Find the Function Rule table[[x,y],[1,5],[2,12],[3,31]]

$\begin{array}{c} x & y \\ 1 & 5 \\ 2 & 12 \\ 3 & 31 \end{array}$

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 $y = a x + b$ .

$5 = a (1) + b 12 = a (2) + b 31 = a (3) + b$

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

Solve for $a$ in $5 = a + b$ .

Rewrite the equation as $a + b = 5$ .

$a + b = 5$

$12 = a (2) + b$

$31 = a (3) + b$

Subtract $b$ from both sides of the equation.

$a = 5 - b$

$12 = a (2) + b$

$31 = a (3) + b$

$a = 5 - b$

$12 = a (2) + b$

$31 = a (3) + b$

Replace all occurrences of $a$ with $5 - b$ in each equation.

Replace all occurrences of $a$ in $12 = a (2) + b$ with $5 - b$ .

$12 = (5 - b) (2) + b$

$a = 5 - b$

$31 = a (3) + b$

Simplify the right side.

Simplify $(5 - b) (2) + b$ .

Simplify each term.

Apply the distributive property.

$12 = 5 \cdot 2 - b \cdot 2 + b$

$a = 5 - b$

$31 = a (3) + b$

Multiply $5$ by $2$ .

$12 = 10 - b \cdot 2 + b$

$a = 5 - b$

$31 = a (3) + b$

Multiply $2$ by $- 1$ .

$12 = 10 - 2 b + b$

$a = 5 - b$

$31 = a (3) + b$

$12 = 10 - 2 b + b$

$a = 5 - b$

$31 = a (3) + b$

Add $- 2 b$ and $b$ .

$12 = 10 - b$

$a = 5 - b$

$31 = a (3) + b$

$12 = 10 - b$

$a = 5 - b$

$31 = a (3) + b$

$12 = 10 - b$

$a = 5 - b$

$31 = a (3) + b$

Replace all occurrences of $a$ in $31 = a (3) + b$ with $5 - b$ .

$31 = (5 - b) (3) + b$

$12 = 10 - b$

$a = 5 - b$

Simplify the right side.

Simplify $(5 - b) (3) + b$ .

Simplify each term.

Apply the distributive property.

$31 = 5 \cdot 3 - b \cdot 3 + b$

$12 = 10 - b$

$a = 5 - b$

Multiply $5$ by $3$ .

$31 = 15 - b \cdot 3 + b$

$12 = 10 - b$

$a = 5 - b$

Multiply $3$ by $- 1$ .

$31 = 15 - 3 b + b$

$12 = 10 - b$

$a = 5 - b$

$31 = 15 - 3 b + b$

$12 = 10 - b$

$a = 5 - b$

Add $- 3 b$ and $b$ .

$31 = 15 - 2 b$

$12 = 10 - b$

$a = 5 - b$

$31 = 15 - 2 b$

$12 = 10 - b$

$a = 5 - b$

$31 = 15 - 2 b$

$12 = 10 - b$

$a = 5 - b$

$31 = 15 - 2 b$

$12 = 10 - b$

$a = 5 - b$

Solve for $b$ in $31 = 15 - 2 b$ .

Rewrite the equation as $15 - 2 b = 31$ .

$15 - 2 b = 31$

$12 = 10 - b$

$a = 5 - b$

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

Subtract $15$ from both sides of the equation.

$- 2 b = 31 - 15$

$12 = 10 - b$

$a = 5 - b$

Subtract $15$ from $31$ .

$- 2 b = 16$

$12 = 10 - b$

$a = 5 - b$

$- 2 b = 16$

$12 = 10 - b$

$a = 5 - b$

Divide each term in $- 2 b = 16$ by $- 2$ and simplify.

Divide each term in $- 2 b = 16$ by $- 2$ .

$\frac{- 2 b}{- 2} = \frac{16}{- 2}$

$12 = 10 - b$

$a = 5 - b$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 b}{- 2} = \frac{16}{- 2}$

$12 = 10 - b$

$a = 5 - b$

Divide $b$ by $1$ .

