Precalculus Examples

$x + | y | = 6$

Step 1

Subtract $| y |$ from both sides of the equation.

$x = 6 - | y |$

Step 2

Find the absolute value vertex. In this case, the vertex for $x = 6 - | y |$ is $(6, 0)$ .

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Step 2.1

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

$y = 0$

Step 2.2

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

$x = 6 - | 0 |$

Step 2.3

Simplify $6 - | 0 |$ .

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Step 2.3.1

Simplify each term.

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Step 2.3.1.1

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

$x = 6 - 0$

Step 2.3.1.2

Multiply $- 1$ by $0$ .

$x = 6 + 0$

Step 2.3.2

Add $6$ and $0$ .

$x = 6$

Step 2.4

The absolute value vertex is $(6, 0)$ .

$(6, 0)$

Step 3

The domain is the set of all valid $x$ values. Use the graph to find the domain.

$(- \infty, 6]$

${x | x \leq 6}$

Step 4

For each $x$ value, there is one positive $y$ value and one negative $y$ 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 4.1

Substitute the $x$ value $4$ into $f (x) = 6 - x$ . In this case, the point is $(4, 2)$ .

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Step 4.1.1

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

$f (4) = 6 - (4)$

Step 4.1.2

Simplify the result.

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Step 4.1.2.1

Multiply $- 1$ by $4$ .

$f (4) = 6 - 4$

Step 4.1.2.2

Subtract $4$ from $6$ .

$f (4) = 2$

Step 4.1.2.3

The final answer is $2$ .

$y = 2$

Step 4.2

Substitute the $x$ value $4$ into $f (x) = - (6 - x)$ . In this case, the point is $(4, - 2)$ .

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Step 4.2.1

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

$f (4) = - (6 - (4))$

Step 4.2.2

Simplify the result.

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Step 4.2.2.1

Multiply $- 1$ by $4$ .

$f (4) = - (6 - 4)$

Step 4.2.2.2

Subtract $4$ from $6$ .

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

Step 4.2.2.3

Multiply $- 1$ by $2$ .

$f (4) = - 2$

Step 4.2.2.4

The final answer is $- 2$ .

$y = - 2$

Step 4.3

Substitute the $x$ value $5$ into $f (x) = 6 - x$ . In this case, the point is $(5, 1)$ .

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Step 4.3.1

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

$f (5) = 6 - (5)$

Step 4.3.2

Simplify the result.

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Step 4.3.2.1

Multiply $- 1$ by $5$ .

$f (5) = 6 - 5$

Step 4.3.2.2

Subtract $5$ from $6$ .

$f (5) = 1$

Step 4.3.2.3

The final answer is $1$ .

$y = 1$

Step 4.4

Substitute the $x$ value $5$ into $f (x) = - (6 - x)$ . In this case, the point is $(5, - 1)$ .

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Step 4.4.1

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

$f (5) = - (6 - (5))$

Step 4.4.2

Simplify the result.

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Step 4.4.2.1

Multiply $- 1$ by $5$ .

$f (5) = - (6 - 5)$

Step 4.4.2.2

Subtract $5$ from $6$ .

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

Step 4.4.2.3

Multiply $- 1$ by $1$ .

$f (5) = - 1$

Step 4.4.2.4

The final answer is $- 1$ .

$y = - 1$

Step 4.5

The absolute value can be graphed using the points around the vertex $(6, 0), (4, 2), (4, - 2), (5, 1), (5, - 1)$

$\begin{array}{c} x & y \\ 4 & 2 \\ 4 & - 2 \\ 5 & 1 \\ 5 & - 1 \\ 6 & 0 \end{array}$

Step 5