Find the Function Rule table[[x,y],[-2,2],[-1,1],[1,1]]

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

Step 1

Check if the function rule is linear.

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

$y = a x + b$

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

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

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

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

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

$a + b = 1$

$2 = a (- 2) + b$

$1 = a (- 1) + b$

Subtract $b$ from both sides of the equation.

$a = 1 - b$

$2 = a (- 2) + b$

$1 = a (- 1) + b$

$a = 1 - b$

$2 = a (- 2) + b$

$1 = a (- 1) + b$

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

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

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

$a = 1 - b$

$1 = a (- 1) + b$

Simplify the right side.

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

Simplify each term.

Apply the distributive property.

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

$a = 1 - b$

$1 = a (- 1) + b$

Multiply $- 2$ by $1$ .

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

$a = 1 - b$

$1 = a (- 1) + b$

Multiply $- 2$ by $- 1$ .

$2 = - 2 + 2 b + b$

$a = 1 - b$

$1 = a (- 1) + b$

$2 = - 2 + 2 b + b$

$a = 1 - b$

$1 = a (- 1) + b$

Add $2 b$ and $b$ .

$2 = - 2 + 3 b$

$a = 1 - b$

$1 = a (- 1) + b$

$2 = - 2 + 3 b$

$a = 1 - b$

$1 = a (- 1) + b$

$2 = - 2 + 3 b$

$a = 1 - b$

$1 = a (- 1) + b$

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

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

$2 = - 2 + 3 b$

$a = 1 - b$

Simplify the right side.

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

Simplify each term.

Apply the distributive property.

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

$2 = - 2 + 3 b$

$a = 1 - b$

Multiply $- 1$ by $1$ .

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

$2 = - 2 + 3 b$

$a = 1 - b$

Multiply $- b \cdot - 1$ .

Multiply $- 1$ by $- 1$ .

$1 = - 1 + 1 b + b$

$2 = - 2 + 3 b$

$a = 1 - b$

Multiply $b$ by $1$ .

$1 = - 1 + b + b$

$2 = - 2 + 3 b$

$a = 1 - b$

$1 = - 1 + b + b$

$2 = - 2 + 3 b$

$a = 1 - b$

$1 = - 1 + b + b$

$2 = - 2 + 3 b$

$a = 1 - b$

Add $b$ and $b$ .

$1 = - 1 + 2 b$

$2 = - 2 + 3 b$

$a = 1 - b$

$1 = - 1 + 2 b$

$2 = - 2 + 3 b$

$a = 1 - b$

$1 = - 1 + 2 b$

$2 = - 2 + 3 b$

$a = 1 - b$

$1 = - 1 + 2 b$

$2 = - 2 + 3 b$

$a = 1 - b$

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

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

$- 1 + 2 b = 1$

$2 = - 2 + 3 b$

$a = 1 - b$

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

Add $1$ to both sides of the equation.

$2 b = 1 + 1$

$2 = - 2 + 3 b$

$a = 1 - b$

Add $1$ and $1$ .

$2 b = 2$

$2 = - 2 + 3 b$

$a = 1 - b$

$2 b = 2$

$2 = - 2 + 3 b$

$a = 1 - b$

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

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

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

$2 = - 2 + 3 b$

$a = 1 - b$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

$2 = - 2 + 3 b$

$a = 1 - b$

Divide $b$ by $1$ .

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

$2 = - 2 + 3 b$

$a = 1 - b$

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

$2 = - 2 + 3 b$

$a = 1 - b$

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

$2 = - 2 + 3 b$

$a = 1 - b$

Simplify the right side.

Divide $2$ by $2$ .

$b = 1$

$2 = - 2 + 3 b$

$a = 1 - b$

$b = 1$

$2 = - 2 + 3 b$

$a = 1 - b$

$b = 1$

$2 = - 2 + 3 b$

$a = 1 - b$

$b = 1$

$2 = - 2 + 3 b$

$a = 1 - b$

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

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

$2 = - 2 + 3 (1)$

$b = 1$

$a = 1 - b$

Simplify the right side.

Simplify $- 2 + 3 (1)$ .

Multiply $3$ by $1$ .

$2 = - 2 + 3$

$b = 1$

$a = 1 - b$

Add $- 2$ and $3$ .

$2 = 1$

$b = 1$

$a = 1 - b$

$2 = 1$

$b = 1$

$a = 1 - b$

$2 = 1$

$b = 1$

$a = 1 - b$

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

$a = 1 - (1)$

$2 = 1$

$b = 1$

Simplify the right side.

Simplify $1 - (1)$ .

Multiply $- 1$ by $1$ .

$a = 1 - 1$

$2 = 1$

$b = 1$

Subtract $1$ from $1$ .

$a = 0$

$2 = 1$

$b = 1$

$a = 0$

$2 = 1$

$b = 1$

$a = 0$

$2 = 1$

$b = 1$

$a = 0$

$2 = 1$

$b = 1$

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

No solution

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.

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

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

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

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

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

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

$a + b + c = 1$

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

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

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

Subtract $b$ from both sides of the equation.

$a + c = 1 - b$

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

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

Subtract $c$ from both sides of the equation.