$b = \frac{16}{- 2}$

$12 = 10 - b$

$a = 5 - b$

$b = \frac{16}{- 2}$

$12 = 10 - b$

$a = 5 - b$

$b = \frac{16}{- 2}$

$12 = 10 - b$

$a = 5 - b$

Simplify the right side.

Divide $16$ by $- 2$ .

$b = - 8$

$12 = 10 - b$

$a = 5 - b$

$b = - 8$

$12 = 10 - b$

$a = 5 - b$

$b = - 8$

$12 = 10 - b$

$a = 5 - b$

$b = - 8$

$12 = 10 - b$

$a = 5 - b$

Replace all occurrences of $b$ with $- 8$ in each equation.

Replace all occurrences of $b$ in $12 = 10 - b$ with $- 8$ .

$12 = 10 - (- 8)$

$b = - 8$

$a = 5 - b$

Simplify the right side.

Simplify $10 - (- 8)$ .

Multiply $- 1$ by $- 8$ .

$12 = 10 + 8$

$b = - 8$

$a = 5 - b$

Add $10$ and $8$ .

$12 = 18$

$b = - 8$

$a = 5 - b$

$12 = 18$

$b = - 8$

$a = 5 - b$

$12 = 18$

$b = - 8$

$a = 5 - b$

Replace all occurrences of $b$ in $a = 5 - b$ with $- 8$ .

$a = 5 - (- 8)$

$12 = 18$

$b = - 8$

Simplify the right side.

Simplify $5 - (- 8)$ .

Multiply $- 1$ by $- 8$ .

$a = 5 + 8$

$12 = 18$

$b = - 8$

Add $5$ and $8$ .

$a = 13$

$12 = 18$

$b = - 8$

$a = 13$

$12 = 18$

$b = - 8$

$a = 13$

$12 = 18$

$b = - 8$

$a = 13$

$12 = 18$

$b = - 8$

Since $12 = 18$ 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 $5 = a + b + c$ .

Rewrite the equation as $a + b + c = 5$ .

$a + b + c = 5$

$12 = a \cdot 2^{2} + b (2) + c$

$31 = a \cdot 3^{2} + b (3) + c$

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

Subtract $b$ from both sides of the equation.

$a + c = 5 - b$

$12 = a \cdot 2^{2} + b (2) + c$

$31 = a \cdot 3^{2} + b (3) + c$

Subtract $c$ from both sides of the equation.

$a = 5 - b - c$

$12 = a \cdot 2^{2} + b (2) + c$

$31 = a \cdot 3^{2} + b (3) + c$

$a = 5 - b - c$

$12 = a \cdot 2^{2} + b (2) + c$

$31 = a \cdot 3^{2} + b (3) + c$

$a = 5 - b - c$

$12 = a \cdot 2^{2} + b (2) + c$

$31 = a \cdot 3^{2} + b (3) + c$

Replace all occurrences of $a$ with $5 - b - c$ in each equation.

Replace all occurrences of $a$ in $12 = a \cdot 2^{2} + b (2) + c$ with $5 - b - c$ .

$12 = (5 - b - c) \cdot 2^{2} + b (2) + c$

$a = 5 - b - c$

$31 = a \cdot 3^{2} + b (3) + c$

Simplify the right side.

Simplify $(5 - b - c) \cdot 2^{2} + b (2) + c$ .

Simplify each term.

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

$12 = (5 - b - c) \cdot 4 + b (2) + c$

$a = 5 - b - c$

$31 = a \cdot 3^{2} + b (3) + c$

Apply the distributive property.

$12 = 5 \cdot 4 - b \cdot 4 - c \cdot 4 + b (2) + c$

$a = 5 - b - c$

$31 = a \cdot 3^{2} + b (3) + c$

Simplify.

Multiply $5$ by $4$ .

$12 = 20 - b \cdot 4 - c \cdot 4 + b (2) + c$

$a = 5 - b - c$

$31 = a \cdot 3^{2} + b (3) + c$

Multiply $4$ by $- 1$ .

