Algebra Examples

$y = x^{2} + | x |$

Step 1

Find the absolute value vertex. In this case, the vertex for $y = x^{2} + | x |$ is $(0, 0)$ .

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Step 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 1.2

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

$y = {(0)}^{2} + | 0 |$

Step 1.3

Simplify ${(0)}^{2} + | 0 |$ .

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Step 1.3.1

Simplify each term.

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Step 1.3.1.1

Raising $0$ to any positive power yields $0$ .

$y = 0 + | 0 |$

Step 1.3.1.2

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

$y = 0 + 0$

Step 1.3.2

Add $0$ and $0$ .

$y = 0$

Step 1.4

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

$(0, 0)$

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 $f (x) = x^{2} + | x |$ . In this case, the point is $(- 2, 6)$ .

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Step 3.1.1

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

$f (- 2) = {(- 2)}^{2} + | - 2 |$

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

Raise $- 2$ to the power of $2$ .

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

Step 3.1.2.1.2

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

$f (- 2) = 4 + 2$

Step 3.1.2.2

Add $4$ and $2$ .

$f (- 2) = 6$

Step 3.1.2.3

The final answer is $6$ .

$y = 6$

Step 3.2

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

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Step 3.2.1

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

$f (- 1) = {(- 1)}^{2} + | - 1 |$

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

Raise $- 1$ to the power of $2$ .

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

Step 3.2.2.1.2

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

$f (- 1) = 1 + 1$

Step 3.2.2.2

Add $1$ and $1$ .

$f (- 1) = 2$

Step 3.2.2.3

The final answer is $2$ .

$y = 2$

Step 3.3

Substitute the $x$ value $2$ into $f (x) = x^{2} + | x |$ . In this case, the point is $(2, 6)$ .

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Step 3.3.1

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

$f (2) = {(2)}^{2} + | 2 |$

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

Raise $2$ to the power of $2$ .

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

Step 3.3.2.1.2

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

$f (2) = 4 + 2$

Step 3.3.2.2

Add $4$ and $2$ .

$f (2) = 6$

Step 3.3.2.3

The final answer is $6$ .

$y = 6$

Step 3.4

The absolute value can be graphed using the points around the vertex $(0, 0), (- 2, 6), (- 1, 2), (1, 2), (2, 6)$

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

Step 4