Find the Holes in the Graph f(x)=(-5x^2+20)/(3x^2-6x)

$f (x) = \frac{- 5 x^{2} + 20}{3 x^{2} - 6 x}$

Step 1

Factor $- 5 x^{2} + 20$ .

Factor $5$ out of $- 5 x^{2} + 20$ .

Factor $5$ out of $- 5 x^{2}$ .

$f (x) = \frac{5 (- x^{2}) + 20}{3 x^{2} - 6 x}$

Factor $5$ out of $20$ .

$f (x) = \frac{5 (- x^{2}) + 5 (4)}{3 x^{2} - 6 x}$

Factor $5$ out of $5 (- x^{2}) + 5 (4)$ .

$f (x) = \frac{5 (- x^{2} + 4)}{3 x^{2} - 6 x}$

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

$f (x) = \frac{5 (- x^{2} + 2^{2})}{3 x^{2} - 6 x}$

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

$f (x) = \frac{5 (2^{2} - x^{2})}{3 x^{2} - 6 x}$

Factor.

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

$f (x) = \frac{5 ((2 + x) (2 - x))}{3 x^{2} - 6 x}$

Remove unnecessary parentheses.

$f (x) = \frac{5 (2 + x) (2 - x)}{3 x^{2} - 6 x}$

Step 2

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

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

$f (x) = \frac{5 (2 + x) (2 - x)}{3 x (x) - 6 x}$

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

$f (x) = \frac{5 (2 + x) (2 - x)}{3 x (x) + 3 x (- 2)}$

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

$f (x) = \frac{5 (2 + x) (2 - x)}{3 x (x - 2)}$

Step 3

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

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

$f (x) = \frac{5 (2 + x) (- 1 \cdot - 2 - x)}{3 x (x - 2)}$

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

$f (x) = \frac{5 (2 + x) (- 1 \cdot - 2 - (x))}{3 x (x - 2)}$

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

$f (x) = \frac{5 (2 + x) (- 1 (- 2 + x))}{3 x (x - 2)}$

Reorder terms.

$f (x) = \frac{5 (2 + x) (- 1 (x - 2))}{3 x (x - 2)}$

Cancel the common factor.

$f (x) = \frac{5 (2 + x) (- 1 (x - 2))}{3 x (x - 2)}$

Rewrite the expression.

$f (x) = \frac{5 (2 + x) \cdot (- 1)}{3 x}$

Step 4

Multiply $- 1$ by $5$ .

$f (x) = \frac{- 5 (2 + x)}{3 x}$

Step 5

Move the negative in front of the fraction.

$f (x) = - \frac{(5) (2 + x)}{3 x}$

Step 6

Multiply $5$ by $2 + x$ .

$f (x) = - \frac{5 (2 + x)}{3 x}$

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 $- \frac{5 (2 + x)}{3 x}$ .

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Substitute $2$ for $x$ in $- \frac{5 (2 + x)}{3 x}$ and simplify.

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

$- \frac{5 (2 + 2)}{3 \cdot 2}$

Simplify.

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

Factor $2$ out of $5 (2 + 2)$ .

$- \frac{2 (5 (1 + 1))}{3 \cdot 2}$

Cancel the common factors.

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

$- \frac{2 (5 (1 + 1))}{2 \cdot 3}$

Cancel the common factor.

$- \frac{2 (5 (1 + 1))}{2 \cdot 3}$

Rewrite the expression.

$- \frac{5 (1 + 1)}{3}$

Add $1$ and $1$ .

$- \frac{5 \cdot 2}{3}$

Multiply $5$ by $2$ .

$- \frac{10}{3}$

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

$(2, - \frac{10}{3})$

Step 9

Find the Holes in the Graph f(x)=(-5x^2+20)/(3x^2-6x)

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