Find the Holes in the Graph f(x)=(-5x+20)/(2x^2-32)

$f (x) = \frac{- 5 x + 20}{2 x^{2} - 32}$

Factor $5$ out of $- 5 x + 20$ .

Factor $5$ out of $- 5 x$ .

$f (x) = \frac{5 (- x) + 20}{2 x^{2} - 32}$

Factor $5$ out of $20$ .

$f (x) = \frac{5 (- x) + 5 (4)}{2 x^{2} - 32}$

Factor $5$ out of $5 (- x) + 5 (4)$ .

$f (x) = \frac{5 (- x + 4)}{2 x^{2} - 32}$

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

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

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

$f (x) = \frac{5 (- x + 4)}{2 (x^{2}) - 32}$

Factor $2$ out of $- 32$ .

$f (x) = \frac{5 (- x + 4)}{2 x^{2} + 2 \cdot - 16}$

Factor $2$ out of $2 x^{2} + 2 \cdot - 16$ .

$f (x) = \frac{5 (- x + 4)}{2 (x^{2} - 16)}$

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

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

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 $b = 4$ .

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

Remove unnecessary parentheses.

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

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

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

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

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

$f (x) = \frac{5 (- (x) - 1 \cdot - 4)}{2 (x + 4) (x - 4)}$

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

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

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

$f (x) = \frac{5 (- 1 (x - 4))}{2 (x + 4) (x - 4)}$

Cancel the common factor.

$f (x) = \frac{5 (- 1 (x - 4))}{2 (x + 4) (x - 4)}$

Rewrite the expression.

$f (x) = \frac{5 \cdot (- 1)}{2 (x + 4)}$

Simplify the numerator.

Multiply $5$ by $- 1$ .

$f (x) = \frac{5 (- 1)}{2 (x + 4)}$

Multiply $5$ by $- 1$ .

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

Move the negative in front of the fraction.

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

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 $- \frac{5}{2 (x + 4)}$ .

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Substitute $4$ for $x$ in $- \frac{5}{2 (x + 4)}$ and simplify.

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

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

Simplify.

Add $4$ and $4$ .

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

Multiply $2$ by $8$ .

$- \frac{5}{16}$

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

$(4, - \frac{5}{16})$

Find the Holes in the Graph f(x)=(-5x+20)/(2x^2-32)

Related Questions

Support LCMGCF.com by disabling your adblocker.