Find the x and y Intercepts f(x)=4x-8x^3-x^2

$f (x) = 4 x - 8 x^{3} - x^{2}$

Step 1

Find the x-intercepts.

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

$0 = 4 x - 8 x^{3} - x^{2}$

Solve the equation.

Rewrite the equation as $4 x - 8 x^{3} - x^{2} = 0$ .

$4 x - 8 x^{3} - x^{2} = 0$

Factor $- x$ out of $4 x - 8 x^{3} - x^{2}$ .

Move $4 x$ .

$- 8 x^{3} - x^{2} + 4 x = 0$

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

$- x (8 x^{2}) - x^{2} + 4 x = 0$

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

$- x (8 x^{2}) - x \cdot x + 4 x = 0$

Factor $- x$ out of $4 x$ .

$- x (8 x^{2}) - x \cdot x - x \cdot - 4 = 0$

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

$- x (8 x^{2} + x) - x \cdot - 4 = 0$

Factor $- x$ out of $- x (8 x^{2} + x) - x (- 4)$ .

$- x (8 x^{2} + x - 4) = 0$

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

$x = 0$

$8 x^{2} + x - 4 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $8 x^{2} + x - 4$ equal to $0$ and solve for $x$ .

Set $8 x^{2} + x - 4$ equal to $0$ .

$8 x^{2} + x - 4 = 0$

Solve $8 x^{2} + x - 4 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 8$ , $b = 1$ , and $c = - 4$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (8 \cdot - 4)}}{2 \cdot 8}$

Simplify.

Simplify the numerator.

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 8 \cdot - 4}}{2 \cdot 8}$

Multiply $- 4 \cdot 8 \cdot - 4$ .

Multiply $- 4$ by $8$ .

$x = \frac{- 1 \pm \sqrt{1 - 32 \cdot - 4}}{2 \cdot 8}$

Multiply $- 32$ by $- 4$ .

$x = \frac{- 1 \pm \sqrt{1 + 128}}{2 \cdot 8}$

Add $1$ and $128$ .

$x = \frac{- 1 \pm \sqrt{129}}{2 \cdot 8}$

Multiply $2$ by $8$ .

$x = \frac{- 1 \pm \sqrt{129}}{16}$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Simplify the numerator.

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 8 \cdot - 4}}{2 \cdot 8}$

Multiply $- 4 \cdot 8 \cdot - 4$ .

Multiply $- 4$ by $8$ .

$x = \frac{- 1 \pm \sqrt{1 - 32 \cdot - 4}}{2 \cdot 8}$

Multiply $- 32$ by $- 4$ .

$x = \frac{- 1 \pm \sqrt{1 + 128}}{2 \cdot 8}$

Add $1$ and $128$ .

$x = \frac{- 1 \pm \sqrt{129}}{2 \cdot 8}$

Multiply $2$ by $8$ .

$x = \frac{- 1 \pm \sqrt{129}}{16}$

Change the $\pm$ to $+$ .

$x = \frac{- 1 + \sqrt{129}}{16}$

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

$x = \frac{- 1 \cdot 1 + \sqrt{129}}{16}$

Factor $- 1$ out of $\sqrt{129}$ .

$x = \frac{- 1 \cdot 1 - 1 (- \sqrt{129})}{16}$

Factor $- 1$ out of $- 1 (1) - 1 (- \sqrt{129})$ .

$x = \frac{- 1 (1 - \sqrt{129})}{16}$

Move the negative in front of the fraction.

$x = - \frac{1 - \sqrt{129}}{16}$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Simplify the numerator.

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 8 \cdot - 4}}{2 \cdot 8}$

Multiply $- 4 \cdot 8 \cdot - 4$ .

Multiply $- 4$ by $8$ .

$x = \frac{- 1 \pm \sqrt{1 - 32 \cdot - 4}}{2 \cdot 8}$

Multiply $- 32$ by $- 4$ .

$x = \frac{- 1 \pm \sqrt{1 + 128}}{2 \cdot 8}$

Add $1$ and $128$ .

$x = \frac{- 1 \pm \sqrt{129}}{2 \cdot 8}$

Multiply $2$ by $8$ .

$x = \frac{- 1 \pm \sqrt{129}}{16}$

Change the $\pm$ to $-$ .

$x = \frac{- 1 - \sqrt{129}}{16}$

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

$x = \frac{- 1 \cdot 1 - \sqrt{129}}{16}$

Factor $- 1$ out of $- \sqrt{129}$ .

$x = \frac{- 1 \cdot 1 - (\sqrt{129})}{16}$

Factor $- 1$ out of $- 1 (1) - (\sqrt{129})$ .

$x = \frac{- 1 (1 + \sqrt{129})}{16}$

Move the negative in front of the fraction.

$x = - \frac{1 + \sqrt{129}}{16}$

The final answer is the combination of both solutions.

$x = - \frac{1 - \sqrt{129}}{16}, - \frac{1 + \sqrt{129}}{16}$

The final solution is all the values that make $- x (8 x^{2} + x - 4) = 0$ true.

$x = 0, - \frac{1 - \sqrt{129}}{16}, - \frac{1 + \sqrt{129}}{16}$

x-intercept(s) in point form.

x-intercept(s): $(0, 0), (- \frac{1 - \sqrt{129}}{16}, 0), (- \frac{1 + \sqrt{129}}{16}, 0)$

Step 2

Find the y-intercepts.

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

$y = 4 (0) - 8 {(0)}^{3} - {(0)}^{2}$

Solve the equation.

Remove parentheses.

$y = 4 (0) - 8 \cdot 0^{3} - {(0)}^{2}$

Remove parentheses.

$y = 4 (0) - 8 \cdot 0^{3} - 0^{2}$

Remove parentheses.

$y = 4 (0) - 8 {(0)}^{3} - {(0)}^{2}$

Simplify $4 (0) - 8 {(0)}^{3} - {(0)}^{2}$ .

Simplify each term.

Multiply $4$ by $0$ .

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

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

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

Multiply $- 8$ by $0$ .

$y = 0 + 0 - {(0)}^{2}$

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

$y = 0 + 0 - 0$

Multiply $- 1$ by $0$ .

$y = 0 + 0 + 0$

Simplify by adding numbers.

Add $0$ and $0$ .

$y = 0 + 0$

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): $(0, 0), (- \frac{1 - \sqrt{129}}{16}, 0), (- \frac{1 + \sqrt{129}}{16}, 0)$

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

Step 4

Find the x and y Intercepts f(x)=4x-8x^3-x^2

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