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

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

Step 1

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.

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

Step 2

Cancel the common factor of $x - 4$ .

Cancel the common factor.

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

Divide $x + 2$ by $1$ .

$f (x) = x + 2$

Step 3

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

$x - 4$

Step 4

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

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Substitute $4$ for $x$ in $x + 2$ and simplify.

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

$4 + 2$

Add $4$ and $2$ .

$6$

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

$(4, 6)$

Step 5

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

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