Algebra Examples

$\begin{array}{c} x & y \\ - 2 & 2 \\ - 1 & 1 \\ 1 & 1 \end{array}$

Step 1

Check if the function rule is linear.

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Step 1.1

To find if the table follows a function rule, check to see if the values follow the linear form $y = a x + b$ .

$y = a x + b$

Step 1.2

Build a set of equations from the table such that $y = a x + b$ .

$2 = a (- 2) + b 1 = a (- 1) + b 1 = a (1) + b$

Step 1.3

Calculate the values of $a$ and $b$ .

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Step 1.3.1

Solve for $a$ in $1 = a + b$ .

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Step 1.3.1.1

Rewrite the equation as $a + b = 1$ .

$a + b = 1$

$2 = a (- 2) + b$

$1 = a (- 1) + b$

Step 1.3.1.2

Subtract $b$ from both sides of the equation.

$a = 1 - b$

$2 = a (- 2) + b$

$1 = a (- 1) + b$

$a = 1 - b$

$2 = a (- 2) + b$

$1 = a (- 1) + b$

Step 1.3.2

Replace all occurrences of $a$ with $1 - b$ in each equation.

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Step 1.3.2.1

Replace all occurrences of $a$ in $2 = a (- 2) + b$ with $1 - b$ .

$2 = (1 - b) (- 2) + b$

$a = 1 - b$

$1 = a (- 1) + b$

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

Simplify $(1 - b) (- 2) + b$ .

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Step 1.3.2.2.1.1

Simplify each term.

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Step 1.3.2.2.1.1.1

Apply the distributive property.

$2 = 1 \cdot - 2 - b \cdot - 2 + b$

$a = 1 - b$

$1 = a (- 1) + b$

Step 1.3.2.2.1.1.2

Multiply $- 2$ by $1$ .

$2 = - 2 - b \cdot - 2 + b$

$a = 1 - b$

$1 = a (- 1) + b$

Step 1.3.2.2.1.1.3

Multiply $- 2$ by $- 1$ .

$2 = - 2 + 2 b + b$

$a = 1 - b$

$1 = a (- 1) + b$

$2 = - 2 + 2 b + b$

$a = 1 - b$

$1 = a (- 1) + b$

Step 1.3.2.2.1.2

Add $2 b$ and $b$ .

$2 = - 2 + 3 b$

$a = 1 - b$

$1 = a (- 1) + b$

$2 = - 2 + 3 b$

$a = 1 - b$

$1 = a (- 1) + b$

$2 = - 2 + 3 b$

$a = 1 - b$

$1 = a (- 1) + b$

Step 1.3.2.3

Replace all occurrences of $a$ in $1 = a (- 1) + b$ with $1 - b$ .

$1 = (1 - b) (- 1) + b$

$2 = - 2 + 3 b$

$a = 1 - b$

Step 1.3.2.4

Simplify the right side.

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Step 1.3.2.4.1

Simplify $(1 - b) (- 1) + b$ .

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Step 1.3.2.4.1.1

Simplify each term.

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Step 1.3.2.4.1.1.1

Apply the distributive property.

$1 = 1 \cdot - 1 - b \cdot - 1 + b$

$2 = - 2 + 3 b$

$a = 1 - b$

Step 1.3.2.4.1.1.2

Multiply $- 1$ by $1$ .

$1 = - 1 - b \cdot - 1 + b$

$2 = - 2 + 3 b$

$a = 1 - b$

Step 1.3.2.4.1.1.3

Multiply $- b \cdot - 1$ .

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Step 1.3.2.4.1.1.3.1

Multiply $- 1$ by $- 1$ .

$1 = - 1 + 1 b + b$

$2 = - 2 + 3 b$

$a = 1 - b$

Step 1.3.2.4.1.1.3.2

Multiply $b$ by $1$ .

$1 = - 1 + b + b$

$2 = - 2 + 3 b$

$a = 1 - b$

$1 = - 1 + b + b$

$2 = - 2 + 3 b$

$a = 1 - b$

$1 = - 1 + b + b$

$2 = - 2 + 3 b$

$a = 1 - b$

Step 1.3.2.4.1.2

Add $b$ and $b$ .

