Algebra Examples

$y = x | x | + | x | - 5 x$

Step 1

Simplify $x | x | + | x | - 5 x$ .

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Step 1.1

Move $| x |$ .

$y = x | x | - 5 x + | x |$

Step 1.2

Reorder $x | x |$ and $- 5 x$ .

$y = - 5 x + x | x | + | x |$

Step 2

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

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Step 2.1

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 2.2

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

$y = - 5 \cdot 0 + (0) | 0 | + | 0 |$

Step 2.3

Simplify $- 5 \cdot 0 + (0) | 0 | + | 0 |$ .

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Step 2.3.1

Simplify each term.

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Step 2.3.1.1

Multiply $- 5$ by $0$ .

$y = 0 + (0) | 0 | + | 0 |$

Step 2.3.1.2

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

$y = 0 + 0 \cdot 0 + | 0 |$

Step 2.3.1.3

Multiply $0$ by $0$ .

$y = 0 + 0 + | 0 |$

Step 2.3.1.4

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

$y = 0 + 0 + 0$

Step 2.3.2

Simplify by adding numbers.

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Step 2.3.2.1

Add $0$ and $0$ .

$y = 0 + 0$

Step 2.3.2.2

Add $0$ and $0$ .

$y = 0$

Step 2.4

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

$(0, 0)$

Step 3

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 4

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 4.1

Substitute the $x$ value $- 2$ into $f (x) = - 5 x + x | x | + | x |$ . In this case, the point is $(- 2, 8)$ .

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Step 4.1.1

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

$f (- 2) = - 5 \cdot - 2 + (- 2) | - 2 | + | - 2 |$

Step 4.1.2

Simplify the result.

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Step 4.1.2.1

Simplify each term.

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Step 4.1.2.1.1

Multiply $- 5$ by $- 2$ .

$f (- 2) = 10 + (- 2) | - 2 | + | - 2 |$

Step 4.1.2.1.2

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

$f (- 2) = 10 - 2 \cdot 2 + | - 2 |$

Step 4.1.2.1.3

Multiply $- 2$ by $2$ .

$f (- 2) = 10 - 4 + | - 2 |$

Step 4.1.2.1.4

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

$f (- 2) = 10 - 4 + 2$

Step 4.1.2.2

Simplify by adding and subtracting.

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Step 4.1.2.2.1

Subtract $4$ from $10$ .

$f (- 2) = 6 + 2$

Step 4.1.2.2.2

Add $6$ and $2$ .

$f (- 2) = 8$

Step 4.1.2.3

The final answer is $8$ .

$y = 8$

Step 4.2

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

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Step 4.2.1

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

$f (- 1) = - 5 \cdot - 1 + (- 1) | - 1 | + | - 1 |$

Step 4.2.2

Simplify the result.

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Step 4.2.2.1

Simplify each term.

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Step 4.2.2.1.1

Multiply $- 5$ by $- 1$ .

$f (- 1) = 5 + (- 1) | - 1 | + | - 1 |$

Step 4.2.2.1.2

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

$f (- 1) = 5 - 1 \cdot 1 + | - 1 |$

Step 4.2.2.1.3

Multiply $- 1$ by $1$ .

$f (- 1) = 5 - 1 + | - 1 |$

Step 4.2.2.1.4

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

$f (- 1) = 5 - 1 + 1$

Step 4.2.2.2

Simplify by adding and subtracting.

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Step 4.2.2.2.1

Subtract $1$ from $5$ .

$f (- 1) = 4 + 1$

Step 4.2.2.2.2

Add $4$ and $1$ .

$f (- 1) = 5$

Step 4.2.2.3

The final answer is $5$ .

$y = 5$

Step 4.3

Substitute the $x$ value $1$ into $f (x) = - 5 x + x | x | + | x |$ . In this case, the point is $(1, - 3)$ .

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Step 4.3.1

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

$f (1) = - 5 \cdot 1 + (1) | 1 | + | 1 |$

Step 4.3.2

Simplify the result.

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Step 4.3.2.1

Simplify each term.

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Step 4.3.2.1.1

Multiply $- 5$ by $1$ .

$f (1) = - 5 + (1) | 1 | + | 1 |$

Step 4.3.2.1.2

Multiply $| 1 |$ by $1$ .

$f (1) = - 5 + | 1 | + | 1 |$

Step 4.3.2.1.3

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

$f (1) = - 5 + 1 + | 1 |$

Step 4.3.2.1.4

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

$f (1) = - 5 + 1 + 1$

Step 4.3.2.2

Simplify by adding numbers.

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Step 4.3.2.2.1

Add $- 5$ and $1$ .

$f (1) = - 4 + 1$

Step 4.3.2.2.2

Add $- 4$ and $1$ .

$f (1) = - 3$

Step 4.3.2.3

The final answer is $- 3$ .

$y = - 3$

Step 4.4

Substitute the $x$ value $2$ into $f (x) = - 5 x + x | x | + | x |$ . In this case, the point is $(2, - 4)$ .

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Step 4.4.1

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

$f (2) = - 5 \cdot 2 + (2) | 2 | + | 2 |$

Step 4.4.2

Simplify the result.

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Step 4.4.2.1

Simplify each term.

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Step 4.4.2.1.1

Multiply $- 5$ by $2$ .

$f (2) = - 10 + (2) | 2 | + | 2 |$

Step 4.4.2.1.2

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

$f (2) = - 10 + 2 \cdot 2 + | 2 |$

Step 4.4.2.1.3

Multiply $2$ by $2$ .

$f (2) = - 10 + 4 + | 2 |$

Step 4.4.2.1.4

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

$f (2) = - 10 + 4 + 2$

Step 4.4.2.2

Simplify by adding numbers.

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Step 4.4.2.2.1

Add $- 10$ and $4$ .

$f (2) = - 6 + 2$

Step 4.4.2.2.2

Add $- 6$ and $2$ .

$f (2) = - 4$

Step 4.4.2.3

The final answer is $- 4$ .

$y = - 4$

Step 4.5

The absolute value can be graphed using the points around the vertex $(0, 0), (- 2, 8), (- 1, 5), (1, - 3), (2, - 4)$

$\begin{array}{c} x & y \\ - 2 & 8 \\ - 1 & 5 \\ 0 & 0 \\ 1 & - 3 \\ 2 & - 4 \end{array}$

Step 5