Pre-Algebra Examples

$| 6 x - 3 |$

Step 1

Find the absolute value vertex. In this case, the vertex for $y = | 6 x - 3 |$ is $(\frac{1}{2}, 0)$ .

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Step 1.1

To find the $x$ coordinate of the vertex, set the inside of the absolute value $6 x - 3$ equal to $0$ . In this case, $6 x - 3 = 0$ .

$6 x - 3 = 0$

Step 1.2

Solve the equation $6 x - 3 = 0$ to find the $x$ coordinate for the absolute value vertex.

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Step 1.2.1

Add $3$ to both sides of the equation.

$6 x = 3$

Step 1.2.2

Divide each term in $6 x = 3$ by $6$ and simplify.

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Step 1.2.2.1

Divide each term in $6 x = 3$ by $6$ .

$\frac{6 x}{6} = \frac{3}{6}$

Step 1.2.2.2

Simplify the left side.

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Step 1.2.2.2.1

Cancel the common factor of $6$ .

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Step 1.2.2.2.1.1

Cancel the common factor.

$\frac{6 x}{6} = \frac{3}{6}$

Step 1.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{3}{6}$

Step 1.2.2.3

Simplify the right side.

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Step 1.2.2.3.1

Cancel the common factor of $3$ and $6$ .

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Step 1.2.2.3.1.1

Factor $3$ out of $3$ .

$x = \frac{3 (1)}{6}$

Step 1.2.2.3.1.2

Cancel the common factors.

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Step 1.2.2.3.1.2.1

Factor $3$ out of $6$ .

$x = \frac{3 \cdot 1}{3 \cdot 2}$

Step 1.2.2.3.1.2.2

Cancel the common factor.

$x = \frac{3 \cdot 1}{3 \cdot 2}$

Step 1.2.2.3.1.2.3

Rewrite the expression.

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

Step 1.3

Replace the variable $x$ with $\frac{1}{2}$ in the expression.

$y = | 6 (\frac{1}{2}) - 3 |$

Step 1.4

Simplify $| 6 (\frac{1}{2}) - 3 |$ .

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Step 1.4.1

Cancel the common factor of $2$ .

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Step 1.4.1.1

Factor $2$ out of $6$ .

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

Step 1.4.1.2

Cancel the common factor.

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

Step 1.4.1.3

Rewrite the expression.

$y = | 3 - 3 |$

Step 1.4.2

Subtract $3$ from $3$ .

$y = | 0 |$

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 $(\frac{1}{2}, 0)$ .

$(\frac{1}{2}, 0)$

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 $f (x) = | 6 x - 3 |$ . In this case, the point is $(- 2, 15)$ .

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Step 3.1.1

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

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

Step 3.1.2

Simplify the result.

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Step 3.1.2.1

Multiply $6$ by $- 2$ .

$f (- 2) = | - 12 - 3 |$

Step 3.1.2.2

Subtract $3$ from $- 12$ .

$f (- 2) = | - 15 |$

Step 3.1.2.3

The absolute value is the distance between a number and zero. The distance between $- 15$ and $0$ is $15$ .

$f (- 2) = 15$

Step 3.1.2.4

The final answer is $15$ .

$y = 15$

Step 3.2

Substitute the $x$ value $- 1$ into $f (x) = | 6 x - 3 |$ . In this case, the point is $(- 1, 9)$ .

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Step 3.2.1

Replace the variable $x$ with $- 1$ in the expression.

$f (- 1) = | 6 (- 1) - 3 |$

Step 3.2.2

Simplify the result.

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Step 3.2.2.1

Multiply $6$ by $- 1$ .

$f (- 1) = | - 6 - 3 |$

Step 3.2.2.2

Subtract $3$ from $- 6$ .

$f (- 1) = | - 9 |$

Step 3.2.2.3

The absolute value is the distance between a number and zero. The distance between $- 9$ and $0$ is $9$ .

$f (- 1) = 9$

Step 3.2.2.4

The final answer is $9$ .

$y = 9$

Step 3.3

Substitute the $x$ value $0$ into $f (x) = | 6 x - 3 |$ . In this case, the point is $(0, 3)$ .

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Step 3.3.1

Replace the variable $x$ with $0$ in the expression.

$f (0) = | 6 (0) - 3 |$

Step 3.3.2

Simplify the result.

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Step 3.3.2.1

Multiply $6$ by $0$ .

$f (0) = | 0 - 3 |$

Step 3.3.2.2

Subtract $3$ from $0$ .

$f (0) = | - 3 |$

Step 3.3.2.3

The absolute value is the distance between a number and zero. The distance between $- 3$ and $0$ is $3$ .

$f (0) = 3$

Step 3.3.2.4

The final answer is $3$ .

$y = 3$

Step 3.4

The absolute value can be graphed using the points around the vertex $(\frac{1}{2}, 0), (- 2, 15), (- 1, 9), (0, 3), (1, 3)$

$\begin{array}{c} x & y \\ - 2 & 15 \\ - 1 & 9 \\ 0 & 3 \\ \frac{1}{2} & 0 \\ 1 & 3 \end{array}$

Step 4