Find the Holes in the Graph f(x)=(4x^2-16)/(-3x^2+6x)

$f (x) = \frac{4 x^{2} - 16}{- 3 x^{2} + 6 x}$

Step 1

Factor $4 x^{2} - 16$ .

Factor $4$ out of $4 x^{2} - 16$ .

Factor $4$ out of $4 x^{2}$ .

$f (x) = \frac{4 (x^{2}) - 16}{- 3 x^{2} + 6 x}$

Factor $4$ out of $- 16$ .

$f (x) = \frac{4 x^{2} + 4 \cdot - 4}{- 3 x^{2} + 6 x}$

Factor $4$ out of $4 x^{2} + 4 \cdot - 4$ .

$f (x) = \frac{4 (x^{2} - 4)}{- 3 x^{2} + 6 x}$

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

$f (x) = \frac{4 (x^{2} - 2^{2})}{- 3 x^{2} + 6 x}$

Factor.

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 $b = 2$ .

$f (x) = \frac{4 ((x + 2) (x - 2))}{- 3 x^{2} + 6 x}$

Remove unnecessary parentheses.

$f (x) = \frac{4 (x + 2) (x - 2)}{- 3 x^{2} + 6 x}$

Step 2

Factor $3 x$ out of $- 3 x^{2} + 6 x$ .

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

$f (x) = \frac{4 (x + 2) (x - 2)}{3 x (- x) + 6 x}$

Factor $3 x$ out of $6 x$ .

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

Factor $3 x$ out of $3 x (- x) + 3 x (2)$ .

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

Step 3

Cancel the common factor of $x - 2$ and $- x + 2$ .

Factor $- 1$ out of $x$ .

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

Rewrite $- 2$ as $- 1 (2)$ .

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

Factor $- 1$ out of $- 1 (- x) - 1 (2)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 4

Multiply $- 1$ by $4$ .

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

Step 5

Move the negative in front of the fraction.

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

Step 6

Multiply $4$ by $x + 2$ .

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

Step 7

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

$- x + 2$

Step 8

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

Set $- x + 2$ equal to $0$ .

$- x + 2 = 0$

Solve $- x + 2 = 0$ for $x$ .

Subtract $2$ from both sides of the equation.

$- x = - 2$

Divide each term in $- x = - 2$ by $- 1$ and simplify.

Divide each term in $- x = - 2$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 2}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 2}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 2}{- 1}$

Simplify the right side.

Divide $- 2$ by $- 1$ .

$x = 2$

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

Substitute $2$ for $x$ to find the $y$ coordinate of the hole.

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

Simplify.

Cancel the common factor of $4$ and $2$ .

Factor $2$ out of $4 (2 + 2)$ .

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

Cancel the common factors.

Factor $2$ out of $3 \cdot 2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Add $2$ and $2$ .

$- \frac{2 \cdot 4}{3}$

Multiply $2$ by $4$ .

$- \frac{8}{3}$

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

$(2, - \frac{8}{3})$

Step 9

Find the Holes in the Graph f(x)=(4x^2-16)/(-3x^2+6x)

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