Find the Function Rule table[[x,y],[3,2500],[5,3000],[7,3700]]

$\begin{array}{c} x & y \\ 3 & 2500 \\ 5 & 3000 \\ 7 & 3700 \end{array}$

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 $y = a x + b$ .

$2500 = a (3) + b 3000 = a (5) + b 3700 = a (7) + b$

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

Solve for $b$ in $2500 = a (3) + b$ .

Rewrite the equation as $a (3) + b = 2500$ .

$a (3) + b = 2500$

$3000 = a (5) + b$

$3700 = a (7) + b$

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

$3 a + b = 2500$

$3000 = a (5) + b$

$3700 = a (7) + b$

Subtract $3 a$ from both sides of the equation.

$b = 2500 - 3 a$

$3000 = a (5) + b$

$3700 = a (7) + b$

$b = 2500 - 3 a$

$3000 = a (5) + b$

$3700 = a (7) + b$

Replace all occurrences of $b$ with $2500 - 3 a$ in each equation.

Replace all occurrences of $b$ in $3000 = a (5) + b$ with $2500 - 3 a$ .

$3000 = a (5) + 2500 - 3 a$

$b = 2500 - 3 a$

$3700 = a (7) + b$

Simplify $3000 = a (5) + 2500 - 3 a$ .

Simplify the left side.

Remove parentheses.

$3000 = a (5) + 2500 - 3 a$

$b = 2500 - 3 a$

$3700 = a (7) + b$

$3000 = a (5) + 2500 - 3 a$

$b = 2500 - 3 a$

$3700 = a (7) + b$

Simplify the right side.

Simplify $a (5) + 2500 - 3 a$ .

Move $5$ to the left of $a$ .

$3000 = 5 a + 2500 - 3 a$

$b = 2500 - 3 a$

$3700 = a (7) + b$

Subtract $3 a$ from $5 a$ .

$3000 = 2 a + 2500$

$b = 2500 - 3 a$

$3700 = a (7) + b$

$3000 = 2 a + 2500$

$b = 2500 - 3 a$

$3700 = a (7) + b$

$3000 = 2 a + 2500$

$b = 2500 - 3 a$

$3700 = a (7) + b$

$3000 = 2 a + 2500$

$b = 2500 - 3 a$

$3700 = a (7) + b$

Replace all occurrences of $b$ in $3700 = a (7) + b$ with $2500 - 3 a$ .

$3700 = a (7) + 2500 - 3 a$

$3000 = 2 a + 2500$

$b = 2500 - 3 a$

Simplify $3700 = a (7) + 2500 - 3 a$ .

Simplify the left side.

Remove parentheses.

$3700 = a (7) + 2500 - 3 a$

$3000 = 2 a + 2500$

$b = 2500 - 3 a$

$3700 = a (7) + 2500 - 3 a$

$3000 = 2 a + 2500$

$b = 2500 - 3 a$

Simplify the right side.

Simplify $a (7) + 2500 - 3 a$ .

Move $7$ to the left of $a$ .

$3700 = 7 a + 2500 - 3 a$

$3000 = 2 a + 2500$

$b = 2500 - 3 a$

Subtract $3 a$ from $7 a$ .

$3700 = 4 a + 2500$

$3000 = 2 a + 2500$

$b = 2500 - 3 a$

$3700 = 4 a + 2500$

$3000 = 2 a + 2500$

$b = 2500 - 3 a$

$3700 = 4 a + 2500$

$3000 = 2 a + 2500$

$b = 2500 - 3 a$

$3700 = 4 a + 2500$

$3000 = 2 a + 2500$

$b = 2500 - 3 a$

$3700 = 4 a + 2500$

$3000 = 2 a + 2500$

$b = 2500 - 3 a$

Solve for $a$ in $3700 = 4 a + 2500$ .

Rewrite the equation as $4 a + 2500 = 3700$ .

$4 a + 2500 = 3700$

$3000 = 2 a + 2500$

$b = 2500 - 3 a$

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

Subtract $2500$ from both sides of the equation.

$4 a = 3700 - 2500$

$3000 = 2 a + 2500$

$b = 2500 - 3 a$

Subtract $2500$ from $3700$ .

$4 a = 1200$

$3000 = 2 a + 2500$

$b = 2500 - 3 a$

$4 a = 1200$

$3000 = 2 a + 2500$

$b = 2500 - 3 a$

Divide each term in $4 a = 1200$ by $4$ and simplify.

Divide each term in $4 a = 1200$ by $4$ .

