Algebra Examples

f (x) \geq - | x - 5 | - 3

$f (x) \geq - | x - 5 | - 3$

Step 1

Find the absolute value vertex. In this case, the vertex for

y = - | x - 5 | - 3

$y = - | x - 5 | - 3$ is

(5, - 3)

$(5, - 3)$ .

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Step 1.1

To find the

x

$x$ coordinate of the vertex, set the inside of the absolute value

x - 5

$x - 5$ equal to

0

$0$ . In this case,

x - 5 = 0

$x - 5 = 0$ .

x - 5 = 0

$x - 5 = 0$

Step 1.2

Add

5

$5$ to both sides of the equation.

x = 5

$x = 5$

Step 1.3

Replace the variable

x

$x$ with

5

$5$ in the expression.

y = - | (5) - 5 | - 3

$y = - | (5) - 5 | - 3$

Step 1.4

Simplify

- | (5) - 5 | - 3

$- | (5) - 5 | - 3$ .

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Step 1.4.1

Simplify each term.

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Step 1.4.1.1

Subtract

5

$5$ from

5

$5$ .

y = - | 0 | - 3

$y = - | 0 | - 3$

Step 1.4.1.2

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

0

$0$ and

0

$0$ is

0

$0$ .

y = - 0 - 3

$y = - 0 - 3$

Step 1.4.1.3

Multiply

- 1

$- 1$ by

0

$0$ .

y = 0 - 3

$y = 0 - 3$

y = 0 - 3

$y = 0 - 3$

Step 1.4.2

Subtract

3

$3$ from

0

$0$ .

y = - 3

$y = - 3$

y = - 3

$y = - 3$

Step 1.5

The absolute value vertex is

(5, - 3)

$(5, - 3)$ .

(5, - 3)

$(5, - 3)$

(5, - 3)

$(5, - 3)$

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)

$(- \infty, \infty)$

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.

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Step 3.1

Substitute the

$x$ value

$3$ into

$f (x) = - | x - 5 | - 3$ . In this case, the point is

$(3, - 5)$ .

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Step 3.1.1

Replace the variable

$x$ with

$3$ in the expression.

$f (3) = - | (3) - 5 | - 3$

Step 3.1.2

Simplify the result.

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Step 3.1.2.1

Simplify each term.

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Step 3.1.2.1.1

Subtract

$5$ from

$3$ .

$f (3) = - | - 2 | - 3$

Step 3.1.2.1.2

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

$- 2$ and

$0$ is

$2$ .

$f (3) = - 1 \cdot 2 - 3$

Step 3.1.2.1.3

Multiply

$- 1$ by

$2$ .

$f (3) = - 2 - 3$

Step 3.1.2.2

Subtract

$3$ from

$- 2$ .

$f (3) = - 5$

Step 3.1.2.3

The final answer is

$- 5$ .

$y = - 5$

Step 3.2

Substitute the

$x$ value

$4$ into

$f (x) = - | x - 5 | - 3$ . In this case, the point is

$(4, - 4)$ .

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Step 3.2.1

Replace the variable

$x$ with

$4$ in the expression.

$f (4) = - | (4) - 5 | - 3$

Step 3.2.2

Simplify the result.

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Step 3.2.2.1

Simplify each term.

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Step 3.2.2.1.1

Subtract

$5$ from

$4$ .

$f (4) = - | - 1 | - 3$

Step 3.2.2.1.2

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

$- 1$ and

$0$ is

$1$ .

$f (4) = - 1 \cdot 1 - 3$

Step 3.2.2.1.3

Multiply

$- 1$ by

$1$ .

$f (4) = - 1 - 3$

Step 3.2.2.2

Subtract

$3$ from

$- 1$ .

$f (4) = - 4$

Step 3.2.2.3

The final answer is

$- 4$ .

$y = - 4$

Step 3.3

Substitute the

$x$ value

$7$ into

$f (x) = - | x - 5 | - 3$ . In this case, the point is

$(7, - 5)$ .

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Step 3.3.1

Replace the variable

$x$ with

$7$ in the expression.

$f (7) = - | (7) - 5 | - 3$

Step 3.3.2

Simplify the result.

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Step 3.3.2.1

Simplify each term.

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Step 3.3.2.1.1

Subtract

$5$ from

$7$ .

$f (7) = - | 2 | - 3$

Step 3.3.2.1.2

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

$0$ and

$2$ is

$2$ .

$f (7) = - 1 \cdot 2 - 3$

Step 3.3.2.1.3

Multiply

$- 1$ by

$2$ .

$f (7) = - 2 - 3$

Step 3.3.2.2

Subtract

$3$ from

$- 2$ .

$f (7) = - 5$

Step 3.3.2.3

The final answer is

$- 5$ .

$y = - 5$

Step 3.4

The absolute value can be graphed using the points around the vertex

$(5, - 3), (3, - 5), (4, - 4), (6, - 4), (7, - 5)$

$\begin{array}{c} x & y \\ 3 & - 5 \\ 4 & - 4 \\ 5 & - 3 \\ 6 & - 4 \\ 7 & - 5 \end{array}$

Step 4