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

$\frac{x^{2} + x - 12}{3 x^{2} - 9 x}$

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.

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

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

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

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

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

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

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

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

Cancel the common factor of $x - 3$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

$x - 3$

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 $0$ .

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

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

Related Questions

Support LCMGCF.com by disabling your adblocker.