Find the Holes in the Graph f(x)=(8x^2-16x)/(5x^2-20)

$f (x) = \frac{8 x^{2} - 16 x}{5 x^{2} - 20}$

Step 1

Factor $8 x$ out of $8 x^{2} - 16 x$ .

Factor $8 x$ out of $8 x^{2}$ .

$f (x) = \frac{8 x (x) - 16 x}{5 x^{2} - 20}$

Factor $8 x$ out of $- 16 x$ .

$f (x) = \frac{8 x (x) + 8 x (- 2)}{5 x^{2} - 20}$

Factor $8 x$ out of $8 x (x) + 8 x (- 2)$ .

$f (x) = \frac{8 x (x - 2)}{5 x^{2} - 20}$

Step 2

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

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

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

$f (x) = \frac{8 x (x - 2)}{5 (x^{2}) - 20}$

Factor $5$ out of $- 20$ .

$f (x) = \frac{8 x (x - 2)}{5 x^{2} + 5 \cdot - 4}$

Factor $5$ out of $5 x^{2} + 5 \cdot - 4$ .

$f (x) = \frac{8 x (x - 2)}{5 (x^{2} - 4)}$

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

$f (x) = \frac{8 x (x - 2)}{5 (x^{2} - 2^{2})}$

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 $b = 2$ .

$f (x) = \frac{8 x (x - 2)}{5 ((x + 2) (x - 2))}$

Remove unnecessary parentheses.

$f (x) = \frac{8 x (x - 2)}{5 (x + 2) (x - 2)}$

Step 3

Cancel the common factor of $x - 2$ .

Cancel the common factor.

$f (x) = \frac{8 x (x - 2)}{5 (x + 2) (x - 2)}$

Rewrite the expression.

$f (x) = \frac{8 x}{5 (x + 2)}$

Step 4

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

$x - 2$

Step 5

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{8 x}{5 (x + 2)}$ .

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Substitute $2$ for $x$ in $\frac{8 x}{5 (x + 2)}$ and simplify.

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

$\frac{8 \cdot 2}{5 (2 + 2)}$

Simplify.

Cancel the common factor of $8$ and $2 + 2$ .

Factor $2$ out of $8 \cdot 2$ .

$\frac{2 (4 \cdot 2)}{5 (2 + 2)}$

Cancel the common factors.

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

$\frac{2 (4 \cdot 2)}{2 (5 (1 + 1))}$

Cancel the common factor.

$\frac{2 (4 \cdot 2)}{2 (5 (1 + 1))}$

Rewrite the expression.

$\frac{4 \cdot 2}{5 (1 + 1)}$

Multiply $4$ by $2$ .

$\frac{8}{5 (1 + 1)}$

Add $1$ and $1$ .

$\frac{8}{5 \cdot 2}$

Multiply $5$ by $2$ .

$\frac{8}{10}$

Cancel the common factor of $8$ and $10$ .

Factor $2$ out of $8$ .

$\frac{2 (4)}{10}$

Cancel the common factors.

Factor $2$ out of $10$ .

$\frac{2 \cdot 4}{2 \cdot 5}$

Cancel the common factor.

$\frac{2 \cdot 4}{2 \cdot 5}$

Rewrite the expression.

$\frac{4}{5}$

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

$(2, \frac{4}{5})$

Step 6

Find the Holes in the Graph f(x)=(8x^2-16x)/(5x^2-20)

Related Questions

Support LCMGCF.com by disabling your adblocker.