Graph f(x)=4+|x|

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

Find the absolute value vertex. In this case, the vertex for $y = 4 + | x |$ is $(0, 4)$ .

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.

$y = 4 + | 0 |$

Simplify $4 + | 0 |$ .

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

$y = 4 + 0$

Add $4$ and $0$ .

$y = 4$

The absolute value vertex is $(0, 4)$ .

$(0, 4)$

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 $f (x) = 4 + | x |$ . In this case, the point is $(- 2, 6)$ .

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

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

Simplify the result.

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

$f (- 2) = 4 + 2$

Add $4$ and $2$ .

$f (- 2) = 6$

The final answer is $6$ .

$y = 6$

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

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

$f (- 1) = 4 + | - 1 |$

Simplify the result.

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

$f (- 1) = 4 + 1$

Add $4$ and $1$ .

$f (- 1) = 5$

The final answer is $5$ .

$y = 5$

Substitute the $x$ value $2$ into $f (x) = 4 + | x |$ . In this case, the point is $(2, 6)$ .

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

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

Simplify the result.

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

$f (2) = 4 + 2$

Add $4$ and $2$ .

$f (2) = 6$

The final answer is $6$ .

$y = 6$

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

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

Graph f(x)=4+|x|

Related Questions

Support LCMGCF.com by disabling your adblocker.