Calculus Examples

$f (x) = | x^{2} - 6 x + 5 |$

Step 1

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 2

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 2.1

Substitute the $x$ value $- 2$ into $f (x) = | x^{2} - 6 x + 5 |$ . In this case, the point is $(- 2, 21)$ .

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Step 2.1.1

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

$f (- 2) = | {(- 2)}^{2} - 6 \cdot - 2 + 5 |$

Step 2.1.2

Simplify the result.

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Step 2.1.2.1

Simplify each term.

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Step 2.1.2.1.1

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

$f (- 2) = | 4 - 6 \cdot - 2 + 5 |$

Step 2.1.2.1.2

Multiply $- 6$ by $- 2$ .

$f (- 2) = | 4 + 12 + 5 |$

Step 2.1.2.2

Simplify by adding numbers.

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Step 2.1.2.2.1

Add $4$ and $12$ .

$f (- 2) = | 16 + 5 |$

Step 2.1.2.2.2

Add $16$ and $5$ .

$f (- 2) = | 21 |$

Step 2.1.2.3

The absolute value is the distance between a number and zero. The distance between $0$ and $21$ is $21$ .

$f (- 2) = 21$

Step 2.1.2.4

The final answer is $21$ .

$y = 21$

Step 2.2

Substitute the $x$ value $- 1$ into $f (x) = | x^{2} - 6 x + 5 |$ . In this case, the point is $(- 1, 12)$ .

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Step 2.2.1

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

$f (- 1) = | {(- 1)}^{2} - 6 \cdot - 1 + 5 |$

Step 2.2.2

Simplify the result.

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Step 2.2.2.1

Simplify each term.

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Step 2.2.2.1.1

Raise $- 1$ to the power of $2$ .

$f (- 1) = | 1 - 6 \cdot - 1 + 5 |$

Step 2.2.2.1.2

Multiply $- 6$ by $- 1$ .

$f (- 1) = | 1 + 6 + 5 |$

Step 2.2.2.2

Simplify by adding numbers.

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Step 2.2.2.2.1

Add $1$ and $6$ .

$f (- 1) = | 7 + 5 |$

Step 2.2.2.2.2

Add $7$ and $5$ .

$f (- 1) = | 12 |$

Step 2.2.2.3

The absolute value is the distance between a number and zero. The distance between $0$ and $12$ is $12$ .

$f (- 1) = 12$

Step 2.2.2.4

The final answer is $12$ .

$y = 12$

Step 2.3

Substitute the $x$ value $0$ into $f (x) = | x^{2} - 6 x + 5 |$ . In this case, the point is $(0, 5)$ .

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Step 2.3.1

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

$f (0) = | {(0)}^{2} - 6 \cdot 0 + 5 |$

Step 2.3.2

Simplify the result.

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Step 2.3.2.1

Simplify each term.

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Step 2.3.2.1.1

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

$f (0) = | 0 - 6 \cdot 0 + 5 |$

Step 2.3.2.1.2

Multiply $- 6$ by $0$ .

$f (0) = | 0 + 0 + 5 |$

Step 2.3.2.2

Simplify by adding numbers.

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Step 2.3.2.2.1

Add $0$ and $0$ .

$f (0) = | 0 + 5 |$

Step 2.3.2.2.2

Add $0$ and $5$ .

$f (0) = | 5 |$

Step 2.3.2.3

The absolute value is the distance between a number and zero. The distance between $0$ and $5$ is $5$ .

$f (0) = 5$

Step 2.3.2.4

The final answer is $5$ .

$y = 5$

Step 2.4

Substitute the $x$ value $1$ into $f (x) = | x^{2} - 6 x + 5 |$ . In this case, the point is $(1, 0)$ .

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Step 2.4.1

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

$f (1) = | {(1)}^{2} - 6 \cdot 1 + 5 |$

Step 2.4.2

Simplify the result.

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Step 2.4.2.1

Simplify each term.

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Step 2.4.2.1.1

One to any power is one.

$f (1) = | 1 - 6 \cdot 1 + 5 |$

Step 2.4.2.1.2

Multiply $- 6$ by $1$ .

$f (1) = | 1 - 6 + 5 |$

Step 2.4.2.2

Simplify by adding and subtracting.

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Step 2.4.2.2.1

Subtract $6$ from $1$ .

$f (1) = | - 5 + 5 |$

Step 2.4.2.2.2

Add $- 5$ and $5$ .

$f (1) = | 0 |$

Step 2.4.2.3

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

$f (1) = 0$

Step 2.4.2.4

The final answer is $0$ .

$y = 0$

Step 2.5

The absolute value can be graphed using the points around the vertex $(- 2, 21), (- 1, 12), (0, 5), (1, 0)$

$\begin{array}{c} x & y \\ - 2 & 21 \\ - 1 & 12 \\ 0 & 5 \\ 1 & 0 \end{array}$

Step 3