$\frac{4 a}{4} = \frac{1200}{4}$

$3000 = 2 a + 2500$

$b = 2500 - 3 a$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 a}{4} = \frac{1200}{4}$

$3000 = 2 a + 2500$

$b = 2500 - 3 a$

Divide $a$ by $1$ .

$a = \frac{1200}{4}$

$3000 = 2 a + 2500$

$b = 2500 - 3 a$

$a = \frac{1200}{4}$

$3000 = 2 a + 2500$

$b = 2500 - 3 a$

$a = \frac{1200}{4}$

$3000 = 2 a + 2500$

$b = 2500 - 3 a$

Simplify the right side.

Divide $1200$ by $4$ .

$a = 300$

$3000 = 2 a + 2500$

$b = 2500 - 3 a$

$a = 300$

$3000 = 2 a + 2500$

$b = 2500 - 3 a$

$a = 300$

$3000 = 2 a + 2500$

$b = 2500 - 3 a$

$a = 300$

$3000 = 2 a + 2500$

$b = 2500 - 3 a$

Replace all occurrences of $a$ with $300$ in each equation.

Replace all occurrences of $a$ in $3000 = 2 a + 2500$ with $300$ .

$3000 = 2 (300) + 2500$

$a = 300$

$b = 2500 - 3 a$

Simplify the right side.

Simplify $2 (300) + 2500$ .

Multiply $2$ by $300$ .

$3000 = 600 + 2500$

$a = 300$

$b = 2500 - 3 a$

Add $600$ and $2500$ .

$3000 = 3100$

$a = 300$

$b = 2500 - 3 a$

$3000 = 3100$

$a = 300$

$b = 2500 - 3 a$

$3000 = 3100$

$a = 300$

$b = 2500 - 3 a$

Replace all occurrences of $a$ in $b = 2500 - 3 a$ with $300$ .

$b = 2500 - 3 \cdot 300$

$3000 = 3100$

$a = 300$

Simplify the right side.

Simplify $2500 - 3 \cdot 300$ .

Multiply $- 3$ by $300$ .

$b = 2500 - 900$

$3000 = 3100$

$a = 300$

Subtract $900$ from $2500$ .

$b = 1600$

$3000 = 3100$

$a = 300$

$b = 1600$

$3000 = 3100$

$a = 300$

$b = 1600$

$3000 = 3100$

$a = 300$

$b = 1600$

$3000 = 3100$

$a = 300$

Since $3000 = 3100$ 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 $c$ in $2500 = a \cdot 3^{2} + b (3) + c$ .

Rewrite the equation as $a \cdot 3^{2} + b (3) + c = 2500$ .

$a \cdot 3^{2} + b (3) + c = 2500$

$3000 = a \cdot 5^{2} + b (5) + c$

$3700 = a \cdot 7^{2} + b (7) + c$

Simplify each term.

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

$a \cdot 9 + b (3) + c = 2500$

$3000 = a \cdot 5^{2} + b (5) + c$

$3700 = a \cdot 7^{2} + b (7) + c$

Move $9$ to the left of $a$ .

$9 \cdot a + b (3) + c = 2500$

$3000 = a \cdot 5^{2} + b (5) + c$

$3700 = a \cdot 7^{2} + b (7) + c$

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

$9 a + 3 b + c = 2500$

$3000 = a \cdot 5^{2} + b (5) + c$

$3700 = a \cdot 7^{2} + b (7) + c$

$9 a + 3 b + c = 2500$

$3000 = a \cdot 5^{2} + b (5) + c$

$3700 = a \cdot 7^{2} + b (7) + c$

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

Subtract $9 a$ from both sides of the equation.

$3 b + c = 2500 - 9 a$

$3000 = a \cdot 5^{2} + b (5) + c$

$3700 = a \cdot 7^{2} + b (7) + c$

Subtract $3 b$ from both sides of the equation.

$c = 2500 - 9 a - 3 b$

$3000 = a \cdot 5^{2} + b (5) + c$

$3700 = a \cdot 7^{2} + b (7) + c$

$c = 2500 - 9 a - 3 b$

$3000 = a \cdot 5^{2} + b (5) + c$

$3700 = a \cdot 7^{2} + b (7) + c$

$c = 2500 - 9 a - 3 b$

$3000 = a \cdot 5^{2} + b (5) + c$

$3700 = a \cdot 7^{2} + b (7) + c$

Replace all occurrences of $c$ with $2500 - 9 a - 3 b$ in each equation.

Replace all occurrences of $c$ in $3000 = a \cdot 5^{2} + b (5) + c$ with $2500 - 9 a - 3 b$ .

