Find the Function Rule table[[x,y],[3,10],[4,16],[5,28]]

$\begin{array}{c} x & y \\ 3 & 10 \\ 4 & 16 \\ 5 & 28 \end{array}$

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 $y = a x + b$ .

$10 = a (3) + b 16 = a (4) + b 28 = a (5) + b$

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

Solve for $b$ in $10 = a (3) + b$ .

Rewrite the equation as $a (3) + b = 10$ .

$a (3) + b = 10$

$16 = a (4) + b$

$28 = a (5) + b$

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

$3 a + b = 10$

$16 = a (4) + b$

$28 = a (5) + b$

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

$b = 10 - 3 a$

$16 = a (4) + b$

$28 = a (5) + b$

$b = 10 - 3 a$

$16 = a (4) + b$

$28 = a (5) + b$

Replace all occurrences of $b$ with $10 - 3 a$ in each equation.

Replace all occurrences of $b$ in $16 = a (4) + b$ with $10 - 3 a$ .

$16 = a (4) + 10 - 3 a$

$b = 10 - 3 a$

$28 = a (5) + b$

Simplify $16 = a (4) + 10 - 3 a$ .

Simplify the left side.

Remove parentheses.

$16 = a (4) + 10 - 3 a$

$b = 10 - 3 a$

$28 = a (5) + b$

$16 = a (4) + 10 - 3 a$

$b = 10 - 3 a$

$28 = a (5) + b$

Simplify the right side.

Simplify $a (4) + 10 - 3 a$ .

Move $4$ to the left of $a$ .

$16 = 4 a + 10 - 3 a$

$b = 10 - 3 a$

$28 = a (5) + b$

Subtract $3 a$ from $4 a$ .

$16 = a + 10$

$b = 10 - 3 a$

$28 = a (5) + b$

$16 = a + 10$

$b = 10 - 3 a$

$28 = a (5) + b$

$16 = a + 10$

$b = 10 - 3 a$

$28 = a (5) + b$

$16 = a + 10$

$b = 10 - 3 a$

$28 = a (5) + b$

Replace all occurrences of $b$ in $28 = a (5) + b$ with $10 - 3 a$ .

$28 = a (5) + 10 - 3 a$

$16 = a + 10$

$b = 10 - 3 a$

Simplify $28 = a (5) + 10 - 3 a$ .

Simplify the left side.

Remove parentheses.

$28 = a (5) + 10 - 3 a$

$16 = a + 10$

$b = 10 - 3 a$

$28 = a (5) + 10 - 3 a$

$16 = a + 10$

$b = 10 - 3 a$

Simplify the right side.

Simplify $a (5) + 10 - 3 a$ .

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

$28 = 5 a + 10 - 3 a$

$16 = a + 10$

$b = 10 - 3 a$

Subtract $3 a$ from $5 a$ .

$28 = 2 a + 10$

$16 = a + 10$

$b = 10 - 3 a$

$28 = 2 a + 10$

$16 = a + 10$

$b = 10 - 3 a$

$28 = 2 a + 10$

$16 = a + 10$

$b = 10 - 3 a$

$28 = 2 a + 10$

$16 = a + 10$

$b = 10 - 3 a$

$28 = 2 a + 10$

$16 = a + 10$

$b = 10 - 3 a$

Solve for $a$ in $28 = 2 a + 10$ .

Rewrite the equation as $2 a + 10 = 28$ .

$2 a + 10 = 28$

$16 = a + 10$

$b = 10 - 3 a$

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

Subtract $10$ from both sides of the equation.

$2 a = 28 - 10$

$16 = a + 10$

$b = 10 - 3 a$

Subtract $10$ from $28$ .

$2 a = 18$

$16 = a + 10$

$b = 10 - 3 a$

$2 a = 18$

$16 = a + 10$

$b = 10 - 3 a$

Divide each term in $2 a = 18$ by $2$ and simplify.

Divide each term in $2 a = 18$ by $2$ .

$\frac{2 a}{2} = \frac{18}{2}$

$16 = a + 10$

$b = 10 - 3 a$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 a}{2} = \frac{18}{2}$

$16 = a + 10$

$b = 10 - 3 a$

Divide $a$ by $1$ .

$a = \frac{18}{2}$

$16 = a + 10$

$b = 10 - 3 a$

$a = \frac{18}{2}$

$16 = a + 10$

$b = 10 - 3 a$

$a = \frac{18}{2}$

$16 = a + 10$

$b = 10 - 3 a$

Simplify the right side.

Divide $18$ by $2$ .

