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

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

Step 1

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 $a \cdot c = - 1 \cdot 2 = - 2$ and whose sum is $b = - 1$ .

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

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

Rewrite $- 1$ as $1$ plus $- 2$

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

Apply the distributive property.

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

Multiply $x$ by $1$ .

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

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

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

Step 2

Factor $x^{2} - 4$ .

Rewrite $4$ as $2^{2}$ .

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 2$ .

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

Step 3

Cancel the common factor of $x + 2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 4

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

$x + 2$

Step 5

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

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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

Substitute $- 2$ for $x$ to find the $y$ coordinate of the hole.

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

Simplify.

Add $2$ and $1$ .

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

Subtract $2$ from $- 2$ .

$\frac{3}{- 4}$

Move the negative in front of the fraction.

$- \frac{3}{4}$

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

$(- 2, - \frac{3}{4})$

Step 6

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

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