$3000 = a \cdot 5^{2} + b (5) + 2500 - 9 a - 3 b$

$c = 2500 - 9 a - 3 b$

$3700 = a \cdot 7^{2} + b (7) + c$

Simplify $3000 = a \cdot 5^{2} + b (5) + 2500 - 9 a - 3 b$ .

Simplify the left side.

Remove parentheses.

$3000 = a \cdot 5^{2} + b (5) + 2500 - 9 a - 3 b$

$c = 2500 - 9 a - 3 b$

$3700 = a \cdot 7^{2} + b (7) + c$

$3000 = a \cdot 5^{2} + b (5) + 2500 - 9 a - 3 b$

$c = 2500 - 9 a - 3 b$

$3700 = a \cdot 7^{2} + b (7) + c$

Simplify the right side.

Simplify $a \cdot 5^{2} + b (5) + 2500 - 9 a - 3 b$ .

Simplify each term.

Raise $5$ to the power of $2$ .

$3000 = a \cdot 25 + b (5) + 2500 - 9 a - 3 b$

$c = 2500 - 9 a - 3 b$

$3700 = a \cdot 7^{2} + b (7) + c$

Move $25$ to the left of $a$ .

$3000 = 25 \cdot a + b (5) + 2500 - 9 a - 3 b$

$c = 2500 - 9 a - 3 b$

$3700 = a \cdot 7^{2} + b (7) + c$

Move $5$ to the left of $b$ .

$3000 = 25 a + 5 b + 2500 - 9 a - 3 b$

$c = 2500 - 9 a - 3 b$

$3700 = a \cdot 7^{2} + b (7) + c$

$3000 = 25 a + 5 b + 2500 - 9 a - 3 b$

$c = 2500 - 9 a - 3 b$

$3700 = a \cdot 7^{2} + b (7) + c$

Simplify by adding terms.

Subtract $9 a$ from $25 a$ .

$3000 = 16 a + 5 b + 2500 - 3 b$

$c = 2500 - 9 a - 3 b$

$3700 = a \cdot 7^{2} + b (7) + c$

Subtract $3 b$ from $5 b$ .

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

$3700 = a \cdot 7^{2} + b (7) + c$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

$3700 = a \cdot 7^{2} + b (7) + c$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

$3700 = a \cdot 7^{2} + b (7) + c$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

$3700 = a \cdot 7^{2} + b (7) + c$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

$3700 = a \cdot 7^{2} + b (7) + c$

Replace all occurrences of $c$ in $3700 = a \cdot 7^{2} + b (7) + c$ with $2500 - 9 a - 3 b$ .

$3700 = a \cdot 7^{2} + b (7) + 2500 - 9 a - 3 b$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

Simplify $3700 = a \cdot 7^{2} + b (7) + 2500 - 9 a - 3 b$ .

Simplify the left side.

Remove parentheses.

$3700 = a \cdot 7^{2} + b (7) + 2500 - 9 a - 3 b$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

$3700 = a \cdot 7^{2} + b (7) + 2500 - 9 a - 3 b$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

Simplify the right side.

Simplify $a \cdot 7^{2} + b (7) + 2500 - 9 a - 3 b$ .

Simplify each term.

Raise $7$ to the power of $2$ .

$3700 = a \cdot 49 + b (7) + 2500 - 9 a - 3 b$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

Move $49$ to the left of $a$ .

$3700 = 49 \cdot a + b (7) + 2500 - 9 a - 3 b$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

Move $7$ to the left of $b$ .

$3700 = 49 a + 7 b + 2500 - 9 a - 3 b$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

$3700 = 49 a + 7 b + 2500 - 9 a - 3 b$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

Simplify by adding terms.

Subtract $9 a$ from $49 a$ .

$3700 = 40 a + 7 b + 2500 - 3 b$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

Subtract $3 b$ from $7 b$ .

$3700 = 40 a + 4 b + 2500$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

$3700 = 40 a + 4 b + 2500$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

$3700 = 40 a + 4 b + 2500$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

$3700 = 40 a + 4 b + 2500$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

$3700 = 40 a + 4 b + 2500$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

$3700 = 40 a + 4 b + 2500$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

Solve for $a$ in $3700 = 40 a + 4 b + 2500$ .

Rewrite the equation as $40 a + 4 b + 2500 = 3700$ .

$40 a + 4 b + 2500 = 3700$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

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

Subtract $4 b$ from both sides of the equation.

$40 a + 2500 = 3700 - 4 b$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

Subtract $2500$ from both sides of the equation.

$40 a = 3700 - 4 b - 2500$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

Subtract $2500$ from $3700$ .

$40 a = - 4 b + 1200$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

$40 a = - 4 b + 1200$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

Divide each term in $40 a = - 4 b + 1200$ by $40$ and simplify.