$12 = 20 - 4 b - c \cdot 4 + b (2) + c$

$a = 5 - b - c$

$31 = a \cdot 3^{2} + b (3) + c$

Multiply $4$ by $- 1$ .

$12 = 20 - 4 b - 4 c + b (2) + c$

$a = 5 - b - c$

$31 = a \cdot 3^{2} + b (3) + c$

$12 = 20 - 4 b - 4 c + b (2) + c$

$a = 5 - b - c$

$31 = a \cdot 3^{2} + b (3) + c$

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

$12 = 20 - 4 b - 4 c + 2 b + c$

$a = 5 - b - c$

$31 = a \cdot 3^{2} + b (3) + c$

$12 = 20 - 4 b - 4 c + 2 b + c$

$a = 5 - b - c$

$31 = a \cdot 3^{2} + b (3) + c$

Simplify by adding terms.

Add $- 4 b$ and $2 b$ .

$12 = 20 - 2 b - 4 c + c$

$a = 5 - b - c$

$31 = a \cdot 3^{2} + b (3) + c$

Add $- 4 c$ and $c$ .

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$31 = a \cdot 3^{2} + b (3) + c$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$31 = a \cdot 3^{2} + b (3) + c$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$31 = a \cdot 3^{2} + b (3) + c$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$31 = a \cdot 3^{2} + b (3) + c$

Replace all occurrences of $a$ in $31 = a \cdot 3^{2} + b (3) + c$ with $5 - b - c$ .

$31 = (5 - b - c) \cdot 3^{2} + b (3) + c$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Simplify the right side.

Simplify $(5 - b - c) \cdot 3^{2} + b (3) + c$ .

Simplify each term.

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

$31 = (5 - b - c) \cdot 9 + b (3) + c$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Apply the distributive property.

$31 = 5 \cdot 9 - b \cdot 9 - c \cdot 9 + b (3) + c$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Simplify.

Multiply $5$ by $9$ .

$31 = 45 - b \cdot 9 - c \cdot 9 + b (3) + c$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Multiply $9$ by $- 1$ .

$31 = 45 - 9 b - c \cdot 9 + b (3) + c$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Multiply $9$ by $- 1$ .

$31 = 45 - 9 b - 9 c + b (3) + c$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$31 = 45 - 9 b - 9 c + b (3) + c$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Move $3$ to the left of $b$ .

$31 = 45 - 9 b - 9 c + 3 b + c$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$31 = 45 - 9 b - 9 c + 3 b + c$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Simplify by adding terms.

Add $- 9 b$ and $3 b$ .

$31 = 45 - 6 b - 9 c + c$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Add $- 9 c$ and $c$ .

$31 = 45 - 6 b - 8 c$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$31 = 45 - 6 b - 8 c$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$31 = 45 - 6 b - 8 c$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$31 = 45 - 6 b - 8 c$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$31 = 45 - 6 b - 8 c$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Solve for $b$ in $31 = 45 - 6 b - 8 c$ .

Rewrite the equation as $45 - 6 b - 8 c = 31$ .

$45 - 6 b - 8 c = 31$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

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

Subtract $45$ from both sides of the equation.

$- 6 b - 8 c = 31 - 45$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Add $8 c$ to both sides of the equation.

$- 6 b = 31 - 45 + 8 c$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Subtract $45$ from $31$ .

$- 6 b = - 14 + 8 c$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$- 6 b = - 14 + 8 c$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Divide each term in $- 6 b = - 14 + 8 c$ by $- 6$ and simplify.

Divide each term in $- 6 b = - 14 + 8 c$ by $- 6$ .

$\frac{- 6 b}{- 6} = \frac{- 14}{- 6} + \frac{8 c}{- 6}$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Simplify the left side.

Cancel the common factor of $- 6$ .

Cancel the common factor.

$\frac{- 6 b}{- 6} = \frac{- 14}{- 6} + \frac{8 c}{- 6}$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Divide $b$ by $1$ .

$b = \frac{- 14}{- 6} + \frac{8 c}{- 6}$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$b = \frac{- 14}{- 6} + \frac{8 c}{- 6}$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$b = \frac{- 14}{- 6} + \frac{8 c}{- 6}$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Simplify the right side.

