Precalculus Examples

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

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

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

Solve the equation.

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Rewrite the equation as $x^{3} - 1 = 0$ .

$x^{3} - 1 = 0$

Add $1$ to both sides of the equation.

$x^{3} = 1$

Move $1$ to the left side of the equation by subtracting it from both sides.

$x^{3} - 1 = 0$

Factor the left side of the equation.

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Rewrite $1$ as $1^{3}$ .

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

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = x$ and $b = 1$ .

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

Simplify.

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Multiply $x$ by $1$ .

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

One to any power is one.

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

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

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Set the factor equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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

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Set the factor equal to $0$ .

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

Use the quadratic formula to find the solutions.

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

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

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

Simplify.

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Simplify the numerator.

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One to any power is one.

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

Multiply $1$ by $1$ .

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

Multiply $- 4$ by $1$ .

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

Subtract $4$ from $1$ .

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

Rewrite $- 3$ as $- 1 (3)$ .

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

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

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

Rewrite $\sqrt{- 1}$ as $i$ .

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

Multiply $i$ by $\sqrt{3}$ .

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

Multiply $2$ by $1$ .

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

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

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Simplify the numerator.

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One to any power is one.

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

Multiply $1$ by $1$ .

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

Multiply $- 4$ by $1$ .

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

Subtract $4$ from $1$ .

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

Rewrite $- 3$ as $- 1 (3)$ .

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

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

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

Rewrite $\sqrt{- 1}$ as $i$ .

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

Multiply $i$ by $\sqrt{3}$ .

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

Multiply $2$ by $1$ .

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

Start simplifying.

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

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

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

Factor $- 1$ out of $i \sqrt{3}$ .

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

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

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

Move the negative in front of the fraction.

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

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

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Simplify the numerator.

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One to any power is one.

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

Multiply $1$ by $1$ .

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

Multiply $- 4$ by $1$ .

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

Subtract $4$ from $1$ .

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

Rewrite $- 3$ as $- 1 (3)$ .

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

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

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

Rewrite $\sqrt{- 1}$ as $i$ .

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

Multiply $i$ by $\sqrt{3}$ .

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

Multiply $2$ by $1$ .

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

Start simplifying.

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

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

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

Factor $- 1$ out of $- i \sqrt{3}$ .

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

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

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

Move the negative in front of the fraction.

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

The final answer is the combination of both solutions.

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

The solution is the result of $x - 1 = 0$ and $x^{2} + x + 1 = 0$ .

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

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

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

Simplify the right side.

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Simplify each term.

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Remove parentheses around $0$ .

$y = 0^{3} - 1$

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

$y = 0 - 1$

Subtract $1$ from $0$ .

$y = - 1$

These are the $x$ and $y$ intercepts of the equation $y = x^{3} - 1$ .

x-intercept: $(1, 0)$

y-intercept: $(0, - 1)$

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