Find the Function Rule table[[x,y],[1,40],[2,800],[3,16000]]

$\begin{array}{c} x & y \\ 1 & 40 \\ 2 & 800 \\ 3 & 16000 \end{array}$

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 $y = a x + b$ .

$40 = a (1) + b 800 = a (2) + b 16000 = a (3) + b$

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

Solve for $a$ in $40 = a + b$ .

Rewrite the equation as $a + b = 40$ .

$a + b = 40$

$800 = a (2) + b$

$16000 = a (3) + b$

Subtract $b$ from both sides of the equation.

$a = 40 - b$

$800 = a (2) + b$

$16000 = a (3) + b$

$a = 40 - b$

$800 = a (2) + b$

$16000 = a (3) + b$

Replace all occurrences of $a$ with $40 - b$ in each equation.

Replace all occurrences of $a$ in $800 = a (2) + b$ with $40 - b$ .

$800 = (40 - b) (2) + b$

$a = 40 - b$

$16000 = a (3) + b$

Simplify the right side.

Simplify $(40 - b) (2) + b$ .

Simplify each term.

Apply the distributive property.

$800 = 40 \cdot 2 - b \cdot 2 + b$

$a = 40 - b$

$16000 = a (3) + b$

Multiply $40$ by $2$ .

$800 = 80 - b \cdot 2 + b$

$a = 40 - b$

$16000 = a (3) + b$

Multiply $2$ by $- 1$ .

$800 = 80 - 2 b + b$

$a = 40 - b$

$16000 = a (3) + b$

$800 = 80 - 2 b + b$

$a = 40 - b$

$16000 = a (3) + b$

Add $- 2 b$ and $b$ .

$800 = 80 - b$

$a = 40 - b$

$16000 = a (3) + b$

$800 = 80 - b$

$a = 40 - b$

$16000 = a (3) + b$

$800 = 80 - b$

$a = 40 - b$

$16000 = a (3) + b$

Replace all occurrences of $a$ in $16000 = a (3) + b$ with $40 - b$ .

$16000 = (40 - b) (3) + b$

$800 = 80 - b$

$a = 40 - b$

Simplify the right side.

Simplify $(40 - b) (3) + b$ .

Simplify each term.

Apply the distributive property.

$16000 = 40 \cdot 3 - b \cdot 3 + b$

$800 = 80 - b$

$a = 40 - b$

Multiply $40$ by $3$ .

$16000 = 120 - b \cdot 3 + b$

$800 = 80 - b$

$a = 40 - b$

Multiply $3$ by $- 1$ .

$16000 = 120 - 3 b + b$

$800 = 80 - b$

$a = 40 - b$

$16000 = 120 - 3 b + b$

$800 = 80 - b$

$a = 40 - b$

Add $- 3 b$ and $b$ .

$16000 = 120 - 2 b$

$800 = 80 - b$

$a = 40 - b$

$16000 = 120 - 2 b$

$800 = 80 - b$

$a = 40 - b$

$16000 = 120 - 2 b$

$800 = 80 - b$

$a = 40 - b$

$16000 = 120 - 2 b$

$800 = 80 - b$

$a = 40 - b$

Solve for $b$ in $16000 = 120 - 2 b$ .

Rewrite the equation as $120 - 2 b = 16000$ .

$120 - 2 b = 16000$

$800 = 80 - b$

$a = 40 - b$

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

Subtract $120$ from both sides of the equation.

$- 2 b = 16000 - 120$

$800 = 80 - b$

$a = 40 - b$

Subtract $120$ from $16000$ .

$- 2 b = 15880$

$800 = 80 - b$

$a = 40 - b$

$- 2 b = 15880$

$800 = 80 - b$

$a = 40 - b$

Divide each term in $- 2 b = 15880$ by $- 2$ and simplify.

Divide each term in $- 2 b = 15880$ by $- 2$ .

$\frac{- 2 b}{- 2} = \frac{15880}{- 2}$

$800 = 80 - b$

$a = 40 - b$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 b}{- 2} = \frac{15880}{- 2}$

$800 = 80 - b$

$a = 40 - b$

Divide $b$ by $1$ .

