Examples

f (x) = x^{3} - 1

$f (x) = x^{3} - 1$

Step 1

Find the x-intercepts.

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Step 1.1

To find the x-intercept(s), substitute in

0

$0$ for

y

$y$ and solve for

x

$x$ .

0 = x^{3} - 1

$0 = x^{3} - 1$

Step 1.2

Solve the equation.

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Step 1.2.1

Rewrite the equation as

x^{3} - 1 = 0

$x^{3} - 1 = 0$ .

x^{3} - 1 = 0

$x^{3} - 1 = 0$

Step 1.2.2

Add

1

$1$ to both sides of the equation.

x^{3} = 1

$x^{3} = 1$

Step 1.2.3

Subtract

1

$1$ from both sides of the equation.

x^{3} - 1 = 0

$x^{3} - 1 = 0$

Step 1.2.4

Factor the left side of the equation.

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Step 1.2.4.1

Rewrite

1

$1$ as

1^{3}

$1^{3}$ .

x^{3} - 1^{3} = 0

$x^{3} - 1^{3} = 0$

Step 1.2.4.2

Since both terms are perfect cubes, factor using the difference of cubes formula,

a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})

$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where

a = x

$a = x$ and

b = 1

$b = 1$ .

(x - 1) (x^{2} + x \cdot 1 + 1^{2}) = 0

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

Step 1.2.4.3

Simplify.

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Step 1.2.4.3.1

Multiply

x

$x$ by

1

$1$ .

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

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

Step 1.2.4.3.2

One to any power is one.

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

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

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

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

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

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

Step 1.2.5

If any individual factor on the left side of the equation is equal to

0

$0$ , the entire expression will be equal to

0

$0$ .

x - 1 = 0

$x - 1 = 0$

x^{2} + x + 1 = 0

$x^{2} + x + 1 = 0$

Step 1.2.6

Set

x - 1

$x - 1$ equal to

0

$0$ and solve for

x

$x$ .

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Step 1.2.6.1

Set

x - 1

$x - 1$ equal to

0

$0$ .

x - 1 = 0

$x - 1 = 0$

Step 1.2.6.2

Add

1

$1$ to both sides of the equation.

x = 1

$x = 1$

x = 1

$x = 1$

Step 1.2.7

Set

x^{2} + x + 1

$x^{2} + x + 1$ equal to

0

$0$ and solve for

x

$x$ .

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Step 1.2.7.1

Set

x^{2} + x + 1

$x^{2} + x + 1$ equal to

0

$0$ .

x^{2} + x + 1 = 0

$x^{2} + x + 1 = 0$

Step 1.2.7.2

Solve

x^{2} + x + 1 = 0

$x^{2} + x + 1 = 0$ for

x

$x$ .

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Step 1.2.7.2.1

Use the quadratic formula to find the solutions.

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

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

Step 1.2.7.2.2

Substitute the values

a = 1

$a = 1$ ,

b = 1

$b = 1$ , and

c = 1

$c = 1$ into the quadratic formula and solve for

x

$x$ .

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

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

Step 1.2.7.2.3

Simplify.

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Step 1.2.7.2.3.1

Simplify the numerator.

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Step 1.2.7.2.3.1.1

One to any power is one.

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

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

Step 1.2.7.2.3.1.2

Multiply

- 4 \cdot 1 \cdot 1

$- 4 \cdot 1 \cdot 1$ .

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Step 1.2.7.2.3.1.2.1

Multiply

- 4

$- 4$ by

1

$1$ .

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

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

Step 1.2.7.2.3.1.2.2

Multiply

- 4

$- 4$ by

1

$1$ .

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

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

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

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

Step 1.2.7.2.3.1.3

Subtract

4

$4$ from

1

$1$ .

x = \frac{- 1 \pm \sqrt{- 3}}{2 \cdot 1}

$x = \frac{- 1 \pm \sqrt{- 3}}{2 \cdot 1}$

Step 1.2.7.2.3.1.4

Rewrite

- 3

$- 3$ as

- 1 (3)

$- 1 (3)$ .

x = \frac{- 1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}

$x = \frac{- 1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Step 1.2.7.2.3.1.5

Rewrite

\sqrt{- 1 (3)}

$\sqrt{- 1 (3)}$ as

\sqrt{- 1} \cdot \sqrt{3}

$\sqrt{- 1} \cdot \sqrt{3}$ .

x = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}

$x = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

Step 1.2.7.2.3.1.6

Rewrite

\sqrt{- 1}

$\sqrt{- 1}$ as

i

$i$ .

