Find the Holes in the Graph f(x)=(x^2+x-12)/(3x^2-9x)

$f (x) = \frac{x^{2} + x - 12}{3 x^{2} - 9 x}$

Step 1

Factor $x^{2} + x - 12$ 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 $- 12$ and whose sum is $1$ .

$- 3, 4$

Write the factored form using these integers.

$f (x) = \frac{(x - 3) (x + 4)}{3 x^{2} - 9 x}$

Step 2

Factor $3 x$ out of $3 x^{2} - 9 x$ .

Factor $3 x$ out of $3 x^{2}$ .

$f (x) = \frac{(x - 3) (x + 4)}{3 x (x) - 9 x}$

Factor $3 x$ out of $- 9 x$ .

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

Factor $3 x$ out of $3 x (x) + 3 x (- 3)$ .

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

Step 3

Cancel the common factor of $x - 3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 4

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

$x - 3$

Step 5

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{x + 4}{3 x}$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

Substitute $3$ for $x$ to find the $y$ coordinate of the hole.

$\frac{3 + 4}{3 \cdot 3}$

Simplify.

Add $3$ and $4$ .

$\frac{7}{3 \cdot 3}$

Multiply $3$ by $3$ .

$\frac{7}{9}$

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

$(3, \frac{7}{9})$

Step 6

Find the Holes in the Graph f(x)=(x^2+x-12)/(3x^2-9x)

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