Divide each term in $40 a = - 4 b + 1200$ by $40$ .

$\frac{40 a}{40} = \frac{- 4 b}{40} + \frac{1200}{40}$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

Simplify the left side.

Cancel the common factor of $40$ .

Cancel the common factor.

$\frac{40 a}{40} = \frac{- 4 b}{40} + \frac{1200}{40}$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

Divide $a$ by $1$ .

$a = \frac{- 4 b}{40} + \frac{1200}{40}$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

$a = \frac{- 4 b}{40} + \frac{1200}{40}$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

$a = \frac{- 4 b}{40} + \frac{1200}{40}$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

Simplify the right side.

Simplify each term.

Cancel the common factor of $- 4$ and $40$ .

Factor $4$ out of $- 4 b$ .

$a = \frac{4 (- b)}{40} + \frac{1200}{40}$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

Cancel the common factors.

Factor $4$ out of $40$ .

$a = \frac{4 (- b)}{4 (10)} + \frac{1200}{40}$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

Cancel the common factor.

$a = \frac{4 (- b)}{4 \cdot 10} + \frac{1200}{40}$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

Rewrite the expression.

$a = \frac{- b}{10} + \frac{1200}{40}$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

$a = \frac{- b}{10} + \frac{1200}{40}$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

$a = \frac{- b}{10} + \frac{1200}{40}$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

Move the negative in front of the fraction.

$a = - \frac{b}{10} + \frac{1200}{40}$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

Divide $1200$ by $40$ .

$a = - \frac{b}{10} + 30$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

$a = - \frac{b}{10} + 30$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

$a = - \frac{b}{10} + 30$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

$a = - \frac{b}{10} + 30$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

$a = - \frac{b}{10} + 30$

$3000 = 16 a + 2 b + 2500$

$c = 2500 - 9 a - 3 b$

Replace all occurrences of $a$ with $- \frac{b}{10} + 30$ in each equation.

Replace all occurrences of $a$ in $3000 = 16 a + 2 b + 2500$ with $- \frac{b}{10} + 30$ .

$3000 = 16 (- \frac{b}{10} + 30) + 2 b + 2500$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

Simplify the right side.

Simplify $16 (- \frac{b}{10} + 30) + 2 b + 2500$ .

Simplify each term.

Apply the distributive property.

$3000 = 16 (- \frac{b}{10}) + 16 \cdot 30 + 2 b + 2500$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{b}{10}$ into the numerator.

$3000 = 16 (\frac{- b}{10}) + 16 \cdot 30 + 2 b + 2500$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

Factor $2$ out of $16$ .

$3000 = 2 (8) (\frac{- b}{10}) + 16 \cdot 30 + 2 b + 2500$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

Factor $2$ out of $10$ .

$3000 = 2 \cdot (8 (\frac{- b}{2 \cdot 5})) + 16 \cdot 30 + 2 b + 2500$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

Cancel the common factor.

$3000 = 2 \cdot (8 (\frac{- b}{2 \cdot 5})) + 16 \cdot 30 + 2 b + 2500$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

Rewrite the expression.

$3000 = 8 (\frac{- b}{5}) + 16 \cdot 30 + 2 b + 2500$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

$3000 = 8 (\frac{- b}{5}) + 16 \cdot 30 + 2 b + 2500$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

Combine $8$ and $\frac{- b}{5}$ .

$3000 = \frac{8 (- b)}{5} + 16 \cdot 30 + 2 b + 2500$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

Multiply $- 1$ by $8$ .

$3000 = \frac{- 8 b}{5} + 16 \cdot 30 + 2 b + 2500$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

Multiply $16$ by $30$ .

$3000 = \frac{- 8 b}{5} + 480 + 2 b + 2500$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

Move the negative in front of the fraction.

$3000 = - \frac{8 b}{5} + 480 + 2 b + 2500$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

$3000 = - \frac{8 b}{5} + 480 + 2 b + 2500$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

To write $2 b$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$3000 = - \frac{8 b}{5} + 2 b \cdot \frac{5}{5} + 480 + 2500$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

Combine $2 b$ and $\frac{5}{5}$ .

$3000 = - \frac{8 b}{5} + \frac{2 b \cdot 5}{5} + 480 + 2500$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

Combine the numerators over the common denominator.

$3000 = \frac{- 8 b + 2 b \cdot 5}{5} + 480 + 2500$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

Find the common denominator.

Write $480$ as a fraction with denominator $1$ .

$3000 = \frac{- 8 b + 2 b \cdot 5}{5} + \frac{480}{1} + 2500$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

Multiply $\frac{480}{1}$ by $\frac{5}{5}$ .