$1 = - 1 + 2 b$

$2 = - 2 + 3 b$

$a = 1 - b$

$1 = - 1 + 2 b$

$2 = - 2 + 3 b$

$a = 1 - b$

$1 = - 1 + 2 b$

$2 = - 2 + 3 b$

$a = 1 - b$

$1 = - 1 + 2 b$

$2 = - 2 + 3 b$

$a = 1 - b$

Step 1.3.3

Solve for $b$ in $1 = - 1 + 2 b$ .

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Step 1.3.3.1

Rewrite the equation as $- 1 + 2 b = 1$ .

$- 1 + 2 b = 1$

$2 = - 2 + 3 b$

$a = 1 - b$

Step 1.3.3.2

Move all terms not containing $b$ to the right side of the equation.

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Step 1.3.3.2.1

Add $1$ to both sides of the equation.

$2 b = 1 + 1$

$2 = - 2 + 3 b$

$a = 1 - b$

Step 1.3.3.2.2

Add $1$ and $1$ .

$2 b = 2$

$2 = - 2 + 3 b$

$a = 1 - b$

$2 b = 2$

$2 = - 2 + 3 b$

$a = 1 - b$

Step 1.3.3.3

Divide each term in $2 b = 2$ by $2$ and simplify.

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Step 1.3.3.3.1

Divide each term in $2 b = 2$ by $2$ .

$\frac{2 b}{2} = \frac{2}{2}$

$2 = - 2 + 3 b$

$a = 1 - b$

Step 1.3.3.3.2

Simplify the left side.

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Step 1.3.3.3.2.1

Cancel the common factor of $2$ .

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Step 1.3.3.3.2.1.1

Cancel the common factor.

$\frac{2 b}{2} = \frac{2}{2}$

$2 = - 2 + 3 b$

$a = 1 - b$

Step 1.3.3.3.2.1.2

Divide $b$ by $1$ .

$b = \frac{2}{2}$

$2 = - 2 + 3 b$

$a = 1 - b$

$b = \frac{2}{2}$

$2 = - 2 + 3 b$

$a = 1 - b$

$b = \frac{2}{2}$

$2 = - 2 + 3 b$

$a = 1 - b$

Step 1.3.3.3.3

Simplify the right side.

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Step 1.3.3.3.3.1

Divide $2$ by $2$ .

$b = 1$

$2 = - 2 + 3 b$

$a = 1 - b$

$b = 1$

$2 = - 2 + 3 b$

$a = 1 - b$

$b = 1$

$2 = - 2 + 3 b$

$a = 1 - b$

$b = 1$

$2 = - 2 + 3 b$

$a = 1 - b$

Step 1.3.4

Replace all occurrences of $b$ with $1$ in each equation.

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Step 1.3.4.1

Replace all occurrences of $b$ in $2 = - 2 + 3 b$ with $1$ .

$2 = - 2 + 3 (1)$

$b = 1$

$a = 1 - b$

Step 1.3.4.2

Simplify the right side.

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Step 1.3.4.2.1

Simplify $- 2 + 3 (1)$ .

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Step 1.3.4.2.1.1

Multiply $3$ by $1$ .

$2 = - 2 + 3$

$b = 1$

$a = 1 - b$

Step 1.3.4.2.1.2

Add $- 2$ and $3$ .

$2 = 1$

$b = 1$

$a = 1 - b$

$2 = 1$

$b = 1$

$a = 1 - b$

$2 = 1$

$b = 1$

$a = 1 - b$

Step 1.3.4.3

Replace all occurrences of $b$ in $a = 1 - b$ with $1$ .

$a = 1 - (1)$

$2 = 1$

$b = 1$

Step 1.3.4.4

Simplify the right side.

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Step 1.3.4.4.1

Simplify $1 - (1)$ .

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Step 1.3.4.4.1.1

Multiply $- 1$ by $1$ .

$a = 1 - 1$

$2 = 1$

$b = 1$

Step 1.3.4.4.1.2

Subtract $1$ from $1$ .

$a = 0$

$2 = 1$

$b = 1$

$a = 0$

$2 = 1$

$b = 1$

$a = 0$

$2 = 1$

$b = 1$

$a = 0$

$2 = 1$

$b = 1$

Step 1.3.5

Since $2 = 1$ is not true, there is no solution.

No solution

Step 1.4

Since $y \neq y$ for the corresponding $x$ values, the function is not linear.

The function is not linear

Step 2

Check if the function rule is quadratic.

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Step 2.1

To find if the table follows a function rule, check whether the function rule could follow the form $y = a x^{2} + b x + c$ .