$a = 1 - b - c$

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

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

$a = 1 - b - c$

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

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

$a = 1 - b - c$

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

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

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

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

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

$a = 1 - b - c$

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

Simplify the right side.

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

Simplify each term.

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

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

$a = 1 - b - c$

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

Apply the distributive property.

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

$a = 1 - b - c$

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

Simplify.

Multiply $4$ by $1$ .

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

$a = 1 - b - c$

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

Multiply $4$ by $- 1$ .

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

$a = 1 - b - c$

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

Multiply $4$ by $- 1$ .

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

$a = 1 - b - c$

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

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

$a = 1 - b - c$

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

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

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

$a = 1 - b - c$

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

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

$a = 1 - b - c$

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

Simplify by adding terms.

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

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

$a = 1 - b - c$

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

Add $- 4 c$ and $c$ .

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

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

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

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

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

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

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

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

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

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

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Simplify the right side.

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

Simplify each term.

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

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

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

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

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

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

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

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

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

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

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

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

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

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Simplify by adding terms.

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

Add $- c$ and $c$ .

$1 = 1 - b - b + 0$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

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

$1 = 1 - b - b$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$1 = 1 - b - b$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Subtract $b$ from $- b$ .

$1 = 1 - 2 b$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$1 = 1 - 2 b$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$1 = 1 - 2 b$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$1 = 1 - 2 b$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$1 = 1 - 2 b$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

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

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

$1 - 2 b = 1$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

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

Subtract $1$ from both sides of the equation.

$- 2 b = 1 - 1$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Subtract $1$ from $1$ .

$- 2 b = 0$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$- 2 b = 0$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

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

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

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

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

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

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Divide $b$ by $1$ .

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

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

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

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

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

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Simplify the right side.

Divide $0$ by $- 2$ .

$b = 0$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$b = 0$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$b = 0$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$b = 0$

$2 = 4 - 6 b - 3 c$

$a = 1 - b - c$

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

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

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

$b = 0$

$a = 1 - b - c$

Simplify the right side.

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

Multiply $- 6$ by $0$ .

$2 = 4 + 0 - 3 c$

$b = 0$

$a = 1 - b - c$

Add $4$ and $0$ .

$2 = 4 - 3 c$

$b = 0$

$a = 1 - b - c$

$2 = 4 - 3 c$

$b = 0$

$a = 1 - b - c$

$2 = 4 - 3 c$

$b = 0$

$a = 1 - b - c$

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

$a = 1 - (0) - c$

$2 = 4 - 3 c$

$b = 0$

Simplify the right side.

Simplify $1 - (0) - c$ .

Multiply $- 1$ by $0$ .

$a = 1 + 0 - c$

$2 = 4 - 3 c$

$b = 0$

Add $1$ and $0$ .

$a = 1 - c$

$2 = 4 - 3 c$

$b = 0$

$a = 1 - c$

$2 = 4 - 3 c$

$b = 0$

$a = 1 - c$

$2 = 4 - 3 c$

$b = 0$

$a = 1 - c$

$2 = 4 - 3 c$

$b = 0$

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

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

$4 - 3 c = 2$

$a = 1 - c$

$b = 0$

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

Subtract $4$ from both sides of the equation.

$- 3 c = 2 - 4$

$a = 1 - c$

$b = 0$

Subtract $4$ from $2$ .

$- 3 c = - 2$

$a = 1 - c$

$b = 0$

$- 3 c = - 2$

$a = 1 - c$

$b = 0$

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

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

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

$a = 1 - c$

$b = 0$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

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

$a = 1 - c$

$b = 0$

Divide $c$ by $1$ .

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

$a = 1 - c$

$b = 0$

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

$a = 1 - c$

$b = 0$

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

$a = 1 - c$

$b = 0$

Simplify the right side.

Dividing two negative values results in a positive value.

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

$a = 1 - c$

$b = 0$

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

$a = 1 - c$

$b = 0$

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

$a = 1 - c$

$b = 0$

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

$a = 1 - c$

$b = 0$

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

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

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

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

$b = 0$

Simplify the right side.

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

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

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

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

$b = 0$

Combine the numerators over the common denominator.

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

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

$b = 0$

Subtract $2$ from $3$ .

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

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

$b = 0$

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

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

$b = 0$

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

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

$b = 0$

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

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

$b = 0$

List all of the solutions.

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

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

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

Simplify each term.

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

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

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

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

Multiply $0$ by $- 2$ .

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

Combine fractions.

Combine the numerators over the common denominator.

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

Simplify the expression.

Add $4$ and $2$ .

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

Divide $6$ by $3$ .

$y = 2$

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

$2 = 2$

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

Simplify each term.

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

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

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

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

Multiply $0$ by $- 1$ .

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

Combine fractions.

Combine the numerators over the common denominator.

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

Simplify the expression.

Add $1$ and $2$ .

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

Divide $3$ by $3$ .

$y = 1$

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

$1 = 1$

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

Simplify each term.

One to any power is one.

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

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

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

Multiply $0$ by $1$ .

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

Combine fractions.

Combine the numerators over the common denominator.

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

Simplify the expression.

Add $1$ and $2$ .

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

Divide $3$ by $3$ .

$y = 1$

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

$1 = 1$

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

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

Find the Function Rule table[[x,y],[-2,2],[-1,1],[1,1]]

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