Pre-Algebra Examples

$| (1.4 x^{2} - 2 x + 9.3) - (7.5 x^{2} - 3.3 + 10) + (1.5 x - 6) |$

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 $f (x) = | - 6.1 x^{2} - 0.5 x - 3.4 |$ . In this case, the point is $(- 2, 26.8)$ .

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Step 2.1.1

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

$f (- 2) = | - 6.1 {(- 2)}^{2} - 0.5 \cdot - 2 - 3.4 |$

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 $2$ .

$f (- 2) = | - 6.1 \cdot 4 - 0.5 \cdot - 2 - 3.4 |$

Step 2.1.2.1.2

Multiply $- 6.1$ by $4$ .

$f (- 2) = | - 24.4 - 0.5 \cdot - 2 - 3.4 |$

Step 2.1.2.1.3

Multiply $- 0.5$ by $- 2$ .

$f (- 2) = | - 24.4 + 1 - 3.4 |$

Step 2.1.2.2

Simplify by adding and subtracting.

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Step 2.1.2.2.1

Add $- 24.4$ and $1$ .

$f (- 2) = | - 23.4 - 3.4 |$

Step 2.1.2.2.2

Subtract $3.4$ from $- 23.4$ .

$f (- 2) = | - 26.8 |$

Step 2.1.2.3

The absolute value is the distance between a number and zero. The distance between $- 26.8$ and $0$ is $26.8$ .

$f (- 2) = 26.8$

Step 2.1.2.4

The final answer is $26.8$ .

$y = 26.8$

Step 2.2

Substitute the $x$ value $- 1$ into $f (x) = | - 6.1 x^{2} - 0.5 x - 3.4 |$ . In this case, the point is $(- 1, 9)$ .

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Step 2.2.1

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

$f (- 1) = | - 6.1 {(- 1)}^{2} - 0.5 \cdot - 1 - 3.4 |$

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 $2$ .

$f (- 1) = | - 6.1 \cdot 1 - 0.5 \cdot - 1 - 3.4 |$

Step 2.2.2.1.2

Multiply $- 6.1$ by $1$ .

$f (- 1) = | - 6.1 - 0.5 \cdot - 1 - 3.4 |$

Step 2.2.2.1.3

Multiply $- 0.5$ by $- 1$ .

$f (- 1) = | - 6.1 + 0.5 - 3.4 |$

Step 2.2.2.2

Simplify by adding and subtracting.

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Step 2.2.2.2.1

Add $- 6.1$ and $0.5$ .

$f (- 1) = | - 5.6 - 3.4 |$

Step 2.2.2.2.2

Subtract $3.4$ from $- 5.6$ .

$f (- 1) = | - 9 |$

Step 2.2.2.3

The absolute value is the distance between a number and zero. The distance between $- 9$ and $0$ is $9$ .

$f (- 1) = 9$

Step 2.2.2.4

The final answer is $9$ .

$y = 9$

Step 2.3

Substitute the $x$ value $0$ into $f (x) = | - 6.1 x^{2} - 0.5 x - 3.4 |$ . In this case, the point is $(0, 3.4)$ .

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Step 2.3.1

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

$f (0) = | - 6.1 {(0)}^{2} - 0.5 \cdot 0 - 3.4 |$

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 $0$ .

$f (0) = | - 6.1 \cdot 0 - 0.5 \cdot 0 - 3.4 |$

Step 2.3.2.1.2

Multiply $- 6.1$ by $0$ .

$f (0) = | 0 - 0.5 \cdot 0 - 3.4 |$

Step 2.3.2.1.3

Multiply $- 0.5$ by $0$ .

$f (0) = | 0 + 0 - 3.4 |$

Step 2.3.2.2

Simplify by adding and subtracting.

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Step 2.3.2.2.1

Add $0$ and $0$ .

$f (0) = | 0 - 3.4 |$

Step 2.3.2.2.2

Subtract $3.4$ from $0$ .

$f (0) = | - 3.4 |$

Step 2.3.2.3

The absolute value is the distance between a number and zero. The distance between $- 3.4$ and $0$ is $3.4$ .

$f (0) = 3.4$

Step 2.3.2.4

The final answer is $3.4$ .

$y = 3.4$

Step 2.4

Substitute the $x$ value $1$ into $f (x) = | - 6.1 x^{2} - 0.5 x - 3.4 |$ . In this case, the point is $(1, 10)$ .

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Step 2.4.1

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

$f (1) = | - 6.1 {(1)}^{2} - 0.5 \cdot 1 - 3.4 |$

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.

$f (1) = | - 6.1 \cdot 1 - 0.5 \cdot 1 - 3.4 |$

Step 2.4.2.1.2

Multiply $- 6.1$ by $1$ .

$f (1) = | - 6.1 - 0.5 \cdot 1 - 3.4 |$

Step 2.4.2.1.3

Multiply $- 0.5$ by $1$ .

$f (1) = | - 6.1 - 0.5 - 3.4 |$

Step 2.4.2.2

Simplify by subtracting numbers.

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Step 2.4.2.2.1

Subtract $0.5$ from $- 6.1$ .

$f (1) = | - 6.6 - 3.4 |$

Step 2.4.2.2.2

Subtract $3.4$ from $- 6.6$ .

$f (1) = | - 10 |$

Step 2.4.2.3

The absolute value is the distance between a number and zero. The distance between $- 10$ and $0$ is $10$ .

$f (1) = 10$

Step 2.4.2.4

The final answer is $10$ .

$y = 10$

Step 2.5

The absolute value can be graphed using the points around the vertex $(- 2, 26.8), (- 1, 9), (0, 3.4), (1, 10)$

$\begin{array}{c} x & y \\ - 2 & 26.8 \\ - 1 & 9 \\ 0 & 3.4 \\ 1 & 10 \end{array}$

Step 3