$y = a x^{2} + b x + c$

Step 2.2

Build a set of $3$ equations from the table such that $y = a x^{2} + b x + c$ .

Step 2.3

Calculate the values of $a$ , $b$ , and $c$ .

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Step 2.3.1

Solve for $a$ in $1 = a + b + c$ .

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Step 2.3.1.1

Rewrite the equation as $a + b + c = 1$ .

$a + b + c = 1$

$2 = a {(- 2)}^{2} + b (- 2) + c$

$1 = a {(- 1)}^{2} + b (- 1) + c$

Step 2.3.1.2

Move all terms not containing $a$ to the right side of the equation.

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Step 2.3.1.2.1

Subtract $b$ from both sides of the equation.

$a + c = 1 - b$

$2 = a {(- 2)}^{2} + b (- 2) + c$

$1 = a {(- 1)}^{2} + b (- 1) + c$

Step 2.3.1.2.2

Subtract $c$ from both sides of the equation.

$a = 1 - b - c$

$2 = a {(- 2)}^{2} + b (- 2) + c$

$1 = a {(- 1)}^{2} + b (- 1) + c$

$a = 1 - b - c$

$2 = a {(- 2)}^{2} + b (- 2) + c$

$1 = a {(- 1)}^{2} + b (- 1) + c$

$a = 1 - b - c$

$2 = a {(- 2)}^{2} + b (- 2) + c$

$1 = a {(- 1)}^{2} + b (- 1) + c$

Step 2.3.2

Replace all occurrences of $a$ with $1 - b - c$ in each equation.

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Step 2.3.2.1

Replace all occurrences of $a$ in $2 = a {(- 2)}^{2} + b (- 2) + c$ with $1 - b - c$ .

$2 = (1 - b - c) {(- 2)}^{2} + b (- 2) + c$

$a = 1 - b - c$

$1 = a {(- 1)}^{2} + b (- 1) + c$

Step 2.3.2.2

Simplify the right side.

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Step 2.3.2.2.1

Simplify $(1 - b - c) {(- 2)}^{2} + b (- 2) + c$ .

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Step 2.3.2.2.1.1

Simplify each term.

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Step 2.3.2.2.1.1.1

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

$2 = (1 - b - c) \cdot 4 + b (- 2) + c$

$a = 1 - b - c$

$1 = a {(- 1)}^{2} + b (- 1) + c$

Step 2.3.2.2.1.1.2

Apply the distributive property.

$2 = 1 \cdot 4 - b \cdot 4 - c \cdot 4 + b (- 2) + c$

$a = 1 - b - c$

$1 = a {(- 1)}^{2} + b (- 1) + c$

Step 2.3.2.2.1.1.3

Simplify.

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Step 2.3.2.2.1.1.3.1

Multiply $4$ by $1$ .

$2 = 4 - b \cdot 4 - c \cdot 4 + b (- 2) + c$

$a = 1 - b - c$

$1 = a {(- 1)}^{2} + b (- 1) + c$

Step 2.3.2.2.1.1.3.2

Multiply $4$ by $- 1$ .

$2 = 4 - 4 b - c \cdot 4 + b (- 2) + c$

$a = 1 - b - c$

$1 = a {(- 1)}^{2} + b (- 1) + c$

Step 2.3.2.2.1.1.3.3

Multiply $4$ by $- 1$ .

$2 = 4 - 4 b - 4 c + b (- 2) + c$

$a = 1 - b - c$

$1 = a {(- 1)}^{2} + b (- 1) + c$

$2 = 4 - 4 b - 4 c + b (- 2) + c$

$a = 1 - b - c$

$1 = a {(- 1)}^{2} + b (- 1) + c$

Step 2.3.2.2.1.1.4

Move $- 2$ to the left of $b$ .

$2 = 4 - 4 b - 4 c - 2 b + c$

$a = 1 - b - c$

$1 = a {(- 1)}^{2} + b (- 1) + c$

$2 = 4 - 4 b - 4 c - 2 b + c$

$a = 1 - b - c$

$1 = a {(- 1)}^{2} + b (- 1) + c$

Step 2.3.2.2.1.2

Simplify by adding terms.

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Step 2.3.2.2.1.2.1

Subtract $2 b$ from $- 4 b$ .

$2 = 4 - 6 b - 4 c + c$

$a = 1 - b - c$

$1 = a {(- 1)}^{2} + b (- 1) + c$

Step 2.3.2.2.1.2.2

Add $- 4 c$ and $c$ .

