Algebra Examples

y = x^{2} - | 6 x + 5 |

$y = x^{2} - | 6 x + 5 |$

Step 1

Find the absolute value vertex. In this case, the vertex for

y = x^{2} - | 6 x + 5 |

$y = x^{2} - | 6 x + 5 |$ is

(- \frac{5}{6}, \frac{25}{36})

$(- \frac{5}{6}, \frac{25}{36})$ .

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Step 1.1

To find the

x

$x$ coordinate of the vertex, set the inside of the absolute value

6 x + 5

$6 x + 5$ equal to

0

$0$ . In this case,

6 x + 5 = 0

$6 x + 5 = 0$ .

6 x + 5 = 0

$6 x + 5 = 0$

Step 1.2

Solve the equation

6 x + 5 = 0

$6 x + 5 = 0$ to find the

x

$x$ coordinate for the absolute value vertex.

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Step 1.2.1

Subtract

5

$5$ from both sides of the equation.

6 x = - 5

$6 x = - 5$

Step 1.2.2

Divide each term in

6 x = - 5

$6 x = - 5$ by

6

$6$ and simplify.

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Step 1.2.2.1

Divide each term in

6 x = - 5

$6 x = - 5$ by

6

$6$ .

\frac{6 x}{6} = \frac{- 5}{6}

$\frac{6 x}{6} = \frac{- 5}{6}$

Step 1.2.2.2

Simplify the left side.

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Step 1.2.2.2.1

Cancel the common factor of

6

$6$ .

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Step 1.2.2.2.1.1

Cancel the common factor.

$\frac{6 x}{6} = \frac{- 5}{6}$

Step 1.2.2.2.1.2

Divide

$x$ by

$1$ .

$x = \frac{- 5}{6}$

Step 1.2.2.3

Simplify the right side.

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Step 1.2.2.3.1

Move the negative in front of the fraction.

$x = - \frac{5}{6}$

Step 1.3

Replace the variable

$x$ with

$- \frac{5}{6}$ in the expression.

$y = {(- \frac{5}{6})}^{2} - | 6 (- \frac{5}{6}) + 5 |$

Step 1.4

Simplify

${(- \frac{5}{6})}^{2} - | 6 (- \frac{5}{6}) + 5 |$ .

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Step 1.4.1

Simplify each term.

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Step 1.4.1.1

Use the power rule

${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 1.4.1.1.1

Apply the product rule to

$- \frac{5}{6}$ .

$y = {(- 1)}^{2} {(\frac{5}{6})}^{2} - | 6 (- \frac{5}{6}) + 5 |$

Step 1.4.1.1.2

Apply the product rule to

$\frac{5}{6}$ .

$y = {(- 1)}^{2} (\frac{5^{2}}{6^{2}}) - | 6 (- \frac{5}{6}) + 5 |$

Step 1.4.1.2

Raise

$- 1$ to the power of

$2$ .

$y = 1 (\frac{5^{2}}{6^{2}}) - | 6 (- \frac{5}{6}) + 5 |$

Step 1.4.1.3

Multiply

$\frac{5^{2}}{6^{2}}$ by

$1$ .

$y = \frac{5^{2}}{6^{2}} - | 6 (- \frac{5}{6}) + 5 |$

Step 1.4.1.4

Raise

$5$ to the power of

$2$ .

$y = \frac{25}{6^{2}} - | 6 (- \frac{5}{6}) + 5 |$

Step 1.4.1.5

Raise

$6$ to the power of

$2$ .

$y = \frac{25}{36} - | 6 (- \frac{5}{6}) + 5 |$

Step 1.4.1.6

Cancel the common factor of

$6$ .

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Step 1.4.1.6.1

Move the leading negative in

$- \frac{5}{6}$ into the numerator.

$y = \frac{25}{36} - | 6 (\frac{- 5}{6}) + 5 |$

Step 1.4.1.6.2

Cancel the common factor.

$y = \frac{25}{36} - | 6 (\frac{- 5}{6}) + 5 |$

Step 1.4.1.6.3

Rewrite the expression.

$y = \frac{25}{36} - | - 5 + 5 |$

Step 1.4.1.7

Add

$- 5$ and

$5$ .

$y = \frac{25}{36} - | 0 |$

Step 1.4.1.8

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

$0$ and

$0$ is

$0$ .

$y = \frac{25}{36} - 0$

Step 1.4.1.9

Multiply

$- 1$ by

$0$ .

