Graph f(x)>=-1/2|x|+5

$f (x) \geq - \frac{1}{2} | x | + 5$

Step 1

Find the absolute value vertex. In this case, the vertex for $y = - \frac{| x |}{2} + 5$ is $(0, 5)$ .

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.

$y = - \frac{| 0 |}{2} + 5$

Simplify $- \frac{| 0 |}{2} + 5$ .

Simplify each term.

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

$y = - \frac{0}{2} + 5$

Divide $0$ by $2$ .

$y = - 0 + 5$

Multiply $- 1$ by $0$ .

$y = 0 + 5$

Add $0$ and $5$ .

$y = 5$

The absolute value vertex is $(0, 5)$ .

$(0, 5)$

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 $f (x) = - \frac{| x |}{2} + 5$ . In this case, the point is $(- 2, 4)$ .

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

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

Simplify the result.

Simplify each term.

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

$f (- 2) = - 1 \cdot 1 + 5$

Multiply $- 1$ by $1$ .

$f (- 2) = - 1 + 5$

Add $- 1$ and $5$ .

$f (- 2) = 4$

The final answer is $4$ .

$y = 4$

Substitute the $x$ value $- 1$ into $f (x) = - \frac{| x |}{2} + 5$ . In this case, the point is $(- 1, \frac{9}{2})$ .

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

$f (- 1) = - \frac{| - 1 |}{2} + 5$

Simplify the result.

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

$f (- 1) = - \frac{1}{2} + 5$

To write $5$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f (- 1) = - \frac{1}{2} + 5 \cdot \frac{2}{2}$

Combine $5$ and $\frac{2}{2}$ .

$f (- 1) = - \frac{1}{2} + \frac{5 \cdot 2}{2}$

Combine the numerators over the common denominator.

$f (- 1) = \frac{- 1 + 5 \cdot 2}{2}$

Simplify the numerator.

Multiply $5$ by $2$ .

$f (- 1) = \frac{- 1 + 10}{2}$

Add $- 1$ and $10$ .

$f (- 1) = \frac{9}{2}$

The final answer is $\frac{9}{2}$ .

$y = \frac{9}{2}$

Substitute the $x$ value $2$ into $f (x) = - \frac{| x |}{2} + 5$ . In this case, the point is $(2, 4)$ .

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

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

Simplify the result.

Simplify each term.

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

$f (2) = - 1 \cdot 1 + 5$

Multiply $- 1$ by $1$ .

$f (2) = - 1 + 5$

Add $- 1$ and $5$ .

$f (2) = 4$

The final answer is $4$ .

$y = 4$

The absolute value can be graphed using the points around the vertex $(0, 5), (- 2, 4), (- 1, 4.5), (1, 4.5), (2, 4)$

$\begin{array}{c} x & y \\ - 2 & 4 \\ - 1 & 4.5 \\ 0 & 5 \\ 1 & 4.5 \\ 2 & 4 \end{array}$

Step 4

Graph f(x)>=-1/2|x|+5

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