$b = \frac{15880}{- 2}$

$800 = 80 - b$

$a = 40 - b$

$b = \frac{15880}{- 2}$

$800 = 80 - b$

$a = 40 - b$

$b = \frac{15880}{- 2}$

$800 = 80 - b$

$a = 40 - b$

Simplify the right side.

Divide $15880$ by $- 2$ .

$b = - 7940$

$800 = 80 - b$

$a = 40 - b$

$b = - 7940$

$800 = 80 - b$

$a = 40 - b$

$b = - 7940$

$800 = 80 - b$

$a = 40 - b$

$b = - 7940$

$800 = 80 - b$

$a = 40 - b$

Replace all occurrences of $b$ with $- 7940$ in each equation.

Replace all occurrences of $b$ in $800 = 80 - b$ with $- 7940$ .

$800 = 80 - (- 7940)$

$b = - 7940$

$a = 40 - b$

Simplify the right side.

Simplify $80 - (- 7940)$ .

Multiply $- 1$ by $- 7940$ .

$800 = 80 + 7940$

$b = - 7940$

$a = 40 - b$

Add $80$ and $7940$ .

$800 = 8020$

$b = - 7940$

$a = 40 - b$

$800 = 8020$

$b = - 7940$

$a = 40 - b$

$800 = 8020$

$b = - 7940$

$a = 40 - b$

Replace all occurrences of $b$ in $a = 40 - b$ with $- 7940$ .

$a = 40 - (- 7940)$

$800 = 8020$

$b = - 7940$

Simplify the right side.

Simplify $40 - (- 7940)$ .

Multiply $- 1$ by $- 7940$ .

$a = 40 + 7940$

$800 = 8020$

$b = - 7940$

Add $40$ and $7940$ .

$a = 7980$

$800 = 8020$

$b = - 7940$

$a = 7980$

$800 = 8020$

$b = - 7940$

$a = 7980$

$800 = 8020$

$b = - 7940$

$a = 7980$

$800 = 8020$

$b = - 7940$

Since $800 = 8020$ 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 $40 = a + b + c$ .

Rewrite the equation as $a + b + c = 40$ .

$a + b + c = 40$

$800 = a \cdot 2^{2} + b (2) + c$

$16000 = a \cdot 3^{2} + b (3) + c$

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

Subtract $b$ from both sides of the equation.

$a + c = 40 - b$

$800 = a \cdot 2^{2} + b (2) + c$

$16000 = a \cdot 3^{2} + b (3) + c$

Subtract $c$ from both sides of the equation.

$a = 40 - b - c$

$800 = a \cdot 2^{2} + b (2) + c$

$16000 = a \cdot 3^{2} + b (3) + c$

$a = 40 - b - c$

$800 = a \cdot 2^{2} + b (2) + c$

$16000 = a \cdot 3^{2} + b (3) + c$

$a = 40 - b - c$

$800 = a \cdot 2^{2} + b (2) + c$

$16000 = a \cdot 3^{2} + b (3) + c$

Replace all occurrences of $a$ with $40 - b - c$ in each equation.

Replace all occurrences of $a$ in $800 = a \cdot 2^{2} + b (2) + c$ with $40 - b - c$ .

$800 = (40 - b - c) \cdot 2^{2} + b (2) + c$

$a = 40 - b - c$

$16000 = a \cdot 3^{2} + b (3) + c$

Simplify the right side.

Simplify $(40 - b - c) \cdot 2^{2} + b (2) + c$ .

Simplify each term.

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

$800 = (40 - b - c) \cdot 4 + b (2) + c$

$a = 40 - b - c$

$16000 = a \cdot 3^{2} + b (3) + c$

Apply the distributive property.

$800 = 40 \cdot 4 - b \cdot 4 - c \cdot 4 + b (2) + c$

$a = 40 - b - c$

$16000 = a \cdot 3^{2} + b (3) + c$

Simplify.

Multiply $40$ by $4$ .

$800 = 160 - b \cdot 4 - c \cdot 4 + b (2) + c$

$a = 40 - b - c$

$16000 = a \cdot 3^{2} + b (3) + c$

Multiply $4$ by $- 1$ .