Simplify each term.

Cancel the common factor of $- 14$ and $- 6$ .

Factor $- 2$ out of $- 14$ .

$b = \frac{- 2 \cdot 7}{- 6} + \frac{8 c}{- 6}$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Cancel the common factors.

Factor $- 2$ out of $- 6$ .

$b = \frac{- 2 \cdot 7}{- 2 \cdot 3} + \frac{8 c}{- 6}$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Cancel the common factor.

$b = \frac{- 2 \cdot 7}{- 2 \cdot 3} + \frac{8 c}{- 6}$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Rewrite the expression.

$b = \frac{7}{3} + \frac{8 c}{- 6}$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$b = \frac{7}{3} + \frac{8 c}{- 6}$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$b = \frac{7}{3} + \frac{8 c}{- 6}$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Cancel the common factor of $8$ and $- 6$ .

Factor $2$ out of $8 c$ .

$b = \frac{7}{3} + \frac{2 (4 c)}{- 6}$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Cancel the common factors.

Factor $2$ out of $- 6$ .

$b = \frac{7}{3} + \frac{2 (4 c)}{2 (- 3)}$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Cancel the common factor.

$b = \frac{7}{3} + \frac{2 (4 c)}{2 \cdot - 3}$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Rewrite the expression.

$b = \frac{7}{3} + \frac{4 c}{- 3}$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$b = \frac{7}{3} + \frac{4 c}{- 3}$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$b = \frac{7}{3} + \frac{4 c}{- 3}$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Move the negative in front of the fraction.

$b = \frac{7}{3} - \frac{4 c}{3}$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$b = \frac{7}{3} - \frac{4 c}{3}$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$b = \frac{7}{3} - \frac{4 c}{3}$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$b = \frac{7}{3} - \frac{4 c}{3}$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$b = \frac{7}{3} - \frac{4 c}{3}$

$12 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Replace all occurrences of $b$ with $\frac{7}{3} - \frac{4 c}{3}$ in each equation.

Replace all occurrences of $b$ in $12 = 20 - 2 b - 3 c$ with $\frac{7}{3} - \frac{4 c}{3}$ .

$12 = 20 - 2 (\frac{7}{3} - \frac{4 c}{3}) - 3 c$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = 5 - b - c$

Simplify the right side.

Simplify $20 - 2 (\frac{7}{3} - \frac{4 c}{3}) - 3 c$ .

Simplify each term.

Apply the distributive property.

$12 = 20 - 2 (\frac{7}{3}) - 2 (- \frac{4 c}{3}) - 3 c$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = 5 - b - c$

Multiply $- 2 (\frac{7}{3})$ .

Combine $- 2$ and $\frac{7}{3}$ .

$12 = 20 + \frac{- 2 \cdot 7}{3} - 2 (- \frac{4 c}{3}) - 3 c$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = 5 - b - c$

Multiply $- 2$ by $7$ .

$12 = 20 + \frac{- 14}{3} - 2 (- \frac{4 c}{3}) - 3 c$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = 5 - b - c$

$12 = 20 + \frac{- 14}{3} - 2 (- \frac{4 c}{3}) - 3 c$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = 5 - b - c$

Multiply $- 2 (- \frac{4 c}{3})$ .

Multiply $- 1$ by $- 2$ .

$12 = 20 + \frac{- 14}{3} + 2 (\frac{4 c}{3}) - 3 c$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = 5 - b - c$

Combine $2$ and $\frac{4 c}{3}$ .

$12 = 20 + \frac{- 14}{3} + \frac{2 (4 c)}{3} - 3 c$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = 5 - b - c$

Multiply $4$ by $2$ .

$12 = 20 + \frac{- 14}{3} + \frac{8 c}{3} - 3 c$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = 5 - b - c$

$12 = 20 + \frac{- 14}{3} + \frac{8 c}{3} - 3 c$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = 5 - b - c$

Move the negative in front of the fraction.

