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

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

Factor $x^{2} - 9$ .

Rewrite $9$ as $3^{2}$ .

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

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

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

Factor $3$ out of $3 x$ .

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

Factor $3$ out of $- 9$ .

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

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

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

Cancel the common factor of $x - 3$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

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

$\frac{3 + 3}{3}$

Simplify.

Cancel the common factor of $3 + 3$ and $3$ .

Factor $3$ out of $3$ .

$\frac{3 \cdot 1 + 3}{3}$

Factor $3$ out of $3$ .

$\frac{3 \cdot 1 + 3 \cdot 1}{3}$

Factor $3$ out of $3 \cdot 1 + 3 \cdot 1$ .

$\frac{3 \cdot (1 + 1)}{3}$

Cancel the common factors.

Factor $3$ out of $3$ .

$\frac{3 \cdot (1 + 1)}{3 (1)}$

Cancel the common factor.

$\frac{3 \cdot (1 + 1)}{3 \cdot 1}$

Rewrite the expression.

$\frac{1 + 1}{1}$

Divide $1 + 1$ by $1$ .

$1 + 1$

Add $1$ and $1$ .

$2$

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

$(3, 2)$

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

Related Questions

Support LCMGCF.com by disabling your adblocker.