Algebra Examples

$f (x) = | \frac{1}{2} x - 2 |$

Step 1

Find the absolute value vertex. In this case, the vertex for $y = | \frac{x}{2} - 2 |$ is $(4, 0)$ .

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Step 1.1

To find the $x$ coordinate of the vertex, set the inside of the absolute value $\frac{x}{2} - 2$ equal to $0$ . In this case, $\frac{x}{2} - 2 = 0$ .

$\frac{x}{2} - 2 = 0$

Step 1.2

Solve the equation $\frac{x}{2} - 2 = 0$ to find the $x$ coordinate for the absolute value vertex.

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Step 1.2.1

Add $2$ to both sides of the equation.

$\frac{x}{2} = 2$

Step 1.2.2

Multiply both sides of the equation by $2$ .

$2 \frac{x}{2} = 2 \cdot 2$

Step 1.2.3

Simplify both sides of the equation.

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Step 1.2.3.1

Simplify the left side.

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Step 1.2.3.1.1

Cancel the common factor of $2$ .

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Step 1.2.3.1.1.1

Cancel the common factor.

$2 \frac{x}{2} = 2 \cdot 2$

Step 1.2.3.1.1.2

Rewrite the expression.

$x = 2 \cdot 2$

Step 1.2.3.2

Simplify the right side.

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Step 1.2.3.2.1

Multiply $2$ by $2$ .

$x = 4$

Step 1.3

Replace the variable $x$ with $4$ in the expression.

$y = | \frac{4}{2} - 2 |$

Step 1.4

Simplify $| \frac{4}{2} - 2 |$ .

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Step 1.4.1

Divide $4$ by $2$ .

$y = | 2 - 2 |$

Step 1.4.2

Subtract $2$ from $2$ .

$y = | 0 |$

Step 1.4.3

The absolute value is the distance between a number and zero. The distance between $0$ and $0$ is $0$ .

$y = 0$

Step 1.5

The absolute value vertex is $(4, 0)$ .

$(4, 0)$

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:

$(- \infty, \infty)$

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 $f (x) = | \frac{x}{2} - 2 |$ . In this case, the point is $(2, 1)$ .

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Step 3.1.1

Replace the variable $x$ with $2$ in the expression.

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

Step 3.1.2

Simplify the result.

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Step 3.1.2.1

Divide $2$ by $2$ .

$f (2) = | 1 - 2 |$

Step 3.1.2.2

Subtract $2$ from $1$ .

$f (2) = | - 1 |$

Step 3.1.2.3

The absolute value is the distance between a number and zero. The distance between $- 1$ and $0$ is $1$ .

$f (2) = 1$

Step 3.1.2.4

The final answer is $1$ .

$y = 1$

Step 3.2

Substitute the $x$ value $3$ into $f (x) = | \frac{x}{2} - 2 |$ . In this case, the point is $(3, \frac{1}{2})$ .

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Step 3.2.1

Replace the variable $x$ with $3$ in the expression.

$f (3) = | \frac{3}{2} - 2 |$

Step 3.2.2

Simplify the result.

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Step 3.2.2.1

To write $- 2$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f (3) = | \frac{3}{2} - 2 \cdot \frac{2}{2} |$

Step 3.2.2.2

Combine $- 2$ and $\frac{2}{2}$ .

$f (3) = | \frac{3}{2} + \frac{- 2 \cdot 2}{2} |$

Step 3.2.2.3

Combine the numerators over the common denominator.

$f (3) = | \frac{3 - 2 \cdot 2}{2} |$

Step 3.2.2.4

Simplify the numerator.

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Step 3.2.2.4.1

Multiply $- 2$ by $2$ .

$f (3) = | \frac{3 - 4}{2} |$

Step 3.2.2.4.2

Subtract $4$ from $3$ .

$f (3) = | \frac{- 1}{2} |$

Step 3.2.2.5

Move the negative in front of the fraction.

$f (3) = | - \frac{1}{2} |$

Step 3.2.2.6

$- \frac{1}{2}$ is approximately $- 0.5$ which is negative so negate $- \frac{1}{2}$ and remove the absolute value

$f (3) = \frac{1}{2}$

Step 3.2.2.7

The final answer is $\frac{1}{2}$ .

$y = \frac{1}{2}$

Step 3.3

Substitute the $x$ value $6$ into $f (x) = | \frac{x}{2} - 2 |$ . In this case, the point is $(6, 1)$ .

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Step 3.3.1

Replace the variable $x$ with $6$ in the expression.

$f (6) = | \frac{6}{2} - 2 |$

Step 3.3.2

Simplify the result.

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Step 3.3.2.1

Divide $6$ by $2$ .

$f (6) = | 3 - 2 |$

Step 3.3.2.2

Subtract $2$ from $3$ .

$f (6) = | 1 |$

Step 3.3.2.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$f (6) = 1$

Step 3.3.2.4

The final answer is $1$ .

$y = 1$

Step 3.4

The absolute value can be graphed using the points around the vertex $(4, 0), (2, 1), (3, 0.5), (5, 0.5), (6, 1)$

$\begin{array}{c} x & y \\ 2 & 1 \\ 3 & 0.5 \\ 4 & 0 \\ 5 & 0.5 \\ 6 & 1 \end{array}$

Step 4