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$1 = a {(- 1)}^{2} + b (- 1) + c$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$1 = a {(- 1)}^{2} + b (- 1) + c$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$1 = a {(- 1)}^{2} + b (- 1) + c$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$1 = a {(- 1)}^{2} + b (- 1) + c$

Step 2.3.2.3

Replace all occurrences of $a$ in $1 = a {(- 1)}^{2} + b (- 1) + c$ with $1 - b - c$ .

$1 = (1 - b - c) {(- 1)}^{2} + b (- 1) + c$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Step 2.3.2.4

Simplify the right side.

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Step 2.3.2.4.1

Simplify $(1 - b - c) {(- 1)}^{2} + b (- 1) + c$ .

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Step 2.3.2.4.1.1

Simplify each term.

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Step 2.3.2.4.1.1.1

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

$1 = (1 - b - c) \cdot 1 + b (- 1) + c$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Step 2.3.2.4.1.1.2

Multiply $1 - b - c$ by $1$ .

$1 = 1 - b - c + b (- 1) + c$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Step 2.3.2.4.1.1.3

Move $- 1$ to the left of $b$ .

$1 = 1 - b - c - 1 b + c$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Step 2.3.2.4.1.1.4

Rewrite $- 1 b$ as $- b$ .

$1 = 1 - b - c - b + c$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$1 = 1 - b - c - b + c$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Step 2.3.2.4.1.2

Simplify by adding terms.

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Step 2.3.2.4.1.2.1

Combine the opposite terms in $1 - b - c - b + c$ .

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Step 2.3.2.4.1.2.1.1

Add $- c$ and $c$ .

$1 = 1 - b - b + 0$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Step 2.3.2.4.1.2.1.2

Add $1 - b - b$ and $0$ .

$1 = 1 - b - b$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$1 = 1 - b - b$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Step 2.3.2.4.1.2.2

Subtract $b$ from $- b$ .

$1 = 1 - 2 b$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$1 = 1 - 2 b$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$1 = 1 - 2 b$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$1 = 1 - 2 b$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$1 = 1 - 2 b$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Step 2.3.3

Solve for $b$ in $1 = 1 - 2 b$ .

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Step 2.3.3.1

Rewrite the equation as $1 - 2 b = 1$ .

$1 - 2 b = 1$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Step 2.3.3.2

Move all terms not containing $b$ to the right side of the equation.

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Step 2.3.3.2.1

Subtract $1$ from both sides of the equation.

$- 2 b = 1 - 1$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Step 2.3.3.2.2

Subtract $1$ from $1$ .

$- 2 b = 0$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$- 2 b = 0$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Step 2.3.3.3

Divide each term in $- 2 b = 0$ by $- 2$ and simplify.

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Step 2.3.3.3.1

Divide each term in $- 2 b = 0$ by $- 2$ .

$\frac{- 2 b}{- 2} = \frac{0}{- 2}$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Step 2.3.3.3.2

Simplify the left side.

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Step 2.3.3.3.2.1

Cancel the common factor of $- 2$ .

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Step 2.3.3.3.2.1.1

Cancel the common factor.

$\frac{- 2 b}{- 2} = \frac{0}{- 2}$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Step 2.3.3.3.2.1.2

Divide $b$ by $1$ .

$b = \frac{0}{- 2}$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$b = \frac{0}{- 2}$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$b = \frac{0}{- 2}$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Step 2.3.3.3.3

Simplify the right side.

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Step 2.3.3.3.3.1

Divide $0$ by $- 2$ .

$b = 0$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$b = 0$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$b = 0$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$b = 0$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Step 2.3.4

Replace all occurrences of $b$ with $0$ in each equation.

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Step 2.3.4.1

Replace all occurrences of $b$ in $2 = 4 - 6 b - 3 c$ with $0$ .

$2 = 4 - 6 \cdot 0 - 3 c$

$b = 0$

$a = 1 - b - c$

Step 2.3.4.2

Simplify the right side.

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Step 2.3.4.2.1

Simplify $4 - 6 \cdot 0 - 3 c$ .

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Step 2.3.4.2.1.1

Multiply $- 6$ by $0$ .

$2 = 4 + 0 - 3 c$

$b = 0$

$a = 1 - b - c$

Step 2.3.4.2.1.2

Add $4$ and $0$ .

