Pre-Algebra Examples

${| 1 - \frac{x}{3} |}^{3}$

Step 1

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

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Step 1.1

To find the $x$ coordinate of the vertex, set the inside of the absolute value $1 - \frac{x}{3}$ equal to $0$ . In this case, $1 - \frac{x}{3} = 0$ .

$1 - \frac{x}{3} = 0$

Step 1.2

Solve the equation $1 - \frac{x}{3} = 0$ to find the $x$ coordinate for the absolute value vertex.

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Step 1.2.1

Subtract $1$ from both sides of the equation.

$- \frac{x}{3} = - 1$

Step 1.2.2

Multiply both sides of the equation by $- 3$ .

$- 3 (- \frac{x}{3}) = - 3 \cdot - 1$

Step 1.2.3

Simplify both sides of the equation.

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Step 1.2.3.1

Simplify the left side.

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Step 1.2.3.1.1

Simplify $- 3 (- \frac{x}{3})$ .

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Step 1.2.3.1.1.1

Cancel the common factor of $3$ .

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Step 1.2.3.1.1.1.1

Move the leading negative in $- \frac{x}{3}$ into the numerator.

$- 3 \frac{- x}{3} = - 3 \cdot - 1$

Step 1.2.3.1.1.1.2

Factor $3$ out of $- 3$ .

$3 (- 1) \frac{- x}{3} = - 3 \cdot - 1$

Step 1.2.3.1.1.1.3

Cancel the common factor.

$3 \cdot - 1 \frac{- x}{3} = - 3 \cdot - 1$

Step 1.2.3.1.1.1.4

Rewrite the expression.

$- - x = - 3 \cdot - 1$

Step 1.2.3.1.1.2

Multiply.

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Step 1.2.3.1.1.2.1

Multiply $- 1$ by $- 1$ .

$1 x = - 3 \cdot - 1$

Step 1.2.3.1.1.2.2

Multiply $x$ by $1$ .

$x = - 3 \cdot - 1$

Step 1.2.3.2

Simplify the right side.

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Step 1.2.3.2.1

Multiply $- 3$ by $- 1$ .

$x = 3$

Step 1.3

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

$y = {| 1 - \frac{3}{3} |}^{3}$

Step 1.4

Simplify ${| 1 - \frac{3}{3} |}^{3}$ .

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Step 1.4.1

Simplify each term.

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Step 1.4.1.1

Cancel the common factor of $3$ .

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Step 1.4.1.1.1

Cancel the common factor.

$y = {| 1 - \frac{3}{3} |}^{3}$

Step 1.4.1.1.2

Rewrite the expression.

$y = {| 1 - 1 \cdot 1 |}^{3}$

Step 1.4.1.2

Multiply $- 1$ by $1$ .

$y = {| 1 - 1 |}^{3}$

Step 1.4.2

Subtract $1$ from $1$ .

$y = {| 0 |}^{3}$

Step 1.4.3

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

$y = 0^{3}$

Step 1.4.4

Raising $0$ to any positive power yields $0$ .

$y = 0$

Step 1.5

The absolute value vertex is $(3, 0)$ .

$(3, 0)$

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 $1$ into $f (x) = {| 1 - \frac{x}{3} |}^{3}$ . In this case, the point is $(1, \frac{8}{27})$ .

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Step 3.1.1

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

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

Step 3.1.2

Simplify the result.

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Step 3.1.2.1

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

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

Step 3.1.2.2

Combine the numerators over the common denominator.

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

Step 3.1.2.3

Subtract $1$ from $3$ .

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

Step 3.1.2.4

$\frac{2}{3}$ is approximately $0.\overline{6}$ which is positive so remove the absolute value

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

Step 3.1.2.5

Apply the product rule to $\frac{2}{3}$ .

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

Step 3.1.2.6

Raise $2$ to the power of $3$ .

$f (1) = \frac{8}{3^{3}}$

Step 3.1.2.7

Raise $3$ to the power of $3$ .

$f (1) = \frac{8}{27}$

Step 3.1.2.8

The final answer is $\frac{8}{27}$ .

$y = \frac{8}{27}$

Step 3.2

Substitute the $x$ value $2$ into $f (x) = {| 1 - \frac{x}{3} |}^{3}$ . In this case, the point is $(2, \frac{1}{27})$ .

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Step 3.2.1

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

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

Step 3.2.2

Simplify the result.

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Step 3.2.2.1

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

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

Step 3.2.2.2

Combine the numerators over the common denominator.

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

Step 3.2.2.3

Subtract $2$ from $3$ .

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

Step 3.2.2.4

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

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

Step 3.2.2.5

Apply the product rule to $\frac{1}{3}$ .

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

Step 3.2.2.6

One to any power is one.

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

Step 3.2.2.7

Raise $3$ to the power of $3$ .

$f (2) = \frac{1}{27}$

Step 3.2.2.8

The final answer is $\frac{1}{27}$ .

$y = \frac{1}{27}$

Step 3.3

Substitute the $x$ value $5$ into $f (x) = {| 1 - \frac{x}{3} |}^{3}$ . In this case, the point is $(5, \frac{8}{27})$ .

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Step 3.3.1

Replace the variable $x$ with $5$ in the expression.

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

Step 3.3.2

Simplify the result.

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Step 3.3.2.1

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

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

Step 3.3.2.2

Combine the numerators over the common denominator.

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

Step 3.3.2.3

Subtract $5$ from $3$ .

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

Step 3.3.2.4

Move the negative in front of the fraction.

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

Step 3.3.2.5

$- \frac{2}{3}$ is approximately $- 0.\overline{6}$ which is negative so negate $- \frac{2}{3}$ and remove the absolute value

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

Step 3.3.2.6

Apply the product rule to $\frac{2}{3}$ .

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

Step 3.3.2.7

Raise $2$ to the power of $3$ .

$f (5) = \frac{8}{3^{3}}$

Step 3.3.2.8

Raise $3$ to the power of $3$ .

$f (5) = \frac{8}{27}$

Step 3.3.2.9

The final answer is $\frac{8}{27}$ .

$y = \frac{8}{27}$

Step 3.4

The absolute value can be graphed using the points around the vertex $(3, 0), (1, 0.3), (2, 0.04), (4, 0.04), (5, 0.3)$

$\begin{array}{c} x & y \\ 1 & 0.3 \\ 2 & 0.04 \\ 3 & 0 \\ 4 & 0.04 \\ 5 & 0.3 \end{array}$

Step 4