Pre-Algebra Examples

$h (x) = | x - 6 |$

Step 1

Find the absolute value vertex. In this case, the vertex for $y = | x - 6 |$ is $(6, 0)$ .

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Step 1.1

To find the $x$ coordinate of the vertex, set the inside of the absolute value $x - 6$ equal to $0$ . In this case, $x - 6 = 0$ .

$x - 6 = 0$

Step 1.2

Add $6$ to both sides of the equation.

$x = 6$

Step 1.3

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

$y = | (6) - 6 |$

Step 1.4

Simplify $| (6) - 6 |$ .

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Step 1.4.1

Subtract $6$ from $6$ .

$y = | 0 |$

Step 1.4.2

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 $(6, 0)$ .

$(6, 0)$

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 $4$ into $f (x) = | x - 6 |$ . In this case, the point is $(4, 2)$ .

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Step 3.1.1

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

$f (4) = | (4) - 6 |$

Step 3.1.2

Simplify the result.

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Step 3.1.2.1

Subtract $6$ from $4$ .

$f (4) = | - 2 |$

Step 3.1.2.2

The absolute value is the distance between a number and zero. The distance between $- 2$ and $0$ is $2$ .

$f (4) = 2$

Step 3.1.2.3

The final answer is $2$ .

$y = 2$

Step 3.2

Substitute the $x$ value $5$ into $f (x) = | x - 6 |$ . In this case, the point is $(5, 1)$ .

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Step 3.2.1

Replace the variable $x$ with $5$ in the expression.

$f (5) = | (5) - 6 |$

Step 3.2.2

Simplify the result.

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Step 3.2.2.1

Subtract $6$ from $5$ .

$f (5) = | - 1 |$

Step 3.2.2.2

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

$f (5) = 1$

Step 3.2.2.3

The final answer is $1$ .

$y = 1$

Step 3.3

Substitute the $x$ value $8$ into $f (x) = | x - 6 |$ . In this case, the point is $(8, 2)$ .

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Step 3.3.1

Replace the variable $x$ with $8$ in the expression.

$f (8) = | (8) - 6 |$

Step 3.3.2

Simplify the result.

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Step 3.3.2.1

Subtract $6$ from $8$ .

$f (8) = | 2 |$

Step 3.3.2.2

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$f (8) = 2$

Step 3.3.2.3

The final answer is $2$ .

$y = 2$

Step 3.4

The absolute value can be graphed using the points around the vertex $(6, 0), (4, 2), (5, 1), (7, 1), (8, 2)$

$\begin{array}{c} x & y \\ 4 & 2 \\ 5 & 1 \\ 6 & 0 \\ 7 & 1 \\ 8 & 2 \end{array}$

Step 4