Find the Holes in the Graph f(x)=(3x+18)/(x^2-2x-48)

$f (x) = \frac{3 x + 18}{x^{2} - 2 x - 48}$

Step 1

Factor $3$ out of $3 x + 18$ .

Factor $3$ out of $3 x$ .

$f (x) = \frac{3 (x) + 18}{x^{2} - 2 x - 48}$

Factor $3$ out of $18$ .

$f (x) = \frac{3 x + 3 \cdot 6}{x^{2} - 2 x - 48}$

Factor $3$ out of $3 x + 3 \cdot 6$ .

$f (x) = \frac{3 (x + 6)}{x^{2} - 2 x - 48}$

Step 2

Factor $x^{2} - 2 x - 48$ 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 $- 48$ and whose sum is $- 2$ .

$- 8, 6$

Write the factored form using these integers.

$f (x) = \frac{3 (x + 6)}{(x - 8) (x + 6)}$

Step 3

Cancel the common factor of $x + 6$ .

Cancel the common factor.

$f (x) = \frac{3 (x + 6)}{(x - 8) (x + 6)}$

Rewrite the expression.

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

Step 4

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

$x + 6$

Step 5

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{3}{x - 8}$ .

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

Substitute $- 6$ for $x$ in $\frac{3}{x - 8}$ and simplify.

Substitute $- 6$ for $x$ to find the $y$ coordinate of the hole.

$\frac{3}{- 6 - 8}$

Simplify.

Subtract $8$ from $- 6$ .

$\frac{3}{- 14}$

Move the negative in front of the fraction.

$- \frac{3}{14}$

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

$(- 6, - \frac{3}{14})$

Step 6

Find the Holes in the Graph f(x)=(3x+18)/(x^2-2x-48)

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