Trigonometry Examples

$| x - 2 | + 3$

Step 1

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

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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

Add $2$ to both sides of the equation.

$x = 2$

Step 1.3

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

$y = | (2) - 2 | + 3$

Step 1.4

Simplify $| (2) - 2 | + 3$ .

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Step 1.4.1

Simplify each term.

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Step 1.4.1.1

Subtract $2$ from $2$ .

$y = | 0 | + 3$

Step 1.4.1.2

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

$y = 0 + 3$

Step 1.4.2

Add $0$ and $3$ .

$y = 3$

Step 1.5

The absolute value vertex is $(2, 3)$ .

$(2, 3)$

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 $0$ into $f (x) = | x - 2 | + 3$ . In this case, the point is $(0, 5)$ .

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Step 3.1.1

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

$f (0) = | (0) - 2 | + 3$

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

Subtract $2$ from $0$ .

$f (0) = | - 2 | + 3$

Step 3.1.2.1.2

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

$f (0) = 2 + 3$

Step 3.1.2.2

Add $2$ and $3$ .

$f (0) = 5$

Step 3.1.2.3

The final answer is $5$ .

$y = 5$

Step 3.2

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

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Step 3.2.1

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

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

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

Subtract $2$ from $1$ .

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

Step 3.2.2.1.2

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

$f (1) = 1 + 3$

Step 3.2.2.2

Add $1$ and $3$ .

$f (1) = 4$

Step 3.2.2.3

The final answer is $4$ .

$y = 4$

Step 3.3

Substitute the $x$ value $4$ into $f (x) = | x - 2 | + 3$ . In this case, the point is $(4, 5)$ .

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Step 3.3.1

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

$f (4) = | (4) - 2 | + 3$

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

Subtract $2$ from $4$ .

$f (4) = | 2 | + 3$

Step 3.3.2.1.2

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

$f (4) = 2 + 3$

Step 3.3.2.2

Add $2$ and $3$ .

$f (4) = 5$

Step 3.3.2.3

The final answer is $5$ .

$y = 5$

Step 3.4

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

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

Step 4