$3000 = \frac{- 8 b + 2 b \cdot 5}{5} + \frac{480}{1} \cdot \frac{5}{5} + 2500$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

Multiply $\frac{480}{1}$ by $\frac{5}{5}$ .

$3000 = \frac{- 8 b + 2 b \cdot 5}{5} + \frac{480 \cdot 5}{5} + 2500$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

Write $2500$ as a fraction with denominator $1$ .

$3000 = \frac{- 8 b + 2 b \cdot 5}{5} + \frac{480 \cdot 5}{5} + \frac{2500}{1}$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

Multiply $\frac{2500}{1}$ by $\frac{5}{5}$ .

$3000 = \frac{- 8 b + 2 b \cdot 5}{5} + \frac{480 \cdot 5}{5} + \frac{2500}{1} \cdot \frac{5}{5}$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

Multiply $\frac{2500}{1}$ by $\frac{5}{5}$ .

$3000 = \frac{- 8 b + 2 b \cdot 5}{5} + \frac{480 \cdot 5}{5} + \frac{2500 \cdot 5}{5}$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

$3000 = \frac{- 8 b + 2 b \cdot 5}{5} + \frac{480 \cdot 5}{5} + \frac{2500 \cdot 5}{5}$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

Combine the numerators over the common denominator.

$3000 = \frac{- 8 b + 2 b \cdot 5 + 480 \cdot 5 + 2500 \cdot 5}{5}$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

Simplify each term.

Multiply $5$ by $2$ .

$3000 = \frac{- 8 b + 10 b + 480 \cdot 5 + 2500 \cdot 5}{5}$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

Multiply $480$ by $5$ .

$3000 = \frac{- 8 b + 10 b + 2400 + 2500 \cdot 5}{5}$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

Multiply $2500$ by $5$ .

$3000 = \frac{- 8 b + 10 b + 2400 + 12500}{5}$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

$3000 = \frac{- 8 b + 10 b + 2400 + 12500}{5}$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

Simplify terms.

Add $- 8 b$ and $10 b$ .

$3000 = \frac{2 b + 2400 + 12500}{5}$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

Add $2400$ and $12500$ .

$3000 = \frac{2 b + 14900}{5}$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

Factor $2$ out of $2 b + 14900$ .

Factor $2$ out of $2 b$ .

$3000 = \frac{2 (b) + 14900}{5}$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

Factor $2$ out of $14900$ .

$3000 = \frac{2 b + 2 \cdot 7450}{5}$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

Factor $2$ out of $2 b + 2 \cdot 7450$ .

$3000 = \frac{2 (b + 7450)}{5}$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

$3000 = \frac{2 (b + 7450)}{5}$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

$3000 = \frac{2 (b + 7450)}{5}$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

$3000 = \frac{2 (b + 7450)}{5}$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

$3000 = \frac{2 (b + 7450)}{5}$

$a = - \frac{b}{10} + 30$

$c = 2500 - 9 a - 3 b$

Replace all occurrences of $a$ in $c = 2500 - 9 a - 3 b$ with $- \frac{b}{10} + 30$ .

$c = 2500 - 9 (- \frac{b}{10} + 30) - 3 b$

$3000 = \frac{2 (b + 7450)}{5}$

$a = - \frac{b}{10} + 30$

Simplify the right side.

Simplify $2500 - 9 (- \frac{b}{10} + 30) - 3 b$ .

Simplify each term.

Apply the distributive property.

$c = 2500 - 9 (- \frac{b}{10}) - 9 \cdot 30 - 3 b$

$3000 = \frac{2 (b + 7450)}{5}$

$a = - \frac{b}{10} + 30$

Multiply $- 9 (- \frac{b}{10})$ .

Multiply $- 1$ by $- 9$ .

$c = 2500 + 9 (\frac{b}{10}) - 9 \cdot 30 - 3 b$

$3000 = \frac{2 (b + 7450)}{5}$

$a = - \frac{b}{10} + 30$

Combine $9$ and $\frac{b}{10}$ .

$c = 2500 + \frac{9 b}{10} - 9 \cdot 30 - 3 b$

$3000 = \frac{2 (b + 7450)}{5}$

$a = - \frac{b}{10} + 30$

$c = 2500 + \frac{9 b}{10} - 9 \cdot 30 - 3 b$

$3000 = \frac{2 (b + 7450)}{5}$

$a = - \frac{b}{10} + 30$

Multiply $- 9$ by $30$ .