x = \frac{- 1 \pm i \sqrt{3}}{2 \cdot 1}

$x = \frac{- 1 \pm i \sqrt{3}}{2 \cdot 1}$

x = \frac{- 1 \pm i \sqrt{3}}{2 \cdot 1}

$x = \frac{- 1 \pm i \sqrt{3}}{2 \cdot 1}$

Step 1.2.7.2.3.2

Multiply

2

$2$ by

1

$1$ .

x = \frac{- 1 \pm i \sqrt{3}}{2}

$x = \frac{- 1 \pm i \sqrt{3}}{2}$

x = \frac{- 1 \pm i \sqrt{3}}{2}

$x = \frac{- 1 \pm i \sqrt{3}}{2}$

Step 1.2.7.2.4

Simplify the expression to solve for the

+

$+$ portion of the

\pm

$\pm$ .

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Step 1.2.7.2.4.1

Simplify the numerator.

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Step 1.2.7.2.4.1.1

One to any power is one.

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

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

Step 1.2.7.2.4.1.2

Multiply

- 4 \cdot 1 \cdot 1

$- 4 \cdot 1 \cdot 1$ .

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Step 1.2.7.2.4.1.2.1

Multiply

- 4

$- 4$ by

1

$1$ .

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

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

Step 1.2.7.2.4.1.2.2

Multiply

- 4

$- 4$ by

1

$1$ .

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

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

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

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

Step 1.2.7.2.4.1.3

Subtract

4

$4$ from

1

$1$ .

x = \frac{- 1 \pm \sqrt{- 3}}{2 \cdot 1}

$x = \frac{- 1 \pm \sqrt{- 3}}{2 \cdot 1}$

Step 1.2.7.2.4.1.4

Rewrite

- 3

$- 3$ as

- 1 (3)

$- 1 (3)$ .

x = \frac{- 1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}

$x = \frac{- 1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Step 1.2.7.2.4.1.5

Rewrite

\sqrt{- 1 (3)}

$\sqrt{- 1 (3)}$ as

\sqrt{- 1} \cdot \sqrt{3}

$\sqrt{- 1} \cdot \sqrt{3}$ .

x = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}

$x = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

Step 1.2.7.2.4.1.6

Rewrite

\sqrt{- 1}

$\sqrt{- 1}$ as

i

$i$ .

x = \frac{- 1 \pm i \sqrt{3}}{2 \cdot 1}

$x = \frac{- 1 \pm i \sqrt{3}}{2 \cdot 1}$

x = \frac{- 1 \pm i \sqrt{3}}{2 \cdot 1}

$x = \frac{- 1 \pm i \sqrt{3}}{2 \cdot 1}$

Step 1.2.7.2.4.2

Multiply

2

$2$ by

1

$1$ .

x = \frac{- 1 \pm i \sqrt{3}}{2}

$x = \frac{- 1 \pm i \sqrt{3}}{2}$

Step 1.2.7.2.4.3

Change the

\pm

$\pm$ to

+

$+$ .

x = \frac{- 1 + i \sqrt{3}}{2}

$x = \frac{- 1 + i \sqrt{3}}{2}$

Step 1.2.7.2.4.4

Rewrite

- 1

$- 1$ as

- 1 (1)

$- 1 (1)$ .

x = \frac{- 1 \cdot 1 + i \sqrt{3}}{2}

$x = \frac{- 1 \cdot 1 + i \sqrt{3}}{2}$

Step 1.2.7.2.4.5

Factor

- 1

$- 1$ out of

i \sqrt{3}

$i \sqrt{3}$ .

x = \frac{- 1 \cdot 1 - (- i \sqrt{3})}{2}

$x = \frac{- 1 \cdot 1 - (- i \sqrt{3})}{2}$

Step 1.2.7.2.4.6

Factor

- 1

$- 1$ out of

- 1 (1) - (- i \sqrt{3})

$- 1 (1) - (- i \sqrt{3})$ .

x = \frac{- 1 (1 - i \sqrt{3})}{2}

$x = \frac{- 1 (1 - i \sqrt{3})}{2}$

Step 1.2.7.2.4.7

Move the negative in front of the fraction.

x = - \frac{1 - i \sqrt{3}}{2}

$x = - \frac{1 - i \sqrt{3}}{2}$

x = - \frac{1 - i \sqrt{3}}{2}

$x = - \frac{1 - i \sqrt{3}}{2}$

Step 1.2.7.2.5

Simplify the expression to solve for the

-

$-$ portion of the

\pm

$\pm$ .

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Step 1.2.7.2.5.1

Simplify the numerator.

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Step 1.2.7.2.5.1.1

One to any power is one.