$12 = 20 - \frac{14}{3} + \frac{8 c}{3} - 3 c$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = 5 - b - c$

$12 = 20 - \frac{14}{3} + \frac{8 c}{3} - 3 c$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = 5 - b - c$

To write $20$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$12 = 20 \cdot \frac{3}{3} - \frac{14}{3} + \frac{8 c}{3} - 3 c$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = 5 - b - c$

Combine $20$ and $\frac{3}{3}$ .

$12 = \frac{20 \cdot 3}{3} - \frac{14}{3} + \frac{8 c}{3} - 3 c$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = 5 - b - c$

Combine the numerators over the common denominator.

$12 = \frac{20 \cdot 3 - 14}{3} + \frac{8 c}{3} - 3 c$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = 5 - b - c$

Simplify the numerator.

Multiply $20$ by $3$ .

$12 = \frac{60 - 14}{3} + \frac{8 c}{3} - 3 c$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = 5 - b - c$

Subtract $14$ from $60$ .

$12 = \frac{46}{3} + \frac{8 c}{3} - 3 c$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = 5 - b - c$

$12 = \frac{46}{3} + \frac{8 c}{3} - 3 c$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = 5 - b - c$

To write $- 3 c$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$12 = \frac{46}{3} + \frac{8 c}{3} - 3 c \cdot \frac{3}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = 5 - b - c$

Combine $- 3 c$ and $\frac{3}{3}$ .

$12 = \frac{46}{3} + \frac{8 c}{3} + \frac{- 3 c \cdot 3}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = 5 - b - c$

Combine the numerators over the common denominator.

$12 = \frac{46}{3} + \frac{8 c - 3 c \cdot 3}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = 5 - b - c$

Combine the numerators over the common denominator.

$12 = \frac{46 + 8 c - 3 c \cdot 3}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = 5 - b - c$

Multiply $3$ by $- 3$ .

$12 = \frac{46 + 8 c - 9 c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = 5 - b - c$

Subtract $9 c$ from $8 c$ .

$12 = \frac{46 - c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = 5 - b - c$

$12 = \frac{46 - c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = 5 - b - c$

$12 = \frac{46 - c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = 5 - b - c$

Replace all occurrences of $b$ in $a = 5 - b - c$ with $\frac{7}{3} - \frac{4 c}{3}$ .

$a = 5 - (\frac{7}{3} - \frac{4 c}{3}) - c$

$12 = \frac{46 - c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

Simplify the right side.

Simplify $5 - (\frac{7}{3} - \frac{4 c}{3}) - c$ .

Simplify each term.

Apply the distributive property.

$a = 5 - \frac{7}{3} + \frac{4 c}{3} - c$

$12 = \frac{46 - c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

Multiply $- - \frac{4 c}{3}$ .

Multiply $- 1$ by $- 1$ .

$a = 5 - \frac{7}{3} + 1 (\frac{4 c}{3}) - c$

$12 = \frac{46 - c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

Multiply $\frac{4 c}{3}$ by $1$ .

$a = 5 - \frac{7}{3} + \frac{4 c}{3} - c$

$12 = \frac{46 - c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = 5 - \frac{7}{3} + \frac{4 c}{3} - c$

$12 = \frac{46 - c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = 5 - \frac{7}{3} + \frac{4 c}{3} - c$

$12 = \frac{46 - c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

To write $5$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$a = 5 \cdot \frac{3}{3} - \frac{7}{3} + \frac{4 c}{3} - c$

$12 = \frac{46 - c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

Combine $5$ and $\frac{3}{3}$ .

$a = \frac{5 \cdot 3}{3} - \frac{7}{3} + \frac{4 c}{3} - c$

$12 = \frac{46 - c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

Combine the numerators over the common denominator.

$a = \frac{5 \cdot 3 - 7}{3} + \frac{4 c}{3} - c$

$12 = \frac{46 - c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

Simplify the numerator.

Multiply $5$ by $3$ .

$a = \frac{15 - 7}{3} + \frac{4 c}{3} - c$

$12 = \frac{46 - c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

Subtract $7$ from $15$ .