$c = 2500 + \frac{9 b}{10} - 270 - 3 b$

$3000 = \frac{2 (b + 7450)}{5}$

$a = - \frac{b}{10} + 30$

$c = 2500 + \frac{9 b}{10} - 270 - 3 b$

$3000 = \frac{2 (b + 7450)}{5}$

$a = - \frac{b}{10} + 30$

Subtract $270$ from $2500$ .

$c = \frac{9 b}{10} + 2230 - 3 b$

$3000 = \frac{2 (b + 7450)}{5}$

$a = - \frac{b}{10} + 30$

To write $- 3 b$ as a fraction with a common denominator, multiply by $\frac{10}{10}$ .

$c = \frac{9 b}{10} - 3 b \cdot \frac{10}{10} + 2230$

$3000 = \frac{2 (b + 7450)}{5}$

$a = - \frac{b}{10} + 30$

Simplify terms.

Combine $- 3 b$ and $\frac{10}{10}$ .

$c = \frac{9 b}{10} + \frac{- 3 b \cdot 10}{10} + 2230$

$3000 = \frac{2 (b + 7450)}{5}$

$a = - \frac{b}{10} + 30$

Combine the numerators over the common denominator.

$c = \frac{9 b - 3 b \cdot 10}{10} + 2230$

$3000 = \frac{2 (b + 7450)}{5}$

$a = - \frac{b}{10} + 30$

$c = \frac{9 b - 3 b \cdot 10}{10} + 2230$

$3000 = \frac{2 (b + 7450)}{5}$

$a = - \frac{b}{10} + 30$

Simplify each term.

Simplify the numerator.

Factor $3 b$ out of $9 b - 3 b \cdot 10$ .

Factor $3 b$ out of $9 b$ .

$c = \frac{3 b (3) - 3 b \cdot 10}{10} + 2230$

$3000 = \frac{2 (b + 7450)}{5}$

$a = - \frac{b}{10} + 30$

Factor $3 b$ out of $- 3 b \cdot 10$ .

$c = \frac{3 b (3) + 3 b (- 1 \cdot 10)}{10} + 2230$

$3000 = \frac{2 (b + 7450)}{5}$

$a = - \frac{b}{10} + 30$

Factor $3 b$ out of $3 b (3) + 3 b (- 1 \cdot 10)$ .

$c = \frac{3 b (3 - 1 \cdot 10)}{10} + 2230$

$3000 = \frac{2 (b + 7450)}{5}$

$a = - \frac{b}{10} + 30$

$c = \frac{3 b (3 - 1 \cdot 10)}{10} + 2230$

$3000 = \frac{2 (b + 7450)}{5}$

$a = - \frac{b}{10} + 30$

Multiply $- 1$ by $10$ .

$c = \frac{3 b (3 - 10)}{10} + 2230$

$3000 = \frac{2 (b + 7450)}{5}$

$a = - \frac{b}{10} + 30$

Subtract $10$ from $3$ .

$c = \frac{3 b \cdot - 7}{10} + 2230$

$3000 = \frac{2 (b + 7450)}{5}$

$a = - \frac{b}{10} + 30$

Multiply $- 7$ by $3$ .

$c = \frac{- 21 b}{10} + 2230$

$3000 = \frac{2 (b + 7450)}{5}$

$a = - \frac{b}{10} + 30$

$c = \frac{- 21 b}{10} + 2230$

$3000 = \frac{2 (b + 7450)}{5}$

$a = - \frac{b}{10} + 30$

Move the negative in front of the fraction.

$c = - \frac{21 b}{10} + 2230$

$3000 = \frac{2 (b + 7450)}{5}$

$a = - \frac{b}{10} + 30$

$c = - \frac{21 b}{10} + 2230$

$3000 = \frac{2 (b + 7450)}{5}$

$a = - \frac{b}{10} + 30$

$c = - \frac{21 b}{10} + 2230$

$3000 = \frac{2 (b + 7450)}{5}$

$a = - \frac{b}{10} + 30$

$c = - \frac{21 b}{10} + 2230$

$3000 = \frac{2 (b + 7450)}{5}$

$a = - \frac{b}{10} + 30$

$c = - \frac{21 b}{10} + 2230$

$3000 = \frac{2 (b + 7450)}{5}$

$a = - \frac{b}{10} + 30$

Solve for $b$ in $3000 = \frac{2 (b + 7450)}{5}$ .

Rewrite the equation as $\frac{2 (b + 7450)}{5} = 3000$ .

$\frac{2 (b + 7450)}{5} = 3000$

$c = - \frac{21 b}{10} + 2230$

$a = - \frac{b}{10} + 30$

Multiply both sides of the equation by $\frac{5}{2}$ .

