Pre-Algebra Examples

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

Step 1

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

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Step 1.1

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

$\frac{x - 2}{x + 2} = 0$

Step 1.2

Solve the equation $\frac{x - 2}{x + 2} = 0$ to find the $x$ coordinate for the absolute value vertex.

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Step 1.2.1

Set the numerator equal to zero.

$x - 2 = 0$

Step 1.2.2

Add $2$ to both sides of the equation.

$x = 2$

Step 1.3

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

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

Step 1.4

Simplify $| \frac{(2) - 2}{(2) + 2} |$ .

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Step 1.4.1

Cancel the common factor of $(2) - 2$ and $(2) + 2$ .

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Step 1.4.1.1

Factor $2$ out of $2$ .

$y = | \frac{2 \cdot 1 - 2}{2 + 2} |$

Step 1.4.1.2

Factor $2$ out of $- 2$ .

$y = | \frac{2 \cdot 1 + 2 \cdot - 1}{(2) + 2} |$

Step 1.4.1.3

Factor $2$ out of $2 \cdot 1 + 2 \cdot - 1$ .

$y = | \frac{2 \cdot (1 - 1)}{(2) + 2} |$

Step 1.4.1.4

Cancel the common factors.

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Step 1.4.1.4.1

Factor $2$ out of $2$ .

$y = | \frac{2 \cdot (1 - 1)}{2 \cdot 1 + 2} |$

Step 1.4.1.4.2

Factor $2$ out of $2$ .

$y = | \frac{2 \cdot (1 - 1)}{2 \cdot 1 + 2 (1)} |$

Step 1.4.1.4.3

Factor $2$ out of $2 \cdot 1 + 2 (1)$ .

$y = | \frac{2 \cdot (1 - 1)}{2 \cdot (1 + 1)} |$

Step 1.4.1.4.4

Cancel the common factor.

$y = | \frac{2 \cdot (1 - 1)}{2 \cdot (1 + 1)} |$

Step 1.4.1.4.5

Rewrite the expression.

$y = | \frac{1 - 1}{1 + 1} |$

Step 1.4.2

Simplify the expression.

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Step 1.4.2.1

Subtract $1$ from $1$ .

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

Step 1.4.2.2

Add $1$ and $1$ .

$y = | \frac{0}{2} |$

Step 1.4.2.3

Divide $0$ by $2$ .

$y = | 0 |$

Step 1.4.3

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

$y = 0$

Step 1.5

The absolute value vertex is $(2, 0)$ .

$(2, 0)$

Step 2

Find the domain for $f (x) = | \frac{x - 2}{x + 2} |$ so that a list of $x$ values can be picked to find a list of points, which will help graphing the absolute value function.

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Step 2.1

Set the denominator in $\frac{x - 2}{x + 2}$ equal to $0$ to find where the expression is undefined.

$x + 2 = 0$

Step 2.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.3

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(- \infty, - 2) \cup (- 2, \infty)$

Set-Builder Notation:

${x | x \neq - 2}$

Interval Notation:

$(- \infty, - 2) \cup (- 2, \infty)$

Set-Builder Notation:

${x | x \neq - 2}$

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 $0$ into $f (x) = | \frac{x - 2}{x + 2} |$ . In this case, the point is $(0, 1)$ .

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Step 3.1.1

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

$f (0) = | \frac{(0) - 2}{(0) + 2} |$

Step 3.1.2

Simplify the result.

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Step 3.1.2.1

Cancel the common factor of $(0) - 2$ and $(0) + 2$ .

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Step 3.1.2.1.1

Factor $2$ out of $0$ .

$f (0) = | \frac{2 (0) - 2}{(0) + 2} |$

Step 3.1.2.1.2

Factor $2$ out of $- 2$ .

$f (0) = | \frac{2 \cdot 0 + 2 \cdot - 1}{(0) + 2} |$

Step 3.1.2.1.3

Factor $2$ out of $2 \cdot 0 + 2 \cdot - 1$ .

$f (0) = | \frac{2 \cdot (0 - 1)}{(0) + 2} |$

Step 3.1.2.1.4

Cancel the common factors.

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Step 3.1.2.1.4.1

Factor $2$ out of $0$ .

$f (0) = | \frac{2 \cdot (0 - 1)}{2 (0) + 2} |$

Step 3.1.2.1.4.2

Factor $2$ out of $2$ .

$f (0) = | \frac{2 \cdot (0 - 1)}{2 \cdot 0 + 2 \cdot 1} |$

Step 3.1.2.1.4.3

Factor $2$ out of $2 \cdot 0 + 2 \cdot 1$ .

$f (0) = | \frac{2 \cdot (0 - 1)}{2 \cdot (0 + 1)} |$

Step 3.1.2.1.4.4

Cancel the common factor.

$f (0) = | \frac{2 \cdot (0 - 1)}{2 \cdot (0 + 1)} |$

Step 3.1.2.1.4.5

Rewrite the expression.

