Find the Function Rule table[[x,y],[2,36],[3,81],[4,144]]

$\begin{array}{c} x & y \\ 2 & 36 \\ 3 & 81 \\ 4 & 144 \end{array}$

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 $y = a x + b$ .

$36 = a (2) + b 81 = a (3) + b 144 = a (4) + b$

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

Solve for $b$ in $36 = a (2) + b$ .

Rewrite the equation as $a (2) + b = 36$ .

$a (2) + b = 36$

$81 = a (3) + b$

$144 = a (4) + b$

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

$2 a + b = 36$

$81 = a (3) + b$

$144 = a (4) + b$

Subtract $2 a$ from both sides of the equation.

$b = 36 - 2 a$

$81 = a (3) + b$

$144 = a (4) + b$

$b = 36 - 2 a$

$81 = a (3) + b$

$144 = a (4) + b$

Replace all occurrences of $b$ with $36 - 2 a$ in each equation.

Replace all occurrences of $b$ in $81 = a (3) + b$ with $36 - 2 a$ .

$81 = a (3) + 36 - 2 a$

$b = 36 - 2 a$

$144 = a (4) + b$

Simplify $81 = a (3) + 36 - 2 a$ .

Simplify the left side.

Remove parentheses.

$81 = a (3) + 36 - 2 a$

$b = 36 - 2 a$

$144 = a (4) + b$

$81 = a (3) + 36 - 2 a$

$b = 36 - 2 a$

$144 = a (4) + b$

Simplify the right side.

Simplify $a (3) + 36 - 2 a$ .

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

$81 = 3 a + 36 - 2 a$

$b = 36 - 2 a$

$144 = a (4) + b$

Subtract $2 a$ from $3 a$ .

$81 = a + 36$

$b = 36 - 2 a$

$144 = a (4) + b$

$81 = a + 36$

$b = 36 - 2 a$

$144 = a (4) + b$

$81 = a + 36$

$b = 36 - 2 a$

$144 = a (4) + b$

$81 = a + 36$

$b = 36 - 2 a$

$144 = a (4) + b$

Replace all occurrences of $b$ in $144 = a (4) + b$ with $36 - 2 a$ .

$144 = a (4) + 36 - 2 a$

$81 = a + 36$

$b = 36 - 2 a$

Simplify $144 = a (4) + 36 - 2 a$ .

Simplify the left side.

Remove parentheses.

$144 = a (4) + 36 - 2 a$

$81 = a + 36$

$b = 36 - 2 a$

$144 = a (4) + 36 - 2 a$

$81 = a + 36$

$b = 36 - 2 a$

Simplify the right side.

Simplify $a (4) + 36 - 2 a$ .

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

$144 = 4 a + 36 - 2 a$

$81 = a + 36$

$b = 36 - 2 a$

Subtract $2 a$ from $4 a$ .

$144 = 2 a + 36$

$81 = a + 36$

$b = 36 - 2 a$

$144 = 2 a + 36$

$81 = a + 36$

$b = 36 - 2 a$

$144 = 2 a + 36$

$81 = a + 36$

$b = 36 - 2 a$

$144 = 2 a + 36$

$81 = a + 36$

$b = 36 - 2 a$

$144 = 2 a + 36$

$81 = a + 36$

$b = 36 - 2 a$

Solve for $a$ in $144 = 2 a + 36$ .

Rewrite the equation as $2 a + 36 = 144$ .

$2 a + 36 = 144$

$81 = a + 36$

$b = 36 - 2 a$

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

Subtract $36$ from both sides of the equation.

$2 a = 144 - 36$

$81 = a + 36$

$b = 36 - 2 a$

Subtract $36$ from $144$ .

$2 a = 108$

$81 = a + 36$

$b = 36 - 2 a$

$2 a = 108$

$81 = a + 36$

$b = 36 - 2 a$

Divide each term in $2 a = 108$ by $2$ and simplify.

Divide each term in $2 a = 108$ by $2$ .

$\frac{2 a}{2} = \frac{108}{2}$

$81 = a + 36$

$b = 36 - 2 a$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 a}{2} = \frac{108}{2}$

$81 = a + 36$

$b = 36 - 2 a$

Divide $a$ by $1$ .

$a = \frac{108}{2}$

$81 = a + 36$

$b = 36 - 2 a$

$a = \frac{108}{2}$

$81 = a + 36$

$b = 36 - 2 a$

$a = \frac{108}{2}$

$81 = a + 36$

$b = 36 - 2 a$

Simplify the right side.