$\frac{5}{2} \cdot \frac{2 (b + 7450)}{5} = \frac{5}{2} \cdot 3000$

$c = - \frac{21 b}{10} + 2230$

$a = - \frac{b}{10} + 30$

Simplify both sides of the equation.

Simplify the left side.

Simplify $\frac{5}{2} \cdot \frac{2 (b + 7450)}{5}$ .

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5}{2} \cdot \frac{2 (b + 7450)}{5} = \frac{5}{2} \cdot 3000$

$c = - \frac{21 b}{10} + 2230$

$a = - \frac{b}{10} + 30$

Rewrite the expression.

$\frac{1}{2} \cdot (2 (b + 7450)) = \frac{5}{2} \cdot 3000$

$c = - \frac{21 b}{10} + 2230$

$a = - \frac{b}{10} + 30$

$\frac{1}{2} \cdot (2 (b + 7450)) = \frac{5}{2} \cdot 3000$

$c = - \frac{21 b}{10} + 2230$

$a = - \frac{b}{10} + 30$

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{1}{2} \cdot (2 (b + 7450)) = \frac{5}{2} \cdot 3000$

$c = - \frac{21 b}{10} + 2230$

$a = - \frac{b}{10} + 30$

Rewrite the expression.

$b + 7450 = \frac{5}{2} \cdot 3000$

$c = - \frac{21 b}{10} + 2230$

$a = - \frac{b}{10} + 30$

$b + 7450 = \frac{5}{2} \cdot 3000$

$c = - \frac{21 b}{10} + 2230$

$a = - \frac{b}{10} + 30$

$b + 7450 = \frac{5}{2} \cdot 3000$

$c = - \frac{21 b}{10} + 2230$

$a = - \frac{b}{10} + 30$

$b + 7450 = \frac{5}{2} \cdot 3000$

$c = - \frac{21 b}{10} + 2230$

$a = - \frac{b}{10} + 30$

Simplify the right side.

Simplify $\frac{5}{2} \cdot 3000$ .

Cancel the common factor of $2$ .

Factor $2$ out of $3000$ .

$b + 7450 = \frac{5}{2} \cdot (2 (1500))$

$c = - \frac{21 b}{10} + 2230$

$a = - \frac{b}{10} + 30$

Cancel the common factor.

$b + 7450 = \frac{5}{2} \cdot (2 \cdot 1500)$

$c = - \frac{21 b}{10} + 2230$

$a = - \frac{b}{10} + 30$

Rewrite the expression.

$b + 7450 = 5 \cdot 1500$

$c = - \frac{21 b}{10} + 2230$

$a = - \frac{b}{10} + 30$

$b + 7450 = 5 \cdot 1500$

$c = - \frac{21 b}{10} + 2230$

$a = - \frac{b}{10} + 30$

Multiply $5$ by $1500$ .

$b + 7450 = 7500$

$c = - \frac{21 b}{10} + 2230$

$a = - \frac{b}{10} + 30$

$b + 7450 = 7500$

$c = - \frac{21 b}{10} + 2230$

$a = - \frac{b}{10} + 30$

$b + 7450 = 7500$

$c = - \frac{21 b}{10} + 2230$

$a = - \frac{b}{10} + 30$

$b + 7450 = 7500$

$c = - \frac{21 b}{10} + 2230$

$a = - \frac{b}{10} + 30$

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

Subtract $7450$ from both sides of the equation.

$b = 7500 - 7450$

$c = - \frac{21 b}{10} + 2230$

$a = - \frac{b}{10} + 30$

Subtract $7450$ from $7500$ .

$b = 50$

$c = - \frac{21 b}{10} + 2230$

$a = - \frac{b}{10} + 30$

$b = 50$

$c = - \frac{21 b}{10} + 2230$

$a = - \frac{b}{10} + 30$

$b = 50$

$c = - \frac{21 b}{10} + 2230$

$a = - \frac{b}{10} + 30$

Replace all occurrences of $b$ with $50$ in each equation.

Replace all occurrences of $b$ in $c = - \frac{21 b}{10} + 2230$ with $50$ .

$c = - \frac{21 (50)}{10} + 2230$

$b = 50$

$a = - \frac{b}{10} + 30$

Simplify the right side.

Simplify $- \frac{21 (50)}{10} + 2230$ .

Simplify each term.

Cancel the common factor of $50$ and $10$ .

Factor $10$ out of $21 (50)$ .

$c = - \frac{10 (21 \cdot (5))}{10} + 2230$

$b = 50$

$a = - \frac{b}{10} + 30$

Cancel the common factors.

Factor $10$ out of $10$ .

$c = - \frac{10 (21 \cdot (5))}{10 (1)} + 2230$

$b = 50$

$a = - \frac{b}{10} + 30$

Cancel the common factor.