$y = \frac{25}{36} + 0$

Step 1.4.2

Add

$\frac{25}{36}$ and

$0$ .

$y = \frac{25}{36}$

Step 1.5

The absolute value vertex is

$(- \frac{5}{6}, \frac{25}{36})$ .

$(- \frac{5}{6}, \frac{25}{36})$

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

$- 3$ into

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

$(- 3, - 4)$ .

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Step 3.1.1

Replace the variable

$x$ with

$- 3$ in the expression.

$f (- 3) = {(- 3)}^{2} - | 6 (- 3) + 5 |$

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

Raise

$- 3$ to the power of

$2$ .

$f (- 3) = 9 - | 6 (- 3) + 5 |$

Step 3.1.2.1.2

Multiply

$6$ by

$- 3$ .

$f (- 3) = 9 - | - 18 + 5 |$

Step 3.1.2.1.3

Add

$- 18$ and

$5$ .

$f (- 3) = 9 - | - 13 |$

Step 3.1.2.1.4

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

$- 13$ and

$0$ is

$13$ .

$f (- 3) = 9 - 1 \cdot 13$

Step 3.1.2.1.5

Multiply

$- 1$ by

$13$ .

$f (- 3) = 9 - 13$

Step 3.1.2.2

Subtract

$13$ from

$9$ .

$f (- 3) = - 4$

Step 3.1.2.3

The final answer is

$- 4$ .

$y = - 4$

Step 3.2

Substitute the

$x$ value

$- 2$ into

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

$(- 2, - 3)$ .

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Step 3.2.1

Replace the variable

$x$ with

$- 2$ in the expression.

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

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

Raise

$- 2$ to the power of

$2$ .

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

Step 3.2.2.1.2

Multiply

$6$ by

$- 2$ .

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

Step 3.2.2.1.3

Add

$- 12$ and

$5$ .

$f (- 2) = 4 - | - 7 |$

Step 3.2.2.1.4

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

$- 7$ and

$0$ is

$7$ .

$f (- 2) = 4 - 1 \cdot 7$

Step 3.2.2.1.5

Multiply

$- 1$ by

$7$ .

$f (- 2) = 4 - 7$

Step 3.2.2.2

Subtract

$7$ from

$4$ .

$f (- 2) = - 3$

Step 3.2.2.3

The final answer is

$- 3$ .

$y = - 3$

Step 3.3

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 3.3.1

Replace the variable

$x$ with

$- 1$ in the expression.

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

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

Raise

$- 1$ to the power of

$2$ .

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

Step 3.3.2.1.2

Multiply

$6$ by

$- 1$ .

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

Step 3.3.2.1.3

Add

$- 6$ and

$5$ .

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

Step 3.3.2.1.4

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

$- 1$ and

$0$ is

$1$ .

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

Step 3.3.2.1.5

Multiply

$- 1$ by

$1$ .

$f (- 1) = 1 - 1$

Step 3.3.2.2

Subtract

$1$ from

$1$ .

$f (- 1) = 0$

Step 3.3.2.3

The final answer is

$0$ .

$y = 0$

Step 3.4

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 3.4.1

Replace the variable

$x$ with

$0$ in the expression.

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

Step 3.4.2

Simplify the result.

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Step 3.4.2.1

Simplify each term.

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Step 3.4.2.1.1

Raising

$0$ to any positive power yields

$0$ .

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

Step 3.4.2.1.2

Multiply

$6$ by

$0$ .

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

Step 3.4.2.1.3

Add

$0$ and

$5$ .

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

Step 3.4.2.1.4

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

$0$ and

$5$ is

$5$ .

$f (0) = 0 - 1 \cdot 5$

Step 3.4.2.1.5

Multiply

$- 1$ by

$5$ .

$f (0) = 0 - 5$

Step 3.4.2.2

Subtract

$5$ from

$0$ .

$f (0) = - 5$

Step 3.4.2.3

The final answer is

$- 5$ .

$y = - 5$

Step 3.5

The absolute value can be graphed using the points around the vertex

$(- \frac{5}{6}, \frac{25}{36}), (- 3, - 4), (- 2, - 3), (- 1, 0), (0, - 5)$

$\begin{array}{c} x & y \\ - 3 & - 4 \\ - 2 & - 3 \\ - 1 & 0 \\ - \frac{5}{6} & \frac{25}{36} \\ 0 & - 5 \end{array}$

Step 4