Trigonometry Examples

$3 - | 2 x |$

Step 1

Find the absolute value vertex. In this case, the vertex for $y = 3 - 2 | x |$ is $(0, 3)$ .

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Step 1.1

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

$x = 0$

Step 1.2

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

$y = 3 - 2 | 0 |$

Step 1.3

Simplify $3 - 2 | 0 |$ .

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Step 1.3.1

Simplify each term.

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Step 1.3.1.1

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

$y = 3 - 2 \cdot 0$

Step 1.3.1.2

Multiply $- 2$ by $0$ .

$y = 3 + 0$

Step 1.3.2

Add $3$ and $0$ .

$y = 3$

Step 1.4

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

$(0, 3)$

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:

$(- \infty, \infty)$

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

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Step 3.1.1

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

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

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

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

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

Step 3.1.2.1.2

Multiply $- 2$ by $2$ .

$f (- 2) = 3 - 4$

Step 3.1.2.2

Subtract $4$ from $3$ .

$f (- 2) = - 1$

Step 3.1.2.3

The final answer is $- 1$ .

$y = - 1$

Step 3.2

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

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Step 3.2.1

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

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

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

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

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

Step 3.2.2.1.2

Multiply $- 2$ by $1$ .

$f (- 1) = 3 - 2$

Step 3.2.2.2

Subtract $2$ from $3$ .

$f (- 1) = 1$

Step 3.2.2.3

The final answer is $1$ .

$y = 1$

Step 3.3

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

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Step 3.3.1

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

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

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

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

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

Step 3.3.2.1.2

Multiply $- 2$ by $2$ .

$f (2) = 3 - 4$

Step 3.3.2.2

Subtract $4$ from $3$ .

$f (2) = - 1$

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 $(0, 3), (- 2, - 1), (- 1, 1), (1, 1), (2, - 1)$

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

Step 4