Mathway

Visit Mathway on the web

Start 7-day free trial on the app

Start 7-day free trial on the app

Download free on Amazon

Download free in Windows Store

Take a photo of your math problem on the app

Precalculus Examples

y = | \frac{1}{3} x + 1 |

$y = | \frac{1}{3} x + 1 |$

Step 1

To find the

x

$x$ coordinate of the vertex, set the inside of the absolute value

\frac{1}{3} x + 1

$\frac{1}{3} x + 1$ equal to

0

$0$ . In this case,

\frac{1}{3} x + 1 = 0

$\frac{1}{3} x + 1 = 0$ .

\frac{1}{3} x + 1 = 0

$\frac{1}{3} x + 1 = 0$

Step 2

Solve the equation

\frac{1}{3} x + 1 = 0

$\frac{1}{3} x + 1 = 0$ to find the

x

$x$ coordinate for the absolute value vertex.

Tap for more steps...

Step 2.1

Combine

\frac{1}{3}

$\frac{1}{3}$ and

x

$x$ .

\frac{x}{3} + 1 = 0

$\frac{x}{3} + 1 = 0$

Step 2.2

Subtract

1

$1$ from both sides of the equation.

\frac{x}{3} = - 1

$\frac{x}{3} = - 1$

Step 2.3

Multiply both sides of the equation by

3

$3$ .

3 \frac{x}{3} = 3 \cdot - 1

$3 \frac{x}{3} = 3 \cdot - 1$

Step 2.4

Simplify both sides of the equation.

Tap for more steps...

Step 2.4.1

Simplify the left side.

Tap for more steps...

Step 2.4.1.1

Cancel the common factor of

3

$3$ .

Tap for more steps...

Step 2.4.1.1.1

Cancel the common factor.

3 \frac{x}{3} = 3 \cdot - 1

$3 \frac{x}{3} = 3 \cdot - 1$

Step 2.4.1.1.2

Rewrite the expression.

x = 3 \cdot - 1

$x = 3 \cdot - 1$

x = 3 \cdot - 1

$x = 3 \cdot - 1$

x = 3 \cdot - 1

$x = 3 \cdot - 1$

Step 2.4.2

Simplify the right side.

Tap for more steps...

Step 2.4.2.1

Multiply

3

$3$ by

- 1

$- 1$ .

x = - 3

$x = - 3$

x = - 3

$x = - 3$

x = - 3

$x = - 3$

x = - 3

$x = - 3$

Step 3

Replace the variable

x

$x$ with

- 3

$- 3$ in the expression.

y = | \frac{1}{3} \cdot (- 3) + 1 |

$y = | \frac{1}{3} \cdot (- 3) + 1 |$

Step 4

Simplify

| \frac{1}{3} \cdot (- 3) + 1 |

$| \frac{1}{3} \cdot (- 3) + 1 |$ .

Tap for more steps...

Step 4.1

Cancel the common factor of

3

$3$ .

Tap for more steps...

Step 4.1.1

Factor

3

$3$ out of

- 3

$- 3$ .

y = | \frac{1}{3} \cdot (3 (- 1)) + 1 |

$y = | \frac{1}{3} \cdot (3 (- 1)) + 1 |$

Step 4.1.2

Cancel the common factor.

y = | \frac{1}{3} \cdot (3 \cdot - 1) + 1 |

$y = | \frac{1}{3} \cdot (3 \cdot - 1) + 1 |$

Step 4.1.3

Rewrite the expression.

y = | - 1 + 1 |

$y = | - 1 + 1 |$

y = | - 1 + 1 |

$y = | - 1 + 1 |$

Step 4.2

Add

- 1

$- 1$ and

1

$1$ .

y = | 0 |

$y = | 0 |$

Step 4.3

The absolute value is the distance between a number and zero. The distance between

0

$0$ and

0

$0$ is

0

$0$ .

y = 0

$y = 0$

y = 0

$y = 0$

Step 5

The absolute value vertex is

(- 3, 0)

$(- 3, 0)$ .

(- 3, 0)

$(- 3, 0)$

Step 6

Enter YOUR Problem

using Amazon.Auth.AccessControlPolicy;

Mathway requires javascript and a modern browser.

[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$