$a = \frac{8}{3} + \frac{4 c}{3} - c$

$12 = \frac{46 - c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = \frac{8}{3} + \frac{4 c}{3} - c$

$12 = \frac{46 - c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

To write $- c$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$a = \frac{8}{3} + \frac{4 c}{3} - c \cdot \frac{3}{3}$

$12 = \frac{46 - c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

Combine $- c$ and $\frac{3}{3}$ .

$a = \frac{8}{3} + \frac{4 c}{3} + \frac{- c \cdot 3}{3}$

$12 = \frac{46 - c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

Combine the numerators over the common denominator.

$a = \frac{8}{3} + \frac{4 c - c \cdot 3}{3}$

$12 = \frac{46 - c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

Combine the numerators over the common denominator.

$a = \frac{8 + 4 c - c \cdot 3}{3}$

$12 = \frac{46 - c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

Multiply $3$ by $- 1$ .

$a = \frac{8 + 4 c - 3 c}{3}$

$12 = \frac{46 - c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

Subtract $3 c$ from $4 c$ .

$a = \frac{8 + c}{3}$

$12 = \frac{46 - c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = \frac{8 + c}{3}$

$12 = \frac{46 - c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = \frac{8 + c}{3}$

$12 = \frac{46 - c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = \frac{8 + c}{3}$

$12 = \frac{46 - c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

Solve for $c$ in $12 = \frac{46 - c}{3}$ .

Rewrite the equation as $\frac{46 - c}{3} = 12$ .

$\frac{46 - c}{3} = 12$

$a = \frac{8 + c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

Multiply both sides by $3$ .

$\frac{46 - c}{3} \cdot 3 = 12 \cdot 3$

$a = \frac{8 + c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

Simplify.

Simplify the left side.

Simplify $\frac{46 - c}{3} \cdot 3$ .

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{46 - c}{3} \cdot 3 = 12 \cdot 3$

$a = \frac{8 + c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

Rewrite the expression.

$46 - c = 12 \cdot 3$

$a = \frac{8 + c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

$46 - c = 12 \cdot 3$

$a = \frac{8 + c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

Reorder $46$ and $- c$ .

$- c + 46 = 12 \cdot 3$

$a = \frac{8 + c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

$- c + 46 = 12 \cdot 3$

$a = \frac{8 + c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

$- c + 46 = 12 \cdot 3$

$a = \frac{8 + c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

Simplify the right side.

Multiply $12$ by $3$ .

$- c + 46 = 36$

$a = \frac{8 + c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

$- c + 46 = 36$

$a = \frac{8 + c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

$- c + 46 = 36$

$a = \frac{8 + c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

Solve for $c$ .

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

Subtract $46$ from both sides of the equation.

$- c = 36 - 46$

$a = \frac{8 + c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

Subtract $46$ from $36$ .

$- c = - 10$

$a = \frac{8 + c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

$- c = - 10$

$a = \frac{8 + c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

Divide each term in $- c = - 10$ by $- 1$ and simplify.

Divide each term in $- c = - 10$ by $- 1$ .

$\frac{- c}{- 1} = \frac{- 10}{- 1}$

$a = \frac{8 + c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{c}{1} = \frac{- 10}{- 1}$

$a = \frac{8 + c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

Divide $c$ by $1$ .

$c = \frac{- 10}{- 1}$

$a = \frac{8 + c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

$c = \frac{- 10}{- 1}$

$a = \frac{8 + c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

Simplify the right side.

Divide $- 10$ by $- 1$ .

$c = 10$

$a = \frac{8 + c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

$c = 10$

$a = \frac{8 + c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

$c = 10$

$a = \frac{8 + c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

$c = 10$

$a = \frac{8 + c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

$c = 10$

$a = \frac{8 + c}{3}$

$b = \frac{7}{3} - \frac{4 c}{3}$

Replace all occurrences of $c$ with $10$ in each equation.

Replace all occurrences of $c$ in $a = \frac{8 + c}{3}$ with $10$ .