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

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

Step 1.2.7.2.5.1.2

Multiply

- 4 \cdot 1 \cdot 1

$- 4 \cdot 1 \cdot 1$ .

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Step 1.2.7.2.5.1.2.1

Multiply

- 4

$- 4$ by

1

$1$ .

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

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

Step 1.2.7.2.5.1.2.2

Multiply

- 4

$- 4$ by

1

$1$ .

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

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

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

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

Step 1.2.7.2.5.1.3

Subtract

4

$4$ from

1

$1$ .

x = \frac{- 1 \pm \sqrt{- 3}}{2 \cdot 1}

$x = \frac{- 1 \pm \sqrt{- 3}}{2 \cdot 1}$

Step 1.2.7.2.5.1.4

Rewrite

- 3

$- 3$ as

- 1 (3)

$- 1 (3)$ .

x = \frac{- 1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}

$x = \frac{- 1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Step 1.2.7.2.5.1.5

Rewrite

\sqrt{- 1 (3)}

$\sqrt{- 1 (3)}$ as

\sqrt{- 1} \cdot \sqrt{3}

$\sqrt{- 1} \cdot \sqrt{3}$ .

x = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}

$x = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

Step 1.2.7.2.5.1.6

Rewrite

\sqrt{- 1}

$\sqrt{- 1}$ as

i

$i$ .

x = \frac{- 1 \pm i \sqrt{3}}{2 \cdot 1}

$x = \frac{- 1 \pm i \sqrt{3}}{2 \cdot 1}$

x = \frac{- 1 \pm i \sqrt{3}}{2 \cdot 1}

$x = \frac{- 1 \pm i \sqrt{3}}{2 \cdot 1}$

Step 1.2.7.2.5.2

Multiply

2

$2$ by

1

$1$ .

x = \frac{- 1 \pm i \sqrt{3}}{2}

$x = \frac{- 1 \pm i \sqrt{3}}{2}$

Step 1.2.7.2.5.3

Change the

\pm

$\pm$ to

-

$-$ .

x = \frac{- 1 - i \sqrt{3}}{2}

$x = \frac{- 1 - i \sqrt{3}}{2}$

Step 1.2.7.2.5.4

Rewrite

- 1

$- 1$ as

- 1 (1)

$- 1 (1)$ .

x = \frac{- 1 \cdot 1 - i \sqrt{3}}{2}

$x = \frac{- 1 \cdot 1 - i \sqrt{3}}{2}$

Step 1.2.7.2.5.5

Factor

- 1

$- 1$ out of

- i \sqrt{3}

$- i \sqrt{3}$ .

x = \frac{- 1 \cdot 1 - (i \sqrt{3})}{2}

$x = \frac{- 1 \cdot 1 - (i \sqrt{3})}{2}$

Step 1.2.7.2.5.6

Factor

- 1

$- 1$ out of

$- 1 (1) - (i \sqrt{3})$ .

$x = \frac{- 1 (1 + i \sqrt{3})}{2}$

Step 1.2.7.2.5.7

Move the negative in front of the fraction.

$x = - \frac{1 + i \sqrt{3}}{2}$

Step 1.2.7.2.6

The final answer is the combination of both solutions.

$x = - \frac{1 - i \sqrt{3}}{2}, - \frac{1 + i \sqrt{3}}{2}$

Step 1.2.8

The final solution is all the values that make

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

$x = 1, - \frac{1 - i \sqrt{3}}{2}, - \frac{1 + i \sqrt{3}}{2}$

Step 1.3

x-intercept(s) in point form.

x-intercept(s):

$(1, 0)$

x-intercept(s):

$(1, 0)$

Step 2

Find the y-intercepts.

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Step 2.1

To find the y-intercept(s), substitute in

$0$ for

$x$ and solve for

$y$ .

$y = {(0)}^{3} - 1$

Step 2.2

Solve the equation.

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Step 2.2.1

Remove parentheses.

$y = 0^{3} - 1$

Step 2.2.2

Remove parentheses.

$y = {(0)}^{3} - 1$

Step 2.2.3

Simplify

${(0)}^{3} - 1$ .

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Step 2.2.3.1

Raising

$0$ to any positive power yields

$0$ .

$y = 0 - 1$

Step 2.2.3.2

Subtract

$1$ from

$0$ .

$y = - 1$

Step 2.3

y-intercept(s) in point form.

y-intercept(s):

$(0, - 1)$

y-intercept(s):

$(0, - 1)$

Step 3

List the intersections.

x-intercept(s):

$(1, 0)$

y-intercept(s):

$(0, - 1)$

Step 4

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