$a = 9$

$16 = a + 10$

$b = 10 - 3 a$

$a = 9$

$16 = a + 10$

$b = 10 - 3 a$

$a = 9$

$16 = a + 10$

$b = 10 - 3 a$

$a = 9$

$16 = a + 10$

$b = 10 - 3 a$

Replace all occurrences of $a$ with $9$ in each equation.

Replace all occurrences of $a$ in $16 = a + 10$ with $9$ .

$16 = (9) + 10$

$a = 9$

$b = 10 - 3 a$

Simplify the right side.

Add $9$ and $10$ .

$16 = 19$

$a = 9$

$b = 10 - 3 a$

$16 = 19$

$a = 9$

$b = 10 - 3 a$

Replace all occurrences of $a$ in $b = 10 - 3 a$ with $9$ .

$b = 10 - 3 \cdot 9$

$16 = 19$

$a = 9$

Simplify the right side.

Simplify $10 - 3 \cdot 9$ .

Multiply $- 3$ by $9$ .

$b = 10 - 27$

$16 = 19$

$a = 9$

Subtract $27$ from $10$ .

$b = - 17$

$16 = 19$

$a = 9$

$b = - 17$

$16 = 19$

$a = 9$

$b = - 17$

$16 = 19$

$a = 9$

$b = - 17$

$16 = 19$

$a = 9$

Since $16 = 19$ 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 $10 = a \cdot 3^{2} + b (3) + c$ .

Rewrite the equation as $a \cdot 3^{2} + b (3) + c = 10$ .

$a \cdot 3^{2} + b (3) + c = 10$

$16 = a \cdot 4^{2} + b (4) + c$

$28 = a \cdot 5^{2} + b (5) + c$

Simplify each term.

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

$a \cdot 9 + b (3) + c = 10$

$16 = a \cdot 4^{2} + b (4) + c$

$28 = a \cdot 5^{2} + b (5) + c$

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

$9 \cdot a + b (3) + c = 10$

$16 = a \cdot 4^{2} + b (4) + c$

$28 = a \cdot 5^{2} + b (5) + c$

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

$9 a + 3 b + c = 10$

$16 = a \cdot 4^{2} + b (4) + c$

$28 = a \cdot 5^{2} + b (5) + c$

$9 a + 3 b + c = 10$

$16 = a \cdot 4^{2} + b (4) + c$

$28 = a \cdot 5^{2} + b (5) + c$

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

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

$3 b + c = 10 - 9 a$

$16 = a \cdot 4^{2} + b (4) + c$

$28 = a \cdot 5^{2} + b (5) + c$

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

$c = 10 - 9 a - 3 b$

$16 = a \cdot 4^{2} + b (4) + c$

$28 = a \cdot 5^{2} + b (5) + c$

$c = 10 - 9 a - 3 b$

$16 = a \cdot 4^{2} + b (4) + c$

$28 = a \cdot 5^{2} + b (5) + c$

$c = 10 - 9 a - 3 b$

$16 = a \cdot 4^{2} + b (4) + c$

$28 = a \cdot 5^{2} + b (5) + c$

Replace all occurrences of $c$ with $10 - 9 a - 3 b$ in each equation.

Replace all occurrences of $c$ in $16 = a \cdot 4^{2} + b (4) + c$ with $10 - 9 a - 3 b$ .

$16 = a \cdot 4^{2} + b (4) + 10 - 9 a - 3 b$

$c = 10 - 9 a - 3 b$

$28 = a \cdot 5^{2} + b (5) + c$

Simplify $16 = a \cdot 4^{2} + b (4) + 10 - 9 a - 3 b$ .

Simplify the left side.

Remove parentheses.

$16 = a \cdot 4^{2} + b (4) + 10 - 9 a - 3 b$

$c = 10 - 9 a - 3 b$

$28 = a \cdot 5^{2} + b (5) + c$

$16 = a \cdot 4^{2} + b (4) + 10 - 9 a - 3 b$

$c = 10 - 9 a - 3 b$

$28 = a \cdot 5^{2} + b (5) + c$

Simplify the right side.

Simplify $a \cdot 4^{2} + b (4) + 10 - 9 a - 3 b$ .

Simplify each term.

Raise $4$ to the power of $2$ .

$16 = a \cdot 16 + b (4) + 10 - 9 a - 3 b$

$c = 10 - 9 a - 3 b$

$28 = a \cdot 5^{2} + b (5) + c$

Move $16$ to the left of $a$ .

$16 = 16 \cdot a + b (4) + 10 - 9 a - 3 b$

$c = 10 - 9 a - 3 b$

$28 = a \cdot 5^{2} + b (5) + c$

Move $4$ to the left of $b$ .

