Algebra Examples

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

Step 1

Find the absolute value vertex. In this case, the vertex for $y = x^{2} - | 6 x + 5 |$ is $(- \frac{5}{6}, \frac{25}{36})$ .

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Step 1.1

To find the $x$ coordinate of the vertex, set the inside of the absolute value $6 x + 5$ equal to $0$ . In this case, $6 x + 5 = 0$ .

$6 x + 5 = 0$

Step 1.2

Solve the equation $6 x + 5 = 0$ to find the $x$ coordinate for the absolute value vertex.

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Step 1.2.1

Subtract $5$ from both sides of the equation.

$6 x = - 5$

Step 1.2.2

Divide each term in $6 x = - 5$ by $6$ and simplify.

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Step 1.2.2.1

Divide each term in $6 x = - 5$ by $6$ .

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

Step 1.2.2.2

Simplify the left side.

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Step 1.2.2.2.1

Cancel the common factor of $6$ .

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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