Pre-Algebra Examples

$f (x) = x^{3} + 2 x^{2} - 3 x$

Step 1

Find the point at $x = - 2$ .

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Step 1.1

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

$f (- 2) = {(- 2)}^{3} + 2 {(- 2)}^{2} - 3 \cdot - 2$

Step 1.2

Simplify the result.

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Step 1.2.1

Simplify each term.

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Step 1.2.1.1

Raise $- 2$ to the power of $3$ .

$f (- 2) = - 8 + 2 {(- 2)}^{2} - 3 \cdot - 2$

Step 1.2.1.2

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

$f (- 2) = - 8 + 2 \cdot 4 - 3 \cdot - 2$

Step 1.2.1.3

Multiply $2$ by $4$ .

$f (- 2) = - 8 + 8 - 3 \cdot - 2$

Step 1.2.1.4

Multiply $- 3$ by $- 2$ .

$f (- 2) = - 8 + 8 + 6$

Step 1.2.2

Simplify by adding numbers.

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Step 1.2.2.1

Add $- 8$ and $8$ .

$f (- 2) = 0 + 6$

Step 1.2.2.2

Add $0$ and $6$ .

$f (- 2) = 6$

Step 1.2.3

The final answer is $6$ .

$6$

Step 1.3

Convert $6$ to decimal.

$y = 6$

Step 2

Find the point at $x = - 1$ .

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Step 2.1

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

$f (- 1) = {(- 1)}^{3} + 2 {(- 1)}^{2} - 3 \cdot - 1$

Step 2.2

Simplify the result.

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Step 2.2.1

Simplify each term.

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Step 2.2.1.1

Raise $- 1$ to the power of $3$ .

$f (- 1) = - 1 + 2 {(- 1)}^{2} - 3 \cdot - 1$

Step 2.2.1.2

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

$f (- 1) = - 1 + 2 \cdot 1 - 3 \cdot - 1$

Step 2.2.1.3

Multiply $2$ by $1$ .

$f (- 1) = - 1 + 2 - 3 \cdot - 1$

Step 2.2.1.4

Multiply $- 3$ by $- 1$ .

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

Step 2.2.2

Simplify by adding numbers.

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Step 2.2.2.1

Add $- 1$ and $2$ .

$f (- 1) = 1 + 3$

Step 2.2.2.2

Add $1$ and $3$ .

$f (- 1) = 4$

Step 2.2.3

The final answer is $4$ .

$4$

Step 2.3

Convert $4$ to decimal.

$y = 4$

Step 3

Find the point at $x = 0$ .

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Step 3.1

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

$f (0) = {(0)}^{3} + 2 {(0)}^{2} - 3 \cdot 0$

Step 3.2

Simplify the result.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

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

$f (0) = 0 + 2 {(0)}^{2} - 3 \cdot 0$

Step 3.2.1.2

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

$f (0) = 0 + 2 \cdot 0 - 3 \cdot 0$

Step 3.2.1.3

Multiply $2$ by $0$ .

$f (0) = 0 + 0 - 3 \cdot 0$

Step 3.2.1.4

Multiply $- 3$ by $0$ .

$f (0) = 0 + 0 + 0$

Step 3.2.2

Simplify by adding numbers.

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Step 3.2.2.1

Add $0$ and $0$ .

$f (0) = 0 + 0$

Step 3.2.2.2

Add $0$ and $0$ .

$f (0) = 0$

Step 3.2.3

The final answer is $0$ .

$0$

Step 3.3

Convert $0$ to decimal.

$y = 0$

Step 4

Find the point at $x = 2$ .

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Step 4.1

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

$f (2) = {(2)}^{3} + 2 {(2)}^{2} - 3 \cdot 2$

Step 4.2

Simplify the result.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Raise $2$ to the power of $3$ .

$f (2) = 8 + 2 {(2)}^{2} - 3 \cdot 2$

Step 4.2.1.2

Multiply $2$ by ${(2)}^{2}$ by adding the exponents.

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Step 4.2.1.2.1

Multiply $2$ by ${(2)}^{2}$ .

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Step 4.2.1.2.1.1

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

$f (2) = 8 + 2 {(2)}^{2} - 3 \cdot 2$

Step 4.2.1.2.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f (2) = 8 + 2^{1 + 2} - 3 \cdot 2$

Step 4.2.1.2.2

Add $1$ and $2$ .

$f (2) = 8 + 2^{3} - 3 \cdot 2$

Step 4.2.1.3

Raise $2$ to the power of $3$ .

$f (2) = 8 + 8 - 3 \cdot 2$

Step 4.2.1.4

Multiply $- 3$ by $2$ .

$f (2) = 8 + 8 - 6$

Step 4.2.2

Simplify by adding and subtracting.

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Step 4.2.2.1

Add $8$ and $8$ .

$f (2) = 16 - 6$

Step 4.2.2.2

Subtract $6$ from $16$ .

$f (2) = 10$

Step 4.2.3

The final answer is $10$ .

$10$

Step 4.3

Convert $10$ to decimal.

$y = 10$

Step 5

The cubic function can be graphed using the function behavior and the points.

$\begin{array}{c} x & y \\ - 2 & 6 \\ - 1 & 4 \\ 0 & 0 \\ 1 & 0 \\ 2 & 10 \end{array}$

Step 6

The cubic function can be graphed using the function behavior and the selected points.

Falls to the left and rises to the right

$\begin{array}{c} x & y \\ - 2 & 6 \\ - 1 & 4 \\ 0 & 0 \\ 1 & 0 \\ 2 & 10 \end{array}$

Step 7