$16 = 16 a + 4 b + 10 - 9 a - 3 b$

$c = 10 - 9 a - 3 b$

$28 = a \cdot 5^{2} + b (5) + c$

$16 = 16 a + 4 b + 10 - 9 a - 3 b$

$c = 10 - 9 a - 3 b$

$28 = a \cdot 5^{2} + b (5) + c$

Simplify by adding terms.

Subtract $9 a$ from $16 a$ .

$16 = 7 a + 4 b + 10 - 3 b$

$c = 10 - 9 a - 3 b$

$28 = a \cdot 5^{2} + b (5) + c$

Subtract $3 b$ from $4 b$ .

$16 = 7 a + b + 10$

$c = 10 - 9 a - 3 b$

$28 = a \cdot 5^{2} + b (5) + c$

$16 = 7 a + b + 10$

$c = 10 - 9 a - 3 b$

$28 = a \cdot 5^{2} + b (5) + c$

$16 = 7 a + b + 10$

$c = 10 - 9 a - 3 b$

$28 = a \cdot 5^{2} + b (5) + c$

$16 = 7 a + b + 10$

$c = 10 - 9 a - 3 b$

$28 = a \cdot 5^{2} + b (5) + c$

$16 = 7 a + b + 10$

$c = 10 - 9 a - 3 b$

$28 = a \cdot 5^{2} + b (5) + c$

Replace all occurrences of $c$ in $28 = a \cdot 5^{2} + b (5) + c$ with $10 - 9 a - 3 b$ .

$28 = a \cdot 5^{2} + b (5) + 10 - 9 a - 3 b$

$16 = 7 a + b + 10$

$c = 10 - 9 a - 3 b$

Simplify $28 = a \cdot 5^{2} + b (5) + 10 - 9 a - 3 b$ .

Simplify the left side.

Remove parentheses.

$28 = a \cdot 5^{2} + b (5) + 10 - 9 a - 3 b$

$16 = 7 a + b + 10$

$c = 10 - 9 a - 3 b$

$28 = a \cdot 5^{2} + b (5) + 10 - 9 a - 3 b$

$16 = 7 a + b + 10$

$c = 10 - 9 a - 3 b$

Simplify the right side.

Simplify $a \cdot 5^{2} + b (5) + 10 - 9 a - 3 b$ .

Simplify each term.

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

$28 = a \cdot 25 + b (5) + 10 - 9 a - 3 b$

$16 = 7 a + b + 10$

$c = 10 - 9 a - 3 b$

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

$28 = 25 \cdot a + b (5) + 10 - 9 a - 3 b$

$16 = 7 a + b + 10$

$c = 10 - 9 a - 3 b$

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

$28 = 25 a + 5 b + 10 - 9 a - 3 b$

$16 = 7 a + b + 10$

$c = 10 - 9 a - 3 b$

$28 = 25 a + 5 b + 10 - 9 a - 3 b$

$16 = 7 a + b + 10$

$c = 10 - 9 a - 3 b$

Simplify by adding terms.

Subtract $9 a$ from $25 a$ .

$28 = 16 a + 5 b + 10 - 3 b$

$16 = 7 a + b + 10$

$c = 10 - 9 a - 3 b$

Subtract $3 b$ from $5 b$ .

$28 = 16 a + 2 b + 10$

$16 = 7 a + b + 10$

$c = 10 - 9 a - 3 b$

$28 = 16 a + 2 b + 10$

$16 = 7 a + b + 10$

$c = 10 - 9 a - 3 b$

$28 = 16 a + 2 b + 10$

$16 = 7 a + b + 10$

$c = 10 - 9 a - 3 b$

$28 = 16 a + 2 b + 10$

$16 = 7 a + b + 10$

$c = 10 - 9 a - 3 b$

$28 = 16 a + 2 b + 10$

$16 = 7 a + b + 10$

$c = 10 - 9 a - 3 b$

$28 = 16 a + 2 b + 10$

$16 = 7 a + b + 10$

$c = 10 - 9 a - 3 b$

Solve for $b$ in $16 = 7 a + b + 10$ .

Rewrite the equation as $7 a + b + 10 = 16$ .

$7 a + b + 10 = 16$

$28 = 16 a + 2 b + 10$

$c = 10 - 9 a - 3 b$

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

Subtract $7 a$ from both sides of the equation.

$b + 10 = 16 - 7 a$

$28 = 16 a + 2 b + 10$

$c = 10 - 9 a - 3 b$

Subtract $10$ from both sides of the equation.

$b = 16 - 7 a - 10$

$28 = 16 a + 2 b + 10$

$c = 10 - 9 a - 3 b$

Subtract $10$ from $16$ .

