Algebra Examples

$y = | x |$ and $y = | x - 5 |$

Step 1

Graph $y = | x |$ .

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Step 1.1

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

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Step 1.1.1

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

$x = 0$

Step 1.1.2

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

$y = | 0 |$

Step 1.1.3

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

$y = 0$

Step 1.1.4

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

$(0, 0)$

Step 1.2

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 1.3

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 1.3.1

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

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Step 1.3.1.1

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

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

Step 1.3.1.2

Simplify the result.

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Step 1.3.1.2.1

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

$f (- 2) = 2$

Step 1.3.1.2.2

The final answer is $2$ .

$y = 2$

Step 1.3.2

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

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Step 1.3.2.1

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

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

Step 1.3.2.2

Simplify the result.

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Step 1.3.2.2.1

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

$f (- 1) = 1$

Step 1.3.2.2.2

The final answer is $1$ .

$y = 1$

Step 1.3.3

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

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Step 1.3.3.1

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

$f (2) = | 2 |$

Step 1.3.3.2

Simplify the result.

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Step 1.3.3.2.1

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

$f (2) = 2$

Step 1.3.3.2.2

The final answer is $2$ .

$y = 2$

Step 1.3.4

The absolute value can be graphed using the points around the vertex $(0, 0), (- 2, 2), (- 1, 1), (1, 1), (2, 2)$

$\begin{array}{c} x & y \\ - 2 & 2 \\ - 1 & 1 \\ 0 & 0 \\ 1 & 1 \\ 2 & 2 \end{array}$

Step 2

Graph $y = | x - 5 |$ .

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Step 2.1

Find the absolute value vertex. In this case, the vertex for $y = | x - 5 |$ is $(5, 0)$ .

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Step 2.1.1

To find the $x$ coordinate of the vertex, set the inside of the absolute value $x - 5$ equal to $0$ . In this case, $x - 5 = 0$ .

$x - 5 = 0$

Step 2.1.2

Add $5$ to both sides of the equation.

$x = 5$

Step 2.1.3

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

$y = | (5) - 5 |$

Step 2.1.4

Simplify $| (5) - 5 |$ .

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Step 2.1.4.1

Subtract $5$ from $5$ .

$y = | 0 |$

Step 2.1.4.2

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

$y = 0$

Step 2.1.5

The absolute value vertex is $(5, 0)$ .

$(5, 0)$

Step 2.2

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 2.3

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 2.3.1

Substitute the $x$ value $3$ into $f (x) = | x - 5 |$ . In this case, the point is $(3, 2)$ .

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Step 2.3.1.1

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

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

Step 2.3.1.2

Simplify the result.

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Step 2.3.1.2.1

Subtract $5$ from $3$ .

$f (3) = | - 2 |$

Step 2.3.1.2.2

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

$f (3) = 2$

Step 2.3.1.2.3

The final answer is $2$ .

$y = 2$

Step 2.3.2

Substitute the $x$ value $4$ into $f (x) = | x - 5 |$ . In this case, the point is $(4, 1)$ .

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Step 2.3.2.1

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

$f (4) = | (4) - 5 |$

Step 2.3.2.2

Simplify the result.

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Step 2.3.2.2.1

Subtract $5$ from $4$ .

$f (4) = | - 1 |$

Step 2.3.2.2.2

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

$f (4) = 1$

Step 2.3.2.2.3

The final answer is $1$ .

$y = 1$

Step 2.3.3

Substitute the $x$ value $7$ into $f (x) = | x - 5 |$ . In this case, the point is $(7, 2)$ .

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Step 2.3.3.1

Replace the variable $x$ with $7$ in the expression.

$f (7) = | (7) - 5 |$

Step 2.3.3.2

Simplify the result.

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Step 2.3.3.2.1

Subtract $5$ from $7$ .

$f (7) = | 2 |$

Step 2.3.3.2.2

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

$f (7) = 2$

Step 2.3.3.2.3

The final answer is $2$ .

$y = 2$

Step 2.3.4

The absolute value can be graphed using the points around the vertex $(5, 0), (3, 2), (4, 1), (6, 1), (7, 2)$

$\begin{array}{c} x & y \\ 3 & 2 \\ 4 & 1 \\ 5 & 0 \\ 6 & 1 \\ 7 & 2 \end{array}$

Step 3

Plot each graph on the same coordinate system.

$y = | x |$

$y = | x - 5 |$

Step 4