Find the Holes in the Graph f(x)=(x^2-2x-8)/(-x^2+3x+4)

$f (x) = \frac{x^{2} - 2 x - 8}{- x^{2} + 3 x + 4}$

Factor $x^{2} - 2 x - 8$ 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 $- 8$ and whose sum is $- 2$ .

$- 4, 2$

Write the factored form using these integers.

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

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

Factor $3$ out of $3 x$ .

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

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

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

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

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

Cancel the common factor of $x - 4$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

$x - 4$

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

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Substitute $4$ for $x$ in $\frac{x + 2}{- x - 1}$ and simplify.

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

$\frac{4 + 2}{- 1 \cdot 4 - 1}$

Simplify.

Add $4$ and $2$ .

$\frac{6}{- 1 \cdot 4 - 1}$

Simplify the denominator.

Multiply $- 1$ by $4$ .

$\frac{6}{- 4 - 1}$

Subtract $1$ from $- 4$ .

$\frac{6}{- 5}$

Move the negative in front of the fraction.

$- \frac{6}{5}$

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

$(4, - \frac{6}{5})$

Find the Holes in the Graph f(x)=(x^2-2x-8)/(-x^2+3x+4)

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