$a = \frac{8 + 10}{3}$

$c = 10$

$b = \frac{7}{3} - \frac{4 c}{3}$

Simplify $a = \frac{8 + 10}{3}$ .

Simplify the left side.

Remove parentheses.

$a = \frac{8 + 10}{3}$

$c = 10$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = \frac{8 + 10}{3}$

$c = 10$

$b = \frac{7}{3} - \frac{4 c}{3}$

Simplify the right side.

Simplify $\frac{8 + 10}{3}$ .

Add $8$ and $10$ .

$a = \frac{18}{3}$

$c = 10$

$b = \frac{7}{3} - \frac{4 c}{3}$

Divide $18$ by $3$ .

$a = 6$

$c = 10$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = 6$

$c = 10$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = 6$

$c = 10$

$b = \frac{7}{3} - \frac{4 c}{3}$

$a = 6$

$c = 10$

$b = \frac{7}{3} - \frac{4 c}{3}$

Replace all occurrences of $c$ in $b = \frac{7}{3} - \frac{4 c}{3}$ with $10$ .

$b = \frac{7}{3} - \frac{4 (10)}{3}$

$a = 6$

$c = 10$

Simplify the right side.

Simplify $\frac{7}{3} - \frac{4 (10)}{3}$ .

Combine the numerators over the common denominator.

$b = \frac{7 - 4 \cdot 10}{3}$

$a = 6$

$c = 10$

Simplify the expression.

Multiply $- 4$ by $10$ .

$b = \frac{7 - 40}{3}$

$a = 6$

$c = 10$

Subtract $40$ from $7$ .

$b = \frac{- 33}{3}$

$a = 6$

$c = 10$

Divide $- 33$ by $3$ .

$b = - 11$

$a = 6$

$c = 10$

$b = - 11$

$a = 6$

$c = 10$

$b = - 11$

$a = 6$

$c = 10$

$b = - 11$

$a = 6$

$c = 10$

$b = - 11$

$a = 6$

$c = 10$

List all of the solutions.

$b = - 11, a = 6, c = 10$

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 $a = 6$ , $b = - 11$ , $c = 10$ , and $x = 1$ .

Simplify each term.

One to any power is one.

$y = 6 \cdot 1 + (- 11) \cdot (1) + 10$

Multiply $6$ by $1$ .

$y = 6 + (- 11) \cdot (1) + 10$

Multiply $- 11$ by $1$ .

$y = 6 - 11 + 10$

Simplify by adding and subtracting.

Subtract $11$ from $6$ .

$y = - 5 + 10$

Add $- 5$ and $10$ .

$y = 5$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 1$ . This check passes since $y = 5$ and $y = 5$ .

$5 = 5$

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = 6$ , $b = - 11$ , $c = 10$ , and $x = 2$ .

Simplify each term.

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

$y = 6 \cdot 4 + (- 11) \cdot (2) + 10$

Multiply $6$ by $4$ .

$y = 24 + (- 11) \cdot (2) + 10$

Multiply $- 11$ by $2$ .

$y = 24 - 22 + 10$

Simplify by adding and subtracting.

Subtract $22$ from $24$ .

$y = 2 + 10$

Add $2$ and $10$ .

$y = 12$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 2$ . This check passes since $y = 12$ and $y = 12$ .

$12 = 12$

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = 6$ , $b = - 11$ , $c = 10$ , and $x = 3$ .

Simplify each term.

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

$y = 6 \cdot 9 + (- 11) \cdot (3) + 10$

Multiply $6$ by $9$ .

$y = 54 + (- 11) \cdot (3) + 10$

Multiply $- 11$ by $3$ .

$y = 54 - 33 + 10$

Simplify by adding and subtracting.

Subtract $33$ from $54$ .

$y = 21 + 10$

Add $21$ and $10$ .

$y = 31$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 3$ . This check passes since $y = 31$ and $y = 31$ .

$31 = 31$

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 $y = 6 x^{2} - 11 x + 10$ .

$y = 6 x^{2} - 11 x + 10$

Find the Function Rule table[[x,y],[1,5],[2,12],[3,31]]

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