Graph y=x+|x|

$y = x + | x |$

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

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

$x = 0$

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

$y = (0) + | 0 |$

Simplify $(0) + | 0 |$ .

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

$y = 0 + 0$

Add $0$ and $0$ .

$y = 0$

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

$(0, 0)$

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:

${x | x \in ℝ}$

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.

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

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

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

Simplify the result.

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

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

Add $- 2$ and $2$ .

$f (- 2) = 0$

The final answer is $0$ .

$y = 0$

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

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

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

Simplify the result.

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

$f (1) = 1 + 1$

Add $1$ and $1$ .

$f (1) = 2$

The final answer is $2$ .

$y = 2$

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

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

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

Simplify the result.

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

$f (2) = 2 + 2$

Add $2$ and $2$ .

$f (2) = 4$

The final answer is $4$ .

$y = 4$

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

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

Graph y=x+|x|

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