Find the Holes in the Graph f(x)=(x^3+2x^2-8x)/(x^2-2x)

$f (x) = \frac{x^{3} + 2 x^{2} - 8 x}{x^{2} - 2 x}$

Step 1

Factor $x^{3} + 2 x^{2} - 8 x$ .

Factor $x$ out of $x^{3} + 2 x^{2} - 8 x$ .

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

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

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

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

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

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

Factor $x$ out of $x \cdot x^{2} + x (2 x)$ .

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

Factor $x$ out of $x (x^{2} + 2 x) + x \cdot - 8$ .

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

Factor.

Factor $x^{2} + 2 x - 8$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 8$ and whose sum is $2$ .

$- 2, 4$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

Step 2

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

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

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

Factor $x$ out of $- 2 x$ .

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

Factor $x$ out of $x \cdot x + x \cdot - 2$ .

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

Step 3

Cancel the common factor of $x$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 4

Cancel the common factor of $x - 2$ .

Cancel the common factor.

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

Divide $x + 4$ by $1$ .

$f (x) = x + 4$

Step 5

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

$x, x - 2$

Step 6

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $x + 4$ .

Set $x$ equal to $0$ .

$x = 0$

Substitute $0$ for $x$ in $x + 4$ and simplify.

Substitute $0$ for $x$ to find the $y$ coordinate of the hole.

$0 + 4$

Add $0$ and $4$ .

$4$

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Substitute $2$ for $x$ in $x + 4$ and simplify.

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

$2 + 4$

Add $2$ and $4$ .

$6$

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

$(0, 4), (2, 6)$

Step 7

Find the Holes in the Graph f(x)=(x^3+2x^2-8x)/(x^2-2x)

Related Questions

Support LCMGCF.com by disabling your adblocker.