$b = - 7 a + 6$

$28 = 16 a + 2 b + 10$

$c = 10 - 9 a - 3 b$

$b = - 7 a + 6$

$28 = 16 a + 2 b + 10$

$c = 10 - 9 a - 3 b$

$b = - 7 a + 6$

$28 = 16 a + 2 b + 10$

$c = 10 - 9 a - 3 b$

Replace all occurrences of $b$ with $- 7 a + 6$ in each equation.

Replace all occurrences of $b$ in $28 = 16 a + 2 b + 10$ with $- 7 a + 6$ .

$28 = 16 a + 2 (- 7 a + 6) + 10$

$b = - 7 a + 6$

$c = 10 - 9 a - 3 b$

Simplify the right side.

Simplify $16 a + 2 (- 7 a + 6) + 10$ .

Simplify each term.

Apply the distributive property.

$28 = 16 a + 2 (- 7 a) + 2 \cdot 6 + 10$

$b = - 7 a + 6$

$c = 10 - 9 a - 3 b$

Multiply $- 7$ by $2$ .

$28 = 16 a - 14 a + 2 \cdot 6 + 10$

$b = - 7 a + 6$

$c = 10 - 9 a - 3 b$

Multiply $2$ by $6$ .

$28 = 16 a - 14 a + 12 + 10$

$b = - 7 a + 6$

$c = 10 - 9 a - 3 b$

$28 = 16 a - 14 a + 12 + 10$

$b = - 7 a + 6$

$c = 10 - 9 a - 3 b$

Simplify by adding terms.

Subtract $14 a$ from $16 a$ .

$28 = 2 a + 12 + 10$

$b = - 7 a + 6$

$c = 10 - 9 a - 3 b$

Add $12$ and $10$ .

$28 = 2 a + 22$

$b = - 7 a + 6$

$c = 10 - 9 a - 3 b$

$28 = 2 a + 22$

$b = - 7 a + 6$

$c = 10 - 9 a - 3 b$

$28 = 2 a + 22$

$b = - 7 a + 6$

$c = 10 - 9 a - 3 b$

$28 = 2 a + 22$

$b = - 7 a + 6$

$c = 10 - 9 a - 3 b$

Replace all occurrences of $b$ in $c = 10 - 9 a - 3 b$ with $- 7 a + 6$ .

$c = 10 - 9 a - 3 (- 7 a + 6)$

$28 = 2 a + 22$

$b = - 7 a + 6$

Simplify the right side.

Simplify $10 - 9 a - 3 (- 7 a + 6)$ .

Simplify each term.

Apply the distributive property.

$c = 10 - 9 a - 3 (- 7 a) - 3 \cdot 6$

$28 = 2 a + 22$

$b = - 7 a + 6$

Multiply $- 7$ by $- 3$ .

$c = 10 - 9 a + 21 a - 3 \cdot 6$

$28 = 2 a + 22$

$b = - 7 a + 6$

Multiply $- 3$ by $6$ .

$c = 10 - 9 a + 21 a - 18$

$28 = 2 a + 22$

$b = - 7 a + 6$

$c = 10 - 9 a + 21 a - 18$

$28 = 2 a + 22$

$b = - 7 a + 6$

Simplify by adding terms.

Subtract $18$ from $10$ .

$c = - 9 a + 21 a - 8$

$28 = 2 a + 22$

$b = - 7 a + 6$

Add $- 9 a$ and $21 a$ .

$c = 12 a - 8$

$28 = 2 a + 22$

$b = - 7 a + 6$

$c = 12 a - 8$

$28 = 2 a + 22$

$b = - 7 a + 6$

$c = 12 a - 8$

$28 = 2 a + 22$

$b = - 7 a + 6$

$c = 12 a - 8$

$28 = 2 a + 22$

$b = - 7 a + 6$

$c = 12 a - 8$

$28 = 2 a + 22$

$b = - 7 a + 6$

Solve for $a$ in $28 = 2 a + 22$ .

Rewrite the equation as $2 a + 22 = 28$ .

$2 a + 22 = 28$

$c = 12 a - 8$

$b = - 7 a + 6$

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

Subtract $22$ from both sides of the equation.

$2 a = 28 - 22$

$c = 12 a - 8$

$b = - 7 a + 6$

Subtract $22$ from $28$ .

$2 a = 6$

$c = 12 a - 8$

$b = - 7 a + 6$

$2 a = 6$

$c = 12 a - 8$

$b = - 7 a + 6$

Divide each term in $2 a = 6$ by $2$ and simplify.

Divide each term in $2 a = 6$ by $2$ .

