Pre-Algebra Examples

$f (x) = | 2 x - 1 | + 3$

Step 1

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

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Step 1.1

To find the $x$ coordinate of the vertex, set the inside of the absolute value $2 x - 1$ equal to $0$ . In this case, $2 x - 1 = 0$ .

$2 x - 1 = 0$

Step 1.2

Solve the equation $2 x - 1 = 0$ to find the $x$ coordinate for the absolute value vertex.

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Step 1.2.1

Add $1$ to both sides of the equation.

$2 x = 1$

Step 1.2.2

Divide each term in $2 x = 1$ by $2$ and simplify.

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Step 1.2.2.1

Divide each term in $2 x = 1$ by $2$ .

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

Step 1.2.2.2

Simplify the left side.

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Step 1.2.2.2.1

Cancel the common factor of $2$ .

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Step 1.2.2.2.1.1

Cancel the common factor.

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

Step 1.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 1.3

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

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

Step 1.4

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

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Step 1.4.1

Simplify each term.

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Step 1.4.1.1

Cancel the common factor of $2$ .

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Step 1.4.1.1.1

Cancel the common factor.

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

Step 1.4.1.1.2

Rewrite the expression.

$y = | 1 - 1 | + 3$

Step 1.4.1.2

Subtract $1$ from $1$ .

$y = | 0 | + 3$

Step 1.4.1.3

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

$y = 0 + 3$

Step 1.4.2

Add $0$ and $3$ .

$y = 3$

Step 1.5

The absolute value vertex is $(\frac{1}{2}, 3)$ .

$(\frac{1}{2}, 3)$

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 $f (x) = | 2 x - 1 | + 3$ . In this case, the point is $(- 2, 8)$ .

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Step 3.1.1

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

$f (- 2) = | 2 (- 2) - 1 | + 3$

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

Multiply $2$ by $- 2$ .

$f (- 2) = | - 4 - 1 | + 3$

Step 3.1.2.1.2

Subtract $1$ from $- 4$ .

$f (- 2) = | - 5 | + 3$

Step 3.1.2.1.3

The absolute value is the distance between a number and zero. The distance between $- 5$ and $0$ is $5$ .

$f (- 2) = 5 + 3$

Step 3.1.2.2

Add $5$ and $3$ .

$f (- 2) = 8$

Step 3.1.2.3

The final answer is $8$ .

$y = 8$

Step 3.2

Substitute the $x$ value $- 1$ into $f (x) = | 2 x - 1 | + 3$ . In this case, the point is $(- 1, 6)$ .

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Step 3.2.1

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

$f (- 1) = | 2 (- 1) - 1 | + 3$

Step 3.2.2

Simplify the result.

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Step 3.2.2.1

Simplify each term.

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Step 3.2.2.1.1

Multiply $2$ by $- 1$ .

$f (- 1) = | - 2 - 1 | + 3$

Step 3.2.2.1.2

Subtract $1$ from $- 2$ .

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

Step 3.2.2.1.3

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

$f (- 1) = 3 + 3$

Step 3.2.2.2

Add $3$ and $3$ .

$f (- 1) = 6$

Step 3.2.2.3

The final answer is $6$ .

$y = 6$

Step 3.3

Substitute the $x$ value $0$ into $f (x) = | 2 x - 1 | + 3$ . In this case, the point is $(0, 4)$ .

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Step 3.3.1

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

$f (0) = | 2 (0) - 1 | + 3$

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

Multiply $2$ by $0$ .

$f (0) = | 0 - 1 | + 3$

Step 3.3.2.1.2

Subtract $1$ from $0$ .

$f (0) = | - 1 | + 3$

Step 3.3.2.1.3

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

$f (0) = 1 + 3$

Step 3.3.2.2

Add $1$ and $3$ .

$f (0) = 4$

Step 3.3.2.3

The final answer is $4$ .

$y = 4$

Step 3.4

The absolute value can be graphed using the points around the vertex $(\frac{1}{2}, 3), (- 2, 8), (- 1, 6), (0, 4), (1, 4)$

$\begin{array}{c} x & y \\ - 2 & 8 \\ - 1 & 6 \\ 0 & 4 \\ \frac{1}{2} & 3 \\ 1 & 4 \end{array}$

Step 4