Divide $108$ by $2$ .

$a = 54$

$81 = a + 36$

$b = 36 - 2 a$

$a = 54$

$81 = a + 36$

$b = 36 - 2 a$

$a = 54$

$81 = a + 36$

$b = 36 - 2 a$

$a = 54$

$81 = a + 36$

$b = 36 - 2 a$

Replace all occurrences of $a$ with $54$ in each equation.

Replace all occurrences of $a$ in $81 = a + 36$ with $54$ .

$81 = (54) + 36$

$a = 54$

$b = 36 - 2 a$

Simplify the right side.

Add $54$ and $36$ .

$81 = 90$

$a = 54$

$b = 36 - 2 a$

$81 = 90$

$a = 54$

$b = 36 - 2 a$

Replace all occurrences of $a$ in $b = 36 - 2 a$ with $54$ .

$b = 36 - 2 \cdot 54$

$81 = 90$

$a = 54$

Simplify the right side.

Simplify $36 - 2 \cdot 54$ .

Multiply $- 2$ by $54$ .

$b = 36 - 108$

$81 = 90$

$a = 54$

Subtract $108$ from $36$ .

$b = - 72$

$81 = 90$

$a = 54$

$b = - 72$

$81 = 90$

$a = 54$

$b = - 72$

$81 = 90$

$a = 54$

$b = - 72$

$81 = 90$

$a = 54$

Since $81 = 90$ 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 $36 = a \cdot 2^{2} + b (2) + c$ .

Rewrite the equation as $a \cdot 2^{2} + b (2) + c = 36$ .

$a \cdot 2^{2} + b (2) + c = 36$

$81 = a \cdot 3^{2} + b (3) + c$

$144 = a \cdot 4^{2} + b (4) + c$

Simplify each term.

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

$a \cdot 4 + b (2) + c = 36$

$81 = a \cdot 3^{2} + b (3) + c$

$144 = a \cdot 4^{2} + b (4) + c$

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

$4 \cdot a + b (2) + c = 36$

$81 = a \cdot 3^{2} + b (3) + c$

$144 = a \cdot 4^{2} + b (4) + c$

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

$4 a + 2 b + c = 36$

$81 = a \cdot 3^{2} + b (3) + c$

$144 = a \cdot 4^{2} + b (4) + c$

$4 a + 2 b + c = 36$

$81 = a \cdot 3^{2} + b (3) + c$

$144 = a \cdot 4^{2} + b (4) + c$

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

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

$2 b + c = 36 - 4 a$

$81 = a \cdot 3^{2} + b (3) + c$

$144 = a \cdot 4^{2} + b (4) + c$

Subtract $2 b$ from both sides of the equation.

$c = 36 - 4 a - 2 b$

$81 = a \cdot 3^{2} + b (3) + c$

$144 = a \cdot 4^{2} + b (4) + c$

$c = 36 - 4 a - 2 b$

$81 = a \cdot 3^{2} + b (3) + c$

$144 = a \cdot 4^{2} + b (4) + c$

$c = 36 - 4 a - 2 b$

$81 = a \cdot 3^{2} + b (3) + c$

$144 = a \cdot 4^{2} + b (4) + c$

Replace all occurrences of $c$ with $36 - 4 a - 2 b$ in each equation.

Replace all occurrences of $c$ in $81 = a \cdot 3^{2} + b (3) + c$ with $36 - 4 a - 2 b$ .

$81 = a \cdot 3^{2} + b (3) + 36 - 4 a - 2 b$

$c = 36 - 4 a - 2 b$

$144 = a \cdot 4^{2} + b (4) + c$

Simplify $81 = a \cdot 3^{2} + b (3) + 36 - 4 a - 2 b$ .

Simplify the left side.

Remove parentheses.

$81 = a \cdot 3^{2} + b (3) + 36 - 4 a - 2 b$

$c = 36 - 4 a - 2 b$

$144 = a \cdot 4^{2} + b (4) + c$

$81 = a \cdot 3^{2} + b (3) + 36 - 4 a - 2 b$

$c = 36 - 4 a - 2 b$

$144 = a \cdot 4^{2} + b (4) + c$

Simplify the right side.

Simplify $a \cdot 3^{2} + b (3) + 36 - 4 a - 2 b$ .

Simplify each term.

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

$81 = a \cdot 9 + b (3) + 36 - 4 a - 2 b$

$c = 36 - