Graph F(x)=1/3|x|+1

$F (x) = \frac{1}{3} | x | + 1$

Find the absolute value vertex. In this case, the vertex for $y = \frac{| x |}{3} + 1$ is $(0, 1)$ .

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

$x = 0$

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

$y = \frac{| 0 |}{3} + 1$

Simplify $\frac{| 0 |}{3} + 1$ .

Simplify each term.

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

$y = \frac{0}{3} + 1$

Divide $0$ by $3$ .

$y = 0 + 1$

Add $0$ and $1$ .

$y = 1$

The absolute value vertex is $(0, 1)$ .

$(0, 1)$

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 $f (x) = \frac{| x |}{3} + 1$ . In this case, the point is $(- 2, \frac{5}{3})$ .

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

$f (- 2) = \frac{| - 2 |}{3} + 1$

Simplify the result.

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

$f (- 2) = \frac{2}{3} + 1$

Write $1$ as a fraction with a common denominator.

$f (- 2) = \frac{2}{3} + \frac{3}{3}$

Combine the numerators over the common denominator.

$f (- 2) = \frac{2 + 3}{3}$

Add $2$ and $3$ .

$f (- 2) = \frac{5}{3}$

The final answer is $\frac{5}{3}$ .

$y = \frac{5}{3}$

Substitute the $x$ value $- 1$ into $f (x) = \frac{| x |}{3} + 1$ . In this case, the point is $(- 1, \frac{4}{3})$ .

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

$f (- 1) = \frac{| - 1 |}{3} + 1$

Simplify the result.

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

$f (- 1) = \frac{1}{3} + 1$

Write $1$ as a fraction with a common denominator.

$f (- 1) = \frac{1}{3} + \frac{3}{3}$

Combine the numerators over the common denominator.

$f (- 1) = \frac{1 + 3}{3}$

Add $1$ and $3$ .

$f (- 1) = \frac{4}{3}$

The final answer is $\frac{4}{3}$ .

$y = \frac{4}{3}$

Substitute the $x$ value $2$ into $f (x) = \frac{| x |}{3} + 1$ . In this case, the point is $(2, \frac{5}{3})$ .

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

$f (2) = \frac{| 2 |}{3} + 1$

Simplify the result.

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

$f (2) = \frac{2}{3} + 1$

Write $1$ as a fraction with a common denominator.

$f (2) = \frac{2}{3} + \frac{3}{3}$

Combine the numerators over the common denominator.

$f (2) = \frac{2 + 3}{3}$

Add $2$ and $3$ .

$f (2) = \frac{5}{3}$

The final answer is $\frac{5}{3}$ .

$y = \frac{5}{3}$

The absolute value can be graphed using the points around the vertex $(0, 1), (- 2, 1.67), (- 1, 1.33), (1, 1.33), (2, 1.67)$

$\begin{array}{c} x & y \\ - 2 & 1.67 \\ - 1 & 1.33 \\ 0 & 1 \\ 1 & 1.33 \\ 2 & 1.67 \end{array}$

Graph F(x)=1/3|x|+1

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