$c = - \frac{10 (21 \cdot (5))}{10 \cdot 1} + 2230$

$b = 50$

$a = - \frac{b}{10} + 30$

Rewrite the expression.

$c = - \frac{21 \cdot (5)}{1} + 2230$

$b = 50$

$a = - \frac{b}{10} + 30$

Divide $21 \cdot (5)$ by $1$ .

$c = - (21 \cdot (5)) + 2230$

$b = 50$

$a = - \frac{b}{10} + 30$

$c = - (21 \cdot (5)) + 2230$

$b = 50$

$a = - \frac{b}{10} + 30$

$c = - (21 \cdot (5)) + 2230$

$b = 50$

$a = - \frac{b}{10} + 30$

Multiply $- (21 \cdot (5))$ .

Multiply $21$ by $5$ .

$c = - 1 \cdot 105 + 2230$

$b = 50$

$a = - \frac{b}{10} + 30$

Multiply $- 1$ by $105$ .

$c = - 105 + 2230$

$b = 50$

$a = - \frac{b}{10} + 30$

$c = - 105 + 2230$

$b = 50$

$a = - \frac{b}{10} + 30$

$c = - 105 + 2230$

$b = 50$

$a = - \frac{b}{10} + 30$

Add $- 105$ and $2230$ .

$c = 2125$

$b = 50$

$a = - \frac{b}{10} + 30$

$c = 2125$

$b = 50$

$a = - \frac{b}{10} + 30$

$c = 2125$

$b = 50$

$a = - \frac{b}{10} + 30$

Replace all occurrences of $b$ in $a = - \frac{b}{10} + 30$ with $50$ .

$a = - \frac{50}{10} + 30$

$c = 2125$

$b = 50$

Simplify the right side.

Simplify $- \frac{50}{10} + 30$ .

Simplify each term.

Divide $50$ by $10$ .

$a = - 1 \cdot 5 + 30$

$c = 2125$

$b = 50$

Multiply $- 1$ by $5$ .

$a = - 5 + 30$

$c = 2125$

$b = 50$

$a = - 5 + 30$

$c = 2125$

$b = 50$

Add $- 5$ and $30$ .

$a = 25$

$c = 2125$

$b = 50$

$a = 25$

$c = 2125$

$b = 50$

$a = 25$

$c = 2125$

$b = 50$

$a = 25$

$c = 2125$

$b = 50$

List all of the solutions.

$a = 25, c = 2125, b = 50$

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 $a = 25$ , $b = 50$ , $c = 2125$ , and $x = 3$ .

Simplify each term.

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

$y = 25 \cdot 9 + (50) \cdot (3) + 2125$

Multiply $25$ by $9$ .

$y = 225 + (50) \cdot (3) + 2125$

Multiply $50$ by $3$ .

$y = 225 + 150 + 2125$

Simplify by adding numbers.

Add $225$ and $150$ .

$y = 375 + 2125$

Add $375$ and $2125$ .

$y = 2500$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 3$ . This check passes since $y = 2500$ and $y = 2500$ .

$2500 = 2500$

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = 25$ , $b = 50$ , $c = 2125$ , and $x = 5$ .

Simplify each term.

Raise $5$ to the power of $2$ .

$y = 25 \cdot 25 + (50) \cdot (5) + 2125$

Multiply $25$ by $25$ .

$y = 625 + (50) \cdot (5) + 2125$

Multiply $50$ by $5$ .

$y = 625 + 250 + 2125$

Simplify by adding numbers.

Add $625$ and $250$ .

$y = 875 + 2125$

Add $875$ and $2125$ .

$y = 3000$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 5$ . This check passes since $y = 3000$ and $y = 3000$ .

$3000 = 3000$

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = 25$ , $b = 50$ , $c = 2125$ , and $x = 7$ .

Simplify each term.

Raise $7$ to the power of $2$ .

$y = 25 \cdot 49 + (50) \cdot (7) + 2125$

Multiply $25$ by $49$ .

$y = 1225 + (50) \cdot (7) + 2125$

Multiply $50$ by $7$ .

$y = 1225 + 350 + 2125$

Simplify by adding numbers.

Add $1225$ and $350$ .

$y = 1575 + 2125$

Add $1575$ and $2125$ .

$y = 3700$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 7$ . This check passes since $y = 3700$ and $y = 3700$ .

$3700 = 3700$

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 $y = 25 x^{2} + 50 x + 2125$ .

$y = 25 x^{2} + 50 x + 2125$

Find the Function Rule table[[x,y],[3,2500],[5,3000],[7,3700]]

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