Find the x and y Intercepts f(x)=8x^3-2x

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

Step 1

Find the x-intercepts.

To find the x-intercept(s), substitute in $0$ for $y$ and solve for $x$ .

$0 = 8 x^{3} - 2 x$

Solve the equation.

Rewrite the equation as $8 x^{3} - 2 x = 0$ .

$8 x^{3} - 2 x = 0$

Factor the left side of the equation.

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

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

$2 x (4 x^{2}) - 2 x = 0$

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

$2 x (4 x^{2}) + 2 x (- 1) = 0$

Factor $2 x$ out of $2 x (4 x^{2}) + 2 x (- 1)$ .

$2 x (4 x^{2} - 1) = 0$

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

$2 x ({(2 x)}^{2} - 1) = 0$

Rewrite $1$ as $1^{2}$ .

$2 x ({(2 x)}^{2} - 1^{2}) = 0$

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 x$ and $b = 1$ .

$2 x ((2 x + 1) (2 x - 1)) = 0$

Remove unnecessary parentheses.

$2 x (2 x + 1) (2 x - 1) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$2 x + 1 = 0$

$2 x - 1 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $2 x + 1$ equal to $0$ and solve for $x$ .

Set $2 x + 1$ equal to $0$ .

$2 x + 1 = 0$

Solve $2 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$2 x = - 1$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Set $2 x - 1$ equal to $0$ and solve for $x$ .

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Solve $2 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$2 x = 1$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

The final solution is all the values that make $2 x (2 x + 1) (2 x - 1) = 0$ true.

$x = 0, - \frac{1}{2}, \frac{1}{2}$

x-intercept(s) in point form.

x-intercept(s): $(0, 0), (- \frac{1}{2}, 0), (\frac{1}{2}, 0)$

Step 2

Find the y-intercepts.

To find the y-intercept(s), substitute in $0$ for $x$ and solve for $y$ .

$y = 8 {(0)}^{3} - 2 \cdot 0$

Solve the equation.

Remove parentheses.

$y = 8 \cdot 0^{3} - 2 \cdot 0$

Remove parentheses.

$y = 8 {(0)}^{3} - 2 \cdot 0$

Simplify $8 {(0)}^{3} - 2 \cdot 0$ .

Simplify each term.

Raising $0$ to any positive power yields $0$ .

$y = 8 \cdot 0 - 2 \cdot 0$

Multiply $8$ by $0$ .

$y = 0 - 2 \cdot 0$

Multiply $- 2$ by $0$ .

$y = 0 + 0$

Add $0$ and $0$ .

$y = 0$

y-intercept(s) in point form.

y-intercept(s): $(0, 0)$

Step 3

List the intersections.

x-intercept(s): $(0, 0), (- \frac{1}{2}, 0), (\frac{1}{2}, 0)$

y-intercept(s): $(0, 0)$

Step 4

Find the x and y Intercepts f(x)=8x^3-2x

Related Questions

Support LCMGCF.com by disabling your adblocker.