$800 = 160 - 4 b - c \cdot 4 + b (2) + c$

$a = 40 - b - c$

$16000 = a \cdot 3^{2} + b (3) + c$

Multiply $4$ by $- 1$ .

$800 = 160 - 4 b - 4 c + b (2) + c$

$a = 40 - b - c$

$16000 = a \cdot 3^{2} + b (3) + c$

$800 = 160 - 4 b - 4 c + b (2) + c$

$a = 40 - b - c$

$16000 = a \cdot 3^{2} + b (3) + c$

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

$800 = 160 - 4 b - 4 c + 2 b + c$

$a = 40 - b - c$

$16000 = a \cdot 3^{2} + b (3) + c$

$800 = 160 - 4 b - 4 c + 2 b + c$

$a = 40 - b - c$

$16000 = a \cdot 3^{2} + b (3) + c$

Simplify by adding terms.

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

$800 = 160 - 2 b - 4 c + c$

$a = 40 - b - c$

$16000 = a \cdot 3^{2} + b (3) + c$

Add $- 4 c$ and $c$ .

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

$16000 = a \cdot 3^{2} + b (3) + c$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

$16000 = a \cdot 3^{2} + b (3) + c$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

$16000 = a \cdot 3^{2} + b (3) + c$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

$16000 = a \cdot 3^{2} + b (3) + c$

Replace all occurrences of $a$ in $16000 = a \cdot 3^{2} + b (3) + c$ with $40 - b - c$ .

$16000 = (40 - b - c) \cdot 3^{2} + b (3) + c$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

Simplify the right side.

Simplify $(40 - b - c) \cdot 3^{2} + b (3) + c$ .

Simplify each term.

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

$16000 = (40 - b - c) \cdot 9 + b (3) + c$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

Apply the distributive property.

$16000 = 40 \cdot 9 - b \cdot 9 - c \cdot 9 + b (3) + c$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

Simplify.

Multiply $40$ by $9$ .

$16000 = 360 - b \cdot 9 - c \cdot 9 + b (3) + c$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

Multiply $9$ by $- 1$ .

$16000 = 360 - 9 b - c \cdot 9 + b (3) + c$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

Multiply $9$ by $- 1$ .

$16000 = 360 - 9 b - 9 c + b (3) + c$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

$16000 = 360 - 9 b - 9 c + b (3) + c$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

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

$16000 = 360 - 9 b - 9 c + 3 b + c$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

$16000 = 360 - 9 b - 9 c + 3 b + c$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

Simplify by adding terms.

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

$16000 = 360 - 6 b - 9 c + c$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

Add $- 9 c$ and $c$ .

$16000 = 360 - 6 b - 8 c$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

$16000 = 360 - 6 b - 8 c$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

$16000 = 360 - 6 b - 8 c$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

$16000 = 360 - 6 b - 8 c$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

$16000 = 360 - 6 b - 8 c$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

Solve for $b$ in $16000 = 360 - 6 b - 8 c$ .

Rewrite the equation as $360 - 6 b - 8 c = 16000$ .

$360 - 6 b - 8 c = 16000$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

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

Subtract $360$ from both sides of the equation.

$- 6 b - 8 c = 16000 - 360$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

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

$- 6 b = 16000 - 360 + 8 c$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

Subtract $360$ from $16000$ .

$- 6 b = 15640 + 8 c$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

$- 6 b = 15640 + 8 c$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

Divide each term in $- 6 b = 15640 + 8 c$ by $- 6$ and simplify.

Divide each term in $- 6 b = 15640 + 8 c$ by $- 6$ .

$\frac{- 6 b}{- 6} = \frac{15640}{- 6} + \frac{8 c}{- 6}$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

Simplify the left side.

Cancel the common factor of $- 6$ .

Cancel the common factor.

$\frac{- 6 b}{- 6} = \frac{15640}{- 6} + \frac{8 c}{- 6}$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

Divide $b$ by $1$ .

$b = \frac{15640}{- 6} + \frac{8 c}{- 6}$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

$b = \frac{15640}{- 6} + \frac{8 c}{- 6}$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

$b = \frac{15640}{- 6} + \frac{8 c}{- 6}$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

Simplify the right side.