$2 = 4 - 3 c$

$b = 0$

$a = 1 - b - c$

$2 = 4 - 3 c$

$b = 0$

$a = 1 - b - c$

$2 = 4 - 3 c$

$b = 0$

$a = 1 - b - c$

Step 2.3.4.3

Replace all occurrences of $b$ in $a = 1 - b - c$ with $0$ .

$a = 1 - (0) - c$

$2 = 4 - 3 c$

$b = 0$

Step 2.3.4.4

Simplify the right side.

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Step 2.3.4.4.1

Simplify $1 - (0) - c$ .

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Step 2.3.4.4.1.1

Multiply $- 1$ by $0$ .

$a = 1 + 0 - c$

$2 = 4 - 3 c$

$b = 0$

Step 2.3.4.4.1.2

Add $1$ and $0$ .

$a = 1 - c$

$2 = 4 - 3 c$

$b = 0$

$a = 1 - c$

$2 = 4 - 3 c$

$b = 0$

$a = 1 - c$

$2 = 4 - 3 c$

$b = 0$

$a = 1 - c$

$2 = 4 - 3 c$

$b = 0$

Step 2.3.5

Solve for $c$ in $2 = 4 - 3 c$ .

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Step 2.3.5.1

Rewrite the equation as $4 - 3 c = 2$ .

$4 - 3 c = 2$

$a = 1 - c$

$b = 0$

Step 2.3.5.2

Move all terms not containing $c$ to the right side of the equation.

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Step 2.3.5.2.1

Subtract $4$ from both sides of the equation.

$- 3 c = 2 - 4$

$a = 1 - c$

$b = 0$

Step 2.3.5.2.2

Subtract $4$ from $2$ .

$- 3 c = - 2$

$a = 1 - c$

$b = 0$

$- 3 c = - 2$

$a = 1 - c$

$b = 0$

Step 2.3.5.3

Divide each term in $- 3 c = - 2$ by $- 3$ and simplify.

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Step 2.3.5.3.1

Divide each term in $- 3 c = - 2$ by $- 3$ .

$\frac{- 3 c}{- 3} = \frac{- 2}{- 3}$

$a = 1 - c$

$b = 0$

Step 2.3.5.3.2

Simplify the left side.

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Step 2.3.5.3.2.1

Cancel the common factor of $- 3$ .

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Step 2.3.5.3.2.1.1

Cancel the common factor.

$\frac{- 3 c}{- 3} = \frac{- 2}{- 3}$

$a = 1 - c$

$b = 0$

Step 2.3.5.3.2.1.2

Divide $c$ by $1$ .

$c = \frac{- 2}{- 3}$

$a = 1 - c$

$b = 0$

$c = \frac{- 2}{- 3}$

$a = 1 - c$

$b = 0$

$c = \frac{- 2}{- 3}$

$a = 1 - c$

$b = 0$

Step 2.3.5.3.3

Simplify the right side.

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Step 2.3.5.3.3.1

Dividing two negative values results in a positive value.

$c = \frac{2}{3}$

$a = 1 - c$

$b = 0$

$c = \frac{2}{3}$

$a = 1 - c$

$b = 0$

$c = \frac{2}{3}$

$a = 1 - c$

$b = 0$

$c = \frac{2}{3}$

$a = 1 - c$

$b = 0$

Step 2.3.6

Replace all occurrences of $c$ with $\frac{2}{3}$ in each equation.

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Step 2.3.6.1

Replace all occurrences of $c$ in $a = 1 - c$ with $\frac{2}{3}$ .

$a = 1 - (\frac{2}{3})$

$c = \frac{2}{3}$

$b = 0$

Step 2.3.6.2

Simplify the right side.

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Step 2.3.6.2.1

Simplify $1 - (\frac{2}{3})$ .

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Step 2.3.6.2.1.1

Write $1$ as a fraction with a common denominator.

$a = \frac{3}{3} - \frac{2}{3}$

$c = \frac{2}{3}$

$b = 0$

Step 2.3.6.2.1.2

Combine the numerators over the common denominator.

$a = \frac{3 - 2}{3}$

$c = \frac{2}{3}$

$b = 0$

Step 2.3.6.2.1.3

Subtract $2$ from $3$ .

