}$

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

$f (x) = \frac{3 x (- (x - 1))}{(x - 1) (2 x - 7)}$

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

$f (x) = \frac{3 x (- 1 (x - 1))}{(x - 1) (2 x - 7)}$

Cancel the common factor.

$f (x) = \frac{3 x (- 1 (x - 1))}{(x - 1) (2 x - 7)}$

Rewrite the expression.

$f (x) = \frac{3 x \cdot (- 1)}{2 x - 7}$

Step 4

Multiply $- 1$ by $3$ .

$f (x) = \frac{- 3 x}{2 x - 7}$

Step 5

Move the negative in front of the fraction.

$f (x) = - \frac{(3) x}{2 x - 7}$

Step 6

Multiply $3$ by $x$ .

$f (x) = - \frac{3 x}{2 x - 7}$

Step 7

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

$x - 1$

Step 8

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $- \frac{3 x}{2 x - 7}$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Substitute $1$ for $x$ in $- \frac{3 x}{2 x - 7}$ and simplify.

Substitute $1$ for $x$ to find the $y$ coordinate of the hole.

$- \frac{3 \cdot 1}{2 \cdot 1 - 7}$

Simplify.

Multiply $3$ by $1$ .

$- \frac{3}{2 \cdot 1 - 7}$

Simplify the denominator.

Multiply $2$ by $1$ .

$- \frac{3}{2 - 7}$

Subtract $7$ from $2$ .

$- \frac{3}{- 5}$

Move the negative in front of the fraction.

$- - \frac{3}{5}$

Multiply $- - \frac{3}{5}$ .

Multiply $- 1$ by $- 1$ .

$1 (\frac{3}{5})$

Multiply $\frac{3}{5}$ by $1$ .

$\frac{3}{5}$

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

$(1, \frac{3}{5})$

Step 9

Find the Holes in the Graph f(x)=(-3x^2+3x)/(2x^2-9x+7)

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