Simplify each term.

Cancel the common factor of $15640$ and $- 6$ .

Factor $2$ out of $15640$ .

$b = \frac{2 (7820)}{- 6} + \frac{8 c}{- 6}$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

Cancel the common factors.

Factor $2$ out of $- 6$ .

$b = \frac{2 \cdot 7820}{2 \cdot - 3} + \frac{8 c}{- 6}$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

Cancel the common factor.

$b = \frac{2 \cdot 7820}{2 \cdot - 3} + \frac{8 c}{- 6}$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

Rewrite the expression.

$b = \frac{7820}{- 3} + \frac{8 c}{- 6}$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

$b = \frac{7820}{- 3} + \frac{8 c}{- 6}$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

$b = \frac{7820}{- 3} + \frac{8 c}{- 6}$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

Move the negative in front of the fraction.

$b = - \frac{7820}{3} + \frac{8 c}{- 6}$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

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

Factor $2$ out of $8 c$ .

$b = - \frac{7820}{3} + \frac{2 (4 c)}{- 6}$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

Cancel the common factors.

Factor $2$ out of $- 6$ .

$b = - \frac{7820}{3} + \frac{2 (4 c)}{2 (- 3)}$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

Cancel the common factor.

$b = - \frac{7820}{3} + \frac{2 (4 c)}{2 \cdot - 3}$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

Rewrite the expression.

$b = - \frac{7820}{3} + \frac{4 c}{- 3}$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

$b = - \frac{7820}{3} + \frac{4 c}{- 3}$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

$b = - \frac{7820}{3} + \frac{4 c}{- 3}$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

Move the negative in front of the fraction.

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$800 = 160 - 2 b - 3 c$

$a = 40 - b - c$

Replace all occurrences of $b$ with $- \frac{7820}{3} - \frac{4 c}{3}$ in each equation.

Replace all occurrences of $b$ in $800 = 160 - 2 b - 3 c$ with $- \frac{7820}{3} - \frac{4 c}{3}$ .

$800 = 160 - 2 (- \frac{7820}{3} - \frac{4 c}{3}) - 3 c$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = 40 - b - c$

Simplify the right side.

Simplify $160 - 2 (- \frac{7820}{3} - \frac{4 c}{3}) - 3 c$ .

Simplify each term.

Apply the distributive property.

$800 = 160 - 2 (- \frac{7820}{3}) - 2 (- \frac{4 c}{3}) - 3 c$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = 40 - b - c$

Multiply $- 2 (- \frac{7820}{3})$ .

Multiply $- 1$ by $- 2$ .

$800 = 160 + 2 (\frac{7820}{3}) - 2 (- \frac{4 c}{3}) - 3 c$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = 40 - b - c$

Combine $2$ and $\frac{7820}{3}$ .

$800 = 160 + \frac{2 \cdot 7820}{3} - 2 (- \frac{4 c}{3}) - 3 c$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = 40 - b - c$

Multiply $2$ by $7820$ .

$800 = 160 + \frac{15640}{3} - 2 (- \frac{4 c}{3}) - 3 c$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = 40 - b - c$

$800 = 160 + \frac{15640}{3} - 2 (- \frac{4 c}{3}) - 3 c$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = 40 - b - c$

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

Multiply $- 1$ by $- 2$ .

$800 = 160 + \frac{15640}{3} + 2 (\frac{4 c}{3}) - 3 c$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = 40 - b - c$

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

$800 = 160 + \frac{15640}{3} + \frac{2 (4 c)}{3} - 3 c$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = 40 - b - c$

Multiply $4$ by $2$ .

$800 = 160 + \frac{15640}{3} + \frac{8 c}{3} - 3 c$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = 40 - b - c$

$800 = 160 + \frac{15640}{3} + \frac{8 c}{3} - 3 c$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = 40 - b - c$

$800 = 160 + \frac{15640}{3} + \frac{8 c}{3} - 3 c$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = 40 - b - c$

To write $160$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$800 = 160 \cdot \frac{3}{3} + \frac{15640}{3} + \frac{8 c}{3} - 3 c$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = 40 - b - c$

Combine $160$ and $\frac{3}{3}$ .

