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

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

Step 1

Factor $x^{2} - x - 6$ 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 $- 6$ and whose sum is $- 1$ .

$- 3, 2$

Write the factored form using these integers.

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

Step 2

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.

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

Step 3

Cancel the common factor of $x + 2$ .

Cancel the common factor.

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

Rewrite the expression.

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

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 $\frac{x - 3}{x - 4}$ .

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Substitute $- 2$ for $x$ in $\frac{x - 3}{x - 4}$ and simplify.

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

$\frac{- 2 - 3}{- 2 - 4}$

Simplify.

Subtract $3$ from $- 2$ .

$\frac{- 5}{- 2 - 4}$

Subtract $4$ from $- 2$ .

$\frac{- 5}{- 6}$

Dividing two negative values results in a positive value.

$\frac{5}{6}$

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

$(- 2, \frac{5}{6})$

Step 6

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

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