$a = \frac{1}{3}$

$c = \frac{2}{3}$

$b = 0$

$a = \frac{1}{3}$

$c = \frac{2}{3}$

$b = 0$

$a = \frac{1}{3}$

$c = \frac{2}{3}$

$b = 0$

$a = \frac{1}{3}$

$c = \frac{2}{3}$

$b = 0$

Step 2.3.7

List all of the solutions.

$a = \frac{1}{3}, c = \frac{2}{3}, b = 0$

Step 2.4

Calculate the value of $y$ using each $x$ value in the table and compare this value to the given $y$ value in the table.

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Step 2.4.1

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = \frac{1}{3}$ , $b = 0$ , $c = \frac{2}{3}$ , and $x = - 2$ .

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Step 2.4.1.1

Simplify each term.

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Step 2.4.1.1.1

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

$y = \frac{1}{3} \cdot 4 + (0) \cdot (- 2) + \frac{2}{3}$

Step 2.4.1.1.2

Combine $\frac{1}{3}$ and $4$ .

$y = \frac{4}{3} + (0) \cdot (- 2) + \frac{2}{3}$

Step 2.4.1.1.3

Multiply $0$ by $- 2$ .

$y = \frac{4}{3} + 0 + \frac{2}{3}$

Step 2.4.1.2

Combine fractions.

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Step 2.4.1.2.1

Combine the numerators over the common denominator.

$y = \frac{4 + 2}{3}$

Step 2.4.1.2.2

Simplify the expression.

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Step 2.4.1.2.2.1

Add $4$ and $2$ .

$y = \frac{6}{3}$

Step 2.4.1.2.2.2

Divide $6$ by $3$ .

$y = 2$

Step 2.4.2

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = - 2$ . This check passes since $y = 2$ and $y = 2$ .

$2 = 2$

Step 2.4.3

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = \frac{1}{3}$ , $b = 0$ , $c = \frac{2}{3}$ , and $x = - 1$ .

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Step 2.4.3.1

Simplify each term.

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Step 2.4.3.1.1

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

$y = \frac{1}{3} \cdot 1 + (0) \cdot (- 1) + \frac{2}{3}$

Step 2.4.3.1.2

Multiply $\frac{1}{3}$ by $1$ .

$y = \frac{1}{3} + (0) \cdot (- 1) + \frac{2}{3}$

Step 2.4.3.1.3

Multiply $0$ by $- 1$ .

$y = \frac{1}{3} + 0 + \frac{2}{3}$

Step 2.4.3.2

Combine fractions.

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Step 2.4.3.2.1

Combine the numerators over the common denominator.

$y = \frac{1 + 2}{3}$

Step 2.4.3.2.2

Simplify the expression.

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Step 2.4.3.2.2.1

Add $1$ and $2$ .

$y = \frac{3}{3}$

Step 2.4.3.2.2.2

Divide $3$ by $3$ .

$y = 1$

Step 2.4.4

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = - 1$ . This check passes since $y = 1$ and $y = 1$ .

$1 = 1$

Step 2.4.5

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = \frac{1}{3}$ , $b = 0$ , $c = \frac{2}{3}$ , and $x = 1$ .

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Step 2.4.5.1

Simplify each term.

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Step 2.4.5.1.1

One to any power is one.

$y = \frac{1}{3} \cdot 1 + (0) \cdot (1) + \frac{2}{3}$

Step 2.4.5.1.2

Multiply $\frac{1}{3}$ by $1$ .

$y = \frac{1}{3} + (0) \cdot (1) + \frac{2}{3}$

Step 2.4.5.1.3

Multiply $0$ by $1$ .

$y = \frac{1}{3} + 0 + \frac{2}{3}$

Step 2.4.5.2

Combine fractions.

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Step 2.4.5.2.1

Combine the numerators over the common denominator.

$y = \frac{1 + 2}{3}$

Step 2.4.5.2.2

Simplify the expression.

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Step 2.4.5.2.2.1

Add $1$ and $2$ .

$y = \frac{3}{3}$

Step 2.4.5.2.2.2

Divide $3$ by $3$ .

$y = 1$

Step 2.4.6

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 1$ . This check passes since $y = 1$ and $y = 1$ .

$1 = 1$

Step 2.4.7

Since $y = y$ for the corresponding $x$ values, the function is quadratic.

The function is quadratic

Step 3

Since all $y = y$ , the function is quadratic and follows the form $y = \frac{x^{2}}{3} + \frac{2}{3}$ .

$y = \frac{x^{2}}{3} + \frac{2}{3}$