$800 = \frac{160 \cdot 3}{3} + \frac{15640}{3} + \frac{8 c}{3} - 3 c$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = 40 - b - c$

Combine the numerators over the common denominator.

$800 = \frac{160 \cdot 3 + 15640}{3} + \frac{8 c}{3} - 3 c$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = 40 - b - c$

Simplify the numerator.

Multiply $160$ by $3$ .

$800 = \frac{480 + 15640}{3} + \frac{8 c}{3} - 3 c$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = 40 - b - c$

Add $480$ and $15640$ .

$800 = \frac{16120}{3} + \frac{8 c}{3} - 3 c$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = 40 - b - c$

$800 = \frac{16120}{3} + \frac{8 c}{3} - 3 c$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = 40 - b - c$

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

$800 = \frac{16120}{3} + \frac{8 c}{3} - 3 c \cdot \frac{3}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = 40 - b - c$

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

$800 = \frac{16120}{3} + \frac{8 c}{3} + \frac{- 3 c \cdot 3}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = 40 - b - c$

Combine the numerators over the common denominator.

$800 = \frac{16120}{3} + \frac{8 c - 3 c \cdot 3}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = 40 - b - c$

Combine the numerators over the common denominator.

$800 = \frac{16120 + 8 c - 3 c \cdot 3}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = 40 - b - c$

Multiply $3$ by $- 3$ .

$800 = \frac{16120 + 8 c - 9 c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = 40 - b - c$

Subtract $9 c$ from $8 c$ .

$800 = \frac{16120 - c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = 40 - b - c$

$800 = \frac{16120 - c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = 40 - b - c$

$800 = \frac{16120 - c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = 40 - b - c$

Replace all occurrences of $b$ in $a = 40 - b - c$ with $- \frac{7820}{3} - \frac{4 c}{3}$ .

$a = 40 - (- \frac{7820}{3} - \frac{4 c}{3}) - c$

$800 = \frac{16120 - c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

Simplify the right side.

Simplify $40 - (- \frac{7820}{3} - \frac{4 c}{3}) - c$ .

Simplify each term.

Apply the distributive property.

$a = 40 + \frac{7820}{3} + \frac{4 c}{3} - c$

$800 = \frac{16120 - c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

Multiply $- - \frac{7820}{3}$ .

Multiply $- 1$ by $- 1$ .

$a = 40 + 1 (\frac{7820}{3}) + \frac{4 c}{3} - c$

$800 = \frac{16120 - c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

Multiply $\frac{7820}{3}$ by $1$ .

$a = 40 + \frac{7820}{3} + \frac{4 c}{3} - c$

$800 = \frac{16120 - c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = 40 + \frac{7820}{3} + \frac{4 c}{3} - c$

$800 = \frac{16120 - c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

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

Multiply $- 1$ by $- 1$ .

$a = 40 + \frac{7820}{3} + 1 (\frac{4 c}{3}) - c$

$800 = \frac{16120 - c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

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

$a = 40 + \frac{7820}{3} + \frac{4 c}{3} - c$

$800 = \frac{16120 - c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = 40 + \frac{7820}{3} + \frac{4 c}{3} - c$

$800 = \frac{16120 - c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = 40 + \frac{7820}{3} + \frac{4 c}{3} - c$

$800 = \frac{16120 - c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

To write $40$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$a = 40 \cdot \frac{3}{3} + \frac{7820}{3} + \frac{4 c}{3} - c$

$800 = \frac{16120 - c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

Combine $40$ and $\frac{3}{3}$ .

$a = \frac{40 \cdot 3}{3} + \frac{7820}{3} + \frac{4 c}{3} - c$

$800 = \frac{16120 - c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

Combine the numerators over the common denominator.

$a = \frac{40 \cdot 3 + 7820}{3} + \frac{4 c}{3} - c$

$800 = \frac{16120 - c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

Simplify the numerator.

Multiply $40$ by $3$ .

$a = \frac{120 + 7820}{3} + \frac{4 c}{3} - c$

$800 = \frac{16120 - c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

Add $120$ and $7820$ .

