Find the Holes in the Graph f(x)=(x-1)/(-2x^2+8x-6)

$f (x) = \frac{x - 1}{- 2 x^{2} + 8 x - 6}$

Step 1

Factor $- 2 x^{2} + 8 x - 6$ .

Factor $2$ out of $- 2 x^{2} + 8 x - 6$ .

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

$f (x) = \frac{x - 1}{2 (- x^{2}) + 8 x - 6}$

Factor $2$ out of $8 x$ .

$f (x) = \frac{x - 1}{2 (- x^{2}) + 2 (4 x) - 6}$

Factor $2$ out of $- 6$ .

$f (x) = \frac{x - 1}{2 (- x^{2}) + 2 (4 x) + 2 (- 3)}$

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

$f (x) = \frac{x - 1}{2 (- x^{2} + 4 x) + 2 (- 3)}$

Factor $2$ out of $2 (- x^{2} + 4 x) + 2 (- 3)$ .

$f (x) = \frac{x - 1}{2 (- x^{2} + 4 x - 3)}$

Factor.

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 $a \cdot c = - 1 \cdot - 3 = 3$ and whose sum is $b = 4$ .

Factor $4$ out of $4 x$ .

$f (x) = \frac{x - 1}{2 (- x^{2} + 4 (x) - 3)}$

Rewrite $4$ as $1$ plus $3$

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

Apply the distributive property.

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

Multiply $x$ by $1$ .

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

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

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

Remove unnecessary parentheses.

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

Step 2

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

Factor $- 1$ out of $x$ .

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

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

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

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 3

Move the negative in front of the fraction.

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

Step 4

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

$- x + 1$

Step 5

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $- \frac{1}{2 (x - 3)}$ .

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

$- x + 1 = 0$

Solve $- x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$- x = - 1$

Divide each term in $- x = - 1$ by $- 1$ and simplify.

Divide each term in $- x = - 1$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 1}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 1}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 1}{- 1}$

Simplify the right side.

Divide $- 1$ by $- 1$ .

$x = 1$

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

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

$- \frac{1}{2 (1 - 3)}$

Simplify.

Subtract $3$ from $1$ .

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

Multiply $2$ by $- 2$ .

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

Move the negative in front of the fraction.

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

Multiply $- - \frac{1}{4}$ .

Multiply $- 1$ by $- 1$ .

$1 (\frac{1}{4})$

Multiply $\frac{1}{4}$ by $1$ .

$\frac{1}{4}$

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

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

Step 6

Find the Holes in the Graph f(x)=(x-1)/(-2x^2+8x-6)

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