$f (0) = | \frac{0 - 1}{0 + 1} |$

Step 3.1.2.2

Cancel the common factor of $0 - 1$ and $0 + 1$ .

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Step 3.1.2.2.1

Rewrite $0$ as $- 1 (0)$ .

$f (0) = | \frac{- 1 \cdot 0 - 1}{0 + 1} |$

Step 3.1.2.2.2

Rewrite $- 1$ as $- 1 (1)$ .

$f (0) = | \frac{- 1 \cdot 0 - 1 \cdot 1}{0 + 1} |$

Step 3.1.2.2.3

Factor $- 1$ out of $- 1 \cdot 0 - 1 \cdot 1$ .

$f (0) = | \frac{- 1 \cdot (0 + 1)}{0 + 1} |$

Step 3.1.2.2.4

Cancel the common factor.

$f (0) = | \frac{- 1 \cdot (0 + 1)}{0 + 1} |$

Step 3.1.2.2.5

Divide $- 1$ by $1$ .

$f (0) = | - 1 |$

Step 3.1.2.3

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

$f (0) = 1$

Step 3.1.2.4

The final answer is $1$ .

$y = 1$

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

Simplify the result.

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Step 3.2.2.1

Subtract $2$ from $1$ .

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

Step 3.2.2.2

Add $1$ and $2$ .

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

Step 3.2.2.3

Move the negative in front of the fraction.

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

Step 3.2.2.4

$- \frac{1}{3}$ is approximately $- 0.\overline{3}$ which is negative so negate $- \frac{1}{3}$ and remove the absolute value

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

Step 3.2.2.5

The final answer is $\frac{1}{3}$ .

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

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

Simplify the result.

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Step 3.3.2.1

Subtract $2$ from $3$ .

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

Step 3.3.2.2

Add $3$ and $2$ .

$f (3) = | \frac{1}{5} |$

Step 3.3.2.3

$\frac{1}{5}$ is approximately $0.2$ which is positive so remove the absolute value

$f (3) = \frac{1}{5}$

Step 3.3.2.4

The final answer is $\frac{1}{5}$ .

$y = \frac{1}{5}$

Step 3.4

Substitute the $x$ value $4$ into $f (x) = | \frac{x - 2}{x + 2} |$ . In this case, the point is $(4, \frac{1}{3})$ .

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Step 3.4.1

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

$f (4) = | \frac{(4) - 2}{(4) + 2} |$

Step 3.4.2

Simplify the result.

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Step 3.4.2.1

Cancel the common factor of $(4) - 2$ and $(4) + 2$ .

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Step 3.4.2.1.1

Factor $2$ out of $4$ .

$f (4) = | \frac{2 \cdot 2 - 2}{4 + 2} |$

Step 3.4.2.1.2

Factor $2$ out of $- 2$ .

$f (4) = | \frac{2 \cdot 2 + 2 \cdot - 1}{(4) + 2} |$

Step 3.4.2.1.3

Factor $2$ out of $2 \cdot 2 + 2 \cdot - 1$ .

$f (4) = | \frac{2 \cdot (2 - 1)}{(4) + 2} |$

Step 3.4.2.1.4

Cancel the common factors.

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Step 3.4.2.1.4.1

Factor $2$ out of $4$ .

$f (4) = | \frac{2 \cdot (2 - 1)}{2 \cdot 2 + 2} |$

Step 3.4.2.1.4.2

Factor $2$ out of $2$ .

$f (4) = | \frac{2 \cdot (2 - 1)}{2 \cdot 2 + 2 (1)} |$

Step 3.4.2.1.4.3

Factor $2$ out of $2 \cdot 2 + 2 (1)$ .

$f (4) = | \frac{2 \cdot (2 - 1)}{2 \cdot (2 + 1)} |$

Step 3.4.2.1.4.4

Cancel the common factor.

$f (4) = | \frac{2 \cdot (2 - 1)}{2 \cdot (2 + 1)} |$

Step 3.4.2.1.4.5

Rewrite the expression.

$f (4) = | \frac{2 - 1}{2 + 1} |$

Step 3.4.2.2

Simplify by adding and subtracting.

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Step 3.4.2.2.1

Subtract $1$ from $2$ .

$f (4) = | \frac{1}{2 + 1} |$

Step 3.4.2.2.2

Add $2$ and $1$ .

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

Step 3.4.2.3

$\frac{1}{3}$ is approximately $0.\overline{3}$ which is positive so remove the absolute value

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

Step 3.4.2.4

The final answer is $\frac{1}{3}$ .

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

Step 3.5

The absolute value can be graphed using the points around the vertex $(2, 0), (0, 1), (1, 0.33), (3, 0.2), (4, 0.33)$

$\begin{array}{c} x & y \\ 0 & 1 \\ 1 & 0.33 \\ 2 & 0 \\ 3 & 0.2 \\ 4 & 0.33 \end{array}$

Step 4