Pre-Algebra Examples

$y = - | x + 2 | + 1$

Step 1

Find the absolute value vertex. In this case, the vertex for $y = - | x + 2 | + 1$ is $(- 2, 1)$ .

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Step 1.1

To find the $x$ coordinate of the vertex, set the inside of the absolute value $x + 2$ equal to $0$ . In this case, $x + 2 = 0$ .

$x + 2 = 0$

Step 1.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 1.3

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

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

Step 1.4

Simplify $- | (- 2) + 2 | + 1$ .

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Step 1.4.1

Simplify each term.

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Step 1.4.1.1

Add $- 2$ and $2$ .

$y = - | 0 | + 1$

Step 1.4.1.2

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

$y = - 0 + 1$

Step 1.4.1.3

Multiply $- 1$ by $0$ .

$y = 0 + 1$

Step 1.4.2

Add $0$ and $1$ .

$y = 1$

Step 1.5

The absolute value vertex is $(- 2, 1)$ .

$(- 2, 1)$

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.

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Step 3.1

Substitute the $x$ value $- 4$ into $f (x) = - | x + 2 | + 1$ . In this case, the point is $(- 4, - 1)$ .

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Step 3.1.1

Replace the variable $x$ with $- 4$ in the expression.

$f (- 4) = - | (- 4) + 2 | + 1$

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

Add $- 4$ and $2$ .

$f (- 4) = - | - 2 | + 1$

Step 3.1.2.1.2

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

$f (- 4) = - 1 \cdot 2 + 1$

Step 3.1.2.1.3

Multiply $- 1$ by $2$ .

$f (- 4) = - 2 + 1$

Step 3.1.2.2

Add $- 2$ and $1$ .

$f (- 4) = - 1$

Step 3.1.2.3

The final answer is $- 1$ .

$y = - 1$

Step 3.2

Substitute the $x$ value $- 3$ into $f (x) = - | x + 2 | + 1$ . In this case, the point is $(- 3, 0)$ .

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Step 3.2.1

Replace the variable $x$ with $- 3$ in the expression.

$f (- 3) = - | (- 3) + 2 | + 1$

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

Add $- 3$ and $2$ .

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

Step 3.2.2.1.2

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

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

Step 3.2.2.1.3

Multiply $- 1$ by $1$ .

$f (- 3) = - 1 + 1$

Step 3.2.2.2

Add $- 1$ and $1$ .

$f (- 3) = 0$

Step 3.2.2.3

The final answer is $0$ .

$y = 0$

Step 3.3

Substitute the $x$ value $0$ into $f (x) = - | x + 2 | + 1$ . In this case, the point is $(0, - 1)$ .

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Step 3.3.1

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

$f (0) = - | (0) + 2 | + 1$

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

Add $0$ and $2$ .

$f (0) = - | 2 | + 1$

Step 3.3.2.1.2

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

$f (0) = - 1 \cdot 2 + 1$

Step 3.3.2.1.3

Multiply $- 1$ by $2$ .

$f (0) = - 2 + 1$

Step 3.3.2.2

Add $- 2$ and $1$ .

$f (0) = - 1$

Step 3.3.2.3

The final answer is $- 1$ .

$y = - 1$

Step 3.4

The absolute value can be graphed using the points around the vertex $(- 2, 1), (- 4, - 1), (- 3, 0), (- 1, 0), (0, - 1)$

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

Step 4