Find the Holes in the Graph f(x)=(-2x^2+32)/(3x-12)

$f (x) = \frac{- 2 x^{2} + 32}{3 x - 12}$

Factor $- 2 x^{2} + 32$ .

Factor $2$ out of $- 2 x^{2} + 32$ .

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

$f (x) = \frac{2 (- x^{2}) + 32}{3 x - 12}$

Factor $2$ out of $32$ .

$f (x) = \frac{2 (- x^{2}) + 2 (16)}{3 x - 12}$

Factor $2$ out of $2 (- x^{2}) + 2 (16)$ .

$f (x) = \frac{2 (- x^{2} + 16)}{3 x - 12}$

Rewrite $16$ as $4^{2}$ .

$f (x) = \frac{2 (- x^{2} + 4^{2})}{3 x - 12}$

Reorder $- x^{2}$ and $4^{2}$ .

$f (x) = \frac{2 (4^{2} - x^{2})}{3 x - 12}$

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 4$ and $b = x$ .

$f (x) = \frac{2 ((4 + x) (4 - x))}{3 x - 12}$

Remove unnecessary parentheses.

$f (x) = \frac{2 (4 + x) (4 - x)}{3 x - 12}$

Factor $3$ out of $3 x - 12$ .

Factor $3$ out of $3 x$ .

$f (x) = \frac{2 (4 + x) (4 - x)}{3 (x) - 12}$

Factor $3$ out of $- 12$ .

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

Factor $3$ out of $3 x + 3 \cdot - 4$ .

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

Cancel the common factor of $4 - x$ and $x - 4$ .

Rewrite $4$ as $- 1 (- 4)$ .

$f (x) = \frac{2 (4 + x) (- 1 \cdot - 4 - x)}{3 (x - 4)}$

Factor $- 1$ out of $- x$ .

$f (x) = \frac{2 (4 + x) (- 1 \cdot - 4 - (x))}{3 (x - 4)}$

Factor $- 1$ out of $- 1 (- 4) - (x)$ .

$f (x) = \frac{2 (4 + x) (- 1 (- 4 + x))}{3 (x - 4)}$

Reorder terms.

$f (x) = \frac{2 (4 + x) (- 1 (x - 4))}{3 (x - 4)}$

Cancel the common factor.

$f (x) = \frac{2 (4 + x) (- 1 (x - 4))}{3 (x - 4)}$

Rewrite the expression.

$f (x) = \frac{2 (4 + x) \cdot (- 1)}{3}$

Multiply $- 1$ by $2$ .

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

Move the negative in front of the fraction.

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

Multiply $2$ by $4 + x$ .

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

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

$x - 4$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $- \frac{2 (4 + x)}{3}$ .

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Substitute $4$ for $x$ in $- \frac{2 (4 + x)}{3}$ and simplify.

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

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

Simplify.

Add $4$ and $4$ .

$- \frac{2 \cdot 8}{3}$

Multiply $2$ by $8$ .

$- \frac{16}{3}$

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

$(4, - \frac{16}{3})$

Find the Holes in the Graph f(x)=(-2x^2+32)/(3x-12)

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