Pre-Algebra Examples

f (x) = \frac{1}{2} \cdot | x | + \frac{4}{5}

$f (x) = \frac{1}{2} \cdot | x | + \frac{4}{5}$

Step 1

Find the absolute value vertex. In this case, the vertex for

y = \frac{| x |}{2} + \frac{4}{5}

$y = \frac{| x |}{2} + \frac{4}{5}$ is

(0, \frac{4}{5})

$(0, \frac{4}{5})$ .

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Step 1.1

To find the

x

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

x

$x$ equal to

0

$0$ . In this case,

x = 0

$x = 0$ .

x = 0

$x = 0$

Step 1.2

Replace the variable

x

$x$ with

0

$0$ in the expression.

y = \frac{| 0 |}{2} + \frac{4}{5}

$y = \frac{| 0 |}{2} + \frac{4}{5}$

Step 1.3

Simplify

\frac{| 0 |}{2} + \frac{4}{5}

$\frac{| 0 |}{2} + \frac{4}{5}$ .

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Step 1.3.1

Simplify each term.

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Step 1.3.1.1

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

0

$0$ and

0

$0$ is

0

$0$ .

y = \frac{0}{2} + \frac{4}{5}

$y = \frac{0}{2} + \frac{4}{5}$

Step 1.3.1.2

Divide

0

$0$ by

2

$2$ .

y = 0 + \frac{4}{5}

$y = 0 + \frac{4}{5}$

y = 0 + \frac{4}{5}

$y = 0 + \frac{4}{5}$

Step 1.3.2

Add

0

$0$ and

\frac{4}{5}

$\frac{4}{5}$ .

y = \frac{4}{5}

$y = \frac{4}{5}$

y = \frac{4}{5}

$y = \frac{4}{5}$

Step 1.4

The absolute value vertex is

(0, \frac{4}{5})

$(0, \frac{4}{5})$ .

(0, \frac{4}{5})

$(0, \frac{4}{5})$

(0, \frac{4}{5})

$(0, \frac{4}{5})$

Step 2

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

(- \infty, \infty)

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 3

For each

$x$ value, there is one

$y$ value. Select a few

$x$ values from the domain. It would be more useful to select the values so that they are around the

$x$ value of the absolute value vertex.

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Step 3.1

Substitute the

$x$ value

$- 2$ into

$f (x) = \frac{| x |}{2} + \frac{4}{5}$ . In this case, the point is

$(- 2, \frac{9}{5})$ .

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Step 3.1.1

Replace the variable

$x$ with

$- 2$ in the expression.

$f (- 2) = \frac{| - 2 |}{2} + \frac{4}{5}$

Step 3.1.2

Simplify the result.

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Step 3.1.2.1

Simplify each term.

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Step 3.1.2.1.1

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

$- 2$ and

$0$ is

$2$ .

$f (- 2) = \frac{2}{2} + \frac{4}{5}$

Step 3.1.2.1.2

Divide

$2$ by

$2$ .

$f (- 2) = 1 + \frac{4}{5}$

Step 3.1.2.2

Simplify the expression.

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Step 3.1.2.2.1

Write

$1$ as a fraction with a common denominator.

$f (- 2) = \frac{5}{5} + \frac{4}{5}$

Step 3.1.2.2.2

Combine the numerators over the common denominator.

$f (- 2) = \frac{5 + 4}{5}$

Step 3.1.2.2.3

Add

$5$ and

$4$ .

$f (- 2) = \frac{9}{5}$

Step 3.1.2.3

The final answer is

$\frac{9}{5}$ .

$y = \frac{9}{5}$

Step 3.2

Substitute the

$x$ value

$- 1$ into

$f (x) = \frac{| x |}{2} + \frac{4}{5}$ . In this case, the point is

$(- 1, \frac{13}{10})$ .

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Step 3.2.1

Replace the variable

$x$ with

$- 1$ in the expression.

$f (- 1) = \frac{| - 1 |}{2} + \frac{4}{5}$

Step 3.2.2

Simplify the result.

