Graph |x|+2

$| x | + 2$

Step 1

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

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.

$y = | 0 | + 2$

Simplify $| 0 | + 2$ .

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

$y = 0 + 2$

Add $0$ and $2$ .

$y = 2$

The absolute value vertex is $(0, 2)$ .

$(0, 2)$

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.

Substitute the $x$ value $- 2$ into $f (x) = | x | + 2$ . 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 $- 2$ and $0$ is $2$ .

$f (- 2) = 2 + 2$

Add $2$ and $2$ .

$f (- 2) = 4$

The final answer is $4$ .

$y = 4$

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

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

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

Simplify the result.

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

$f (- 1) = 1 + 2$

Add $1$ and $2$ .

$f (- 1) = 3$

The final answer is $3$ .

$y = 3$

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

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

Step 4

Graph |x|+2

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