$a = \frac{7940}{3} + \frac{4 c}{3} - c$

$800 = \frac{16120 - c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = \frac{7940}{3} + \frac{4 c}{3} - c$

$800 = \frac{16120 - c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

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

$a = \frac{7940}{3} + \frac{4 c}{3} - c \cdot \frac{3}{3}$

$800 = \frac{16120 - c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

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

$a = \frac{7940}{3} + \frac{4 c}{3} + \frac{- c \cdot 3}{3}$

$800 = \frac{16120 - c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

Combine the numerators over the common denominator.

$a = \frac{7940}{3} + \frac{4 c - c \cdot 3}{3}$

$800 = \frac{16120 - c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

Combine the numerators over the common denominator.

$a = \frac{7940 + 4 c - c \cdot 3}{3}$

$800 = \frac{16120 - c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

Multiply $3$ by $- 1$ .

$a = \frac{7940 + 4 c - 3 c}{3}$

$800 = \frac{16120 - c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

Subtract $3 c$ from $4 c$ .

$a = \frac{7940 + c}{3}$

$800 = \frac{16120 - c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = \frac{7940 + c}{3}$

$800 = \frac{16120 - c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = \frac{7940 + c}{3}$

$800 = \frac{16120 - c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = \frac{7940 + c}{3}$

$800 = \frac{16120 - c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

Solve for $c$ in $800 = \frac{16120 - c}{3}$ .

Rewrite the equation as $\frac{16120 - c}{3} = 800$ .

$\frac{16120 - c}{3} = 800$

$a = \frac{7940 + c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

Multiply both sides by $3$ .

$\frac{16120 - c}{3} \cdot 3 = 800 \cdot 3$

$a = \frac{7940 + c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

Simplify.

Simplify the left side.

Simplify $\frac{16120 - c}{3} \cdot 3$ .

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{16120 - c}{3} \cdot 3 = 800 \cdot 3$

$a = \frac{7940 + c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

Rewrite the expression.

$16120 - c = 800 \cdot 3$

$a = \frac{7940 + c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$16120 - c = 800 \cdot 3$

$a = \frac{7940 + c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

Reorder $16120$ and $- c$ .

$- c + 16120 = 800 \cdot 3$

$a = \frac{7940 + c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$- c + 16120 = 800 \cdot 3$

$a = \frac{7940 + c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$- c + 16120 = 800 \cdot 3$

$a = \frac{7940 + c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

Simplify the right side.

Multiply $800$ by $3$ .

$- c + 16120 = 2400$

$a = \frac{7940 + c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$- c + 16120 = 2400$

$a = \frac{7940 + c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$- c + 16120 = 2400$

$a = \frac{7940 + c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

Solve for $c$ .

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

Subtract $16120$ from both sides of the equation.

$- c = 2400 - 16120$

$a = \frac{7940 + c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

Subtract $16120$ from $2400$ .

$- c = - 13720$

$a = \frac{7940 + c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$- c = - 13720$

$a = \frac{7940 + c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

Divide each term in $- c = - 13720$ by $- 1$ and simplify.

Divide each term in $- c = - 13720$ by $- 1$ .

$\frac{- c}{- 1} = \frac{- 13720}{- 1}$

$a = \frac{7940 + c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{c}{1} = \frac{- 13720}{- 1}$

$a = \frac{7940 + c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

Divide $c$ by $1$ .

$c = \frac{- 13720}{- 1}$

$a = \frac{7940 + c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$c = \frac{- 13720}{- 1}$

$a = \frac{7940 + c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

Simplify the right side.

Divide $- 13720$ by $- 1$ .

$c = 13720$

$a = \frac{7940 + c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$c = 13720$

$a = \frac{7940 + c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$c = 13720$

$a = \frac{7940 + c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$c = 13720$

$a = \frac{7940 + c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$c = 13720$

$a = \frac{7940 + c}{3}$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

Replace all occurrences of $c$ with $13720$ in each equation.

Replace all occurrences of $c$ in $a = \frac{7940 + c}{3}$ with $13720$ .

$a = \frac{7940 + 13720}{3}$

$c = 13720$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

Simplify $a = \frac{7940 + 13720}{3}$ .