$\frac{2 a}{2} = \frac{6}{2}$

$c = 12 a - 8$

$b = - 7 a + 6$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 a}{2} = \frac{6}{2}$

$c = 12 a - 8$

$b = - 7 a + 6$

Divide $a$ by $1$ .

$a = \frac{6}{2}$

$c = 12 a - 8$

$b = - 7 a + 6$

$a = \frac{6}{2}$

$c = 12 a - 8$

$b = - 7 a + 6$

$a = \frac{6}{2}$

$c = 12 a - 8$

$b = - 7 a + 6$

Simplify the right side.

Divide $6$ by $2$ .

$a = 3$

$c = 12 a - 8$

$b = - 7 a + 6$

$a = 3$

$c = 12 a - 8$

$b = - 7 a + 6$

$a = 3$

$c = 12 a - 8$

$b = - 7 a + 6$

$a = 3$

$c = 12 a - 8$

$b = - 7 a + 6$

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

Replace all occurrences of $a$ in $c = 12 a - 8$ with $3$ .

$c = 12 (3) - 8$

$a = 3$

$b = - 7 a + 6$

Simplify the right side.

Simplify $12 (3) - 8$ .

Multiply $12$ by $3$ .

$c = 36 - 8$

$a = 3$

$b = - 7 a + 6$

Subtract $8$ from $36$ .

$c = 28$

$a = 3$

$b = - 7 a + 6$

$c = 28$

$a = 3$

$b = - 7 a + 6$

$c = 28$

$a = 3$

$b = - 7 a + 6$

Replace all occurrences of $a$ in $b = - 7 a + 6$ with $3$ .

$b = - 7 \cdot 3 + 6$

$c = 28$

$a = 3$

Simplify the right side.

Simplify $- 7 \cdot 3 + 6$ .

Multiply $- 7$ by $3$ .

$b = - 21 + 6$

$c = 28$

$a = 3$

Add $- 21$ and $6$ .

$b = - 15$

$c = 28$

$a = 3$

$b = - 15$

$c = 28$

$a = 3$

$b = - 15$

$c = 28$

$a = 3$

$b = - 15$

$c = 28$

$a = 3$

List all of the solutions.

$b = - 15, c = 28, a = 3$

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 $a = 3$ , $b = - 15$ , $c = 28$ , and $x = 3$ .

Simplify each term.

Multiply $3$ by ${(3)}^{2}$ by adding the exponents.

Multiply $3$ by ${(3)}^{2}$ .

Raise $3$ to the power of $1$ .

$y = 3 \cdot {(3)}^{2} + (- 15) \cdot (3) + 28$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$y = 3^{1 + 2} + (- 15) \cdot (3) + 28$

Add $1$ and $2$ .

$y = 3^{3} + (- 15) \cdot (3) + 28$

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

$y = 27 + (- 15) \cdot (3) + 28$

Multiply $- 15$ by $3$ .

$y = 27 - 45 + 28$

Simplify by adding and subtracting.

Subtract $45$ from $27$ .

$y = - 18 + 28$

Add $- 18$ and $28$ .

$y = 10$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 3$ . This check passes since $y = 10$ and $y = 10$ .

$10 = 10$

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = 3$ , $b = - 15$ , $c = 28$ , and $x = 4$ .

Simplify each term.

Raise $4$ to the power of $2$ .

$y = 3 \cdot 16 + (- 15) \cdot (4) + 28$

Multiply $3$ by $16$ .

$y = 48 + (- 15) \cdot (4) + 28$

Multiply $- 15$ by $4$ .

$y = 48 - 60 + 28$

Simplify by adding and subtracting.

Subtract $60$ from $48$ .

$y = - 12 + 28$

Add $- 12$ and $28$ .

$y = 16$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 4$ . This check passes since $y = 16$ and $y = 16$ .

$16 = 16$

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = 3$ , $b = - 15$ , $c = 28$ , and $x = 5$ .

Simplify each term.

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

$y = 3 \cdot 25 + (- 15) \cdot (5) + 28$

Multiply $3$ by $25$ .

$y = 75 + (- 15) \cdot (5) + 28$

Multiply $- 15$ by $5$ .

$y = 75 - 75 + 28$

Simplify by adding and subtracting.

Subtract $75$ from $75$ .

$y = 0 + 28$

Add $0$ and $28$ .

$y = 28$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 5$ . This check passes since $y = 28$ and $y = 28$ .

$28 = 28$

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 $y = 3 x^{2} - 15 x + 28$ .

$y = 3 x^{2} - 15 x + 28$

Find the Function Rule table[[x,y],[3,10],[4,16],[5,28]]

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