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Step 3.2.2.1

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

$- 1$ and

$0$ is

$1$ .

$f (- 1) = \frac{1}{2} + \frac{4}{5}$

Step 3.2.2.2

To write

$\frac{1}{2}$ as a fraction with a common denominator, multiply by

$\frac{5}{5}$ .

$f (- 1) = \frac{1}{2} \cdot \frac{5}{5} + \frac{4}{5}$

Step 3.2.2.3

To write

$\frac{4}{5}$ as a fraction with a common denominator, multiply by

$\frac{2}{2}$ .

$f (- 1) = \frac{1}{2} \cdot \frac{5}{5} + \frac{4}{5} \cdot \frac{2}{2}$

Step 3.2.2.4

Write each expression with a common denominator of

$10$ , by multiplying each by an appropriate factor of

$1$ .

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Step 3.2.2.4.1

Multiply

$\frac{1}{2}$ by

$\frac{5}{5}$ .

$f (- 1) = \frac{5}{2 \cdot 5} + \frac{4}{5} \cdot \frac{2}{2}$

Step 3.2.2.4.2

Multiply

$2$ by

$5$ .

$f (- 1) = \frac{5}{10} + \frac{4}{5} \cdot \frac{2}{2}$

Step 3.2.2.4.3

Multiply

$\frac{4}{5}$ by

$\frac{2}{2}$ .

$f (- 1) = \frac{5}{10} + \frac{4 \cdot 2}{5 \cdot 2}$

Step 3.2.2.4.4

Multiply

$5$ by

$2$ .

$f (- 1) = \frac{5}{10} + \frac{4 \cdot 2}{10}$

Step 3.2.2.5

Combine the numerators over the common denominator.

$f (- 1) = \frac{5 + 4 \cdot 2}{10}$

Step 3.2.2.6

Simplify the numerator.

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Step 3.2.2.6.1

Multiply

$4$ by

$2$ .

$f (- 1) = \frac{5 + 8}{10}$

Step 3.2.2.6.2

Add

$5$ and

$8$ .

$f (- 1) = \frac{13}{10}$

Step 3.2.2.7

The final answer is

$\frac{13}{10}$ .

$y = \frac{13}{10}$

Step 3.3

Substitute the

$x$ value

$2$ into

$f (x) = \frac{| x |}{2} + \frac{4}{5}$ . In this case, the point is

$(2, \frac{9}{5})$ .

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Step 3.3.1

Replace the variable

$x$ with

$2$ in the expression.

$f (2) = \frac{| 2 |}{2} + \frac{4}{5}$

Step 3.3.2

Simplify the result.

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Step 3.3.2.1

Simplify each term.

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Step 3.3.2.1.1

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

$0$ and

$2$ is

$2$ .

$f (2) = \frac{2}{2} + \frac{4}{5}$

Step 3.3.2.1.2

Divide

$2$ by

$2$ .

$f (2) = 1 + \frac{4}{5}$

Step 3.3.2.2

Simplify the expression.

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Step 3.3.2.2.1

Write

$1$ as a fraction with a common denominator.

$f (2) = \frac{5}{5} + \frac{4}{5}$

Step 3.3.2.2.2

Combine the numerators over the common denominator.

$f (2) = \frac{5 + 4}{5}$

Step 3.3.2.2.3

Add

$5$ and

$4$ .

$f (2) = \frac{9}{5}$

Step 3.3.2.3

The final answer is

$\frac{9}{5}$ .

$y = \frac{9}{5}$

Step 3.4

The absolute value can be graphed using the points around the vertex

$(0, \frac{4}{5}), (- 2, 1.8), (- 1, 1.3), (1, 1.3), (2, 1.8)$

$\begin{array}{c} x & y \\ - 2 & 1.8 \\ - 1 & 1.3 \\ 0 & \frac{4}{5} \\ 1 & 1.3 \\ 2 & 1.8 \end{array}$

Step 4