Simplify the left side.

Remove parentheses.

$a = \frac{7940 + 13720}{3}$

$c = 13720$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = \frac{7940 + 13720}{3}$

$c = 13720$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

Simplify the right side.

Simplify $\frac{7940 + 13720}{3}$ .

Add $7940$ and $13720$ .

$a = \frac{21660}{3}$

$c = 13720$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

Divide $21660$ by $3$ .

$a = 7220$

$c = 13720$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = 7220$

$c = 13720$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = 7220$

$c = 13720$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

$a = 7220$

$c = 13720$

$b = - \frac{7820}{3} - \frac{4 c}{3}$

Replace all occurrences of $c$ in $b = - \frac{7820}{3} - \frac{4 c}{3}$ with $13720$ .

$b = - \frac{7820}{3} - \frac{4 (13720)}{3}$

$a = 7220$

$c = 13720$

Simplify the right side.

Simplify $- \frac{7820}{3} - \frac{4 (13720)}{3}$ .

Combine the numerators over the common denominator.

$b = \frac{- 7820 - 4 \cdot 13720}{3}$

$a = 7220$

$c = 13720$

Simplify the expression.

Multiply $- 4$ by $13720$ .

$b = \frac{- 7820 - 54880}{3}$

$a = 7220$

$c = 13720$

Subtract $54880$ from $- 7820$ .

$b = \frac{- 62700}{3}$

$a = 7220$

$c = 13720$

Divide $- 62700$ by $3$ .

$b = - 20900$

$a = 7220$

$c = 13720$

$b = - 20900$

$a = 7220$

$c = 13720$

$b = - 20900$

$a = 7220$

$c = 13720$

$b = - 20900$

$a = 7220$

$c = 13720$

$b = - 20900$

$a = 7220$

$c = 13720$

List all of the solutions.

$b = - 20900, a = 7220, c = 13720$

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 $a = 7220$ , $b = - 20900$ , $c = 13720$ , and $x = 1$ .

Simplify each term.

One to any power is one.

$y = 7220 \cdot 1 + (- 20900) \cdot (1) + 13720$

Multiply $7220$ by $1$ .

$y = 7220 + (- 20900) \cdot (1) + 13720$

Multiply $- 20900$ by $1$ .

$y = 7220 - 20900 + 13720$

Simplify by adding and subtracting.

Subtract $20900$ from $7220$ .

$y = - 13680 + 13720$

Add $- 13680$ and $13720$ .

$y = 40$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 1$ . This check passes since $y = 40$ and $y = 40$ .

$40 = 40$

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = 7220$ , $b = - 20900$ , $c = 13720$ , and $x = 2$ .

Simplify each term.

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

$y = 7220 \cdot 4 + (- 20900) \cdot (2) + 13720$

Multiply $7220$ by $4$ .

$y = 28880 + (- 20900) \cdot (2) + 13720$

Multiply $- 20900$ by $2$ .

$y = 28880 - 41800 + 13720$

Simplify by adding and subtracting.

Subtract $41800$ from $28880$ .

$y = - 12920 + 13720$

Add $- 12920$ and $13720$ .

$y = 800$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 2$ . This check passes since $y = 800$ and $y = 800$ .

$800 = 800$

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = 7220$ , $b = - 20900$ , $c = 13720$ , and $x = 3$ .

Simplify each term.

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

$y = 7220 \cdot 9 + (- 20900) \cdot (3) + 13720$

Multiply $7220$ by $9$ .

$y = 64980 + (- 20900) \cdot (3) + 13720$

Multiply $- 20900$ by $3$ .

$y = 64980 - 62700 + 13720$

Simplify by adding and subtracting.

Subtract $62700$ from $64980$ .

$y = 2280 + 13720$

Add $2280$ and $13720$ .

$y = 16000$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 3$ . This check passes since $y = 16000$ and $y = 16000$ .

$16000 = 16000$

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 $y = 7220 x^{2} - 20900 x + 13720$ .

$y = 7220 x^{2} - 20900 x + 13720$

Find the Function Rule table[[x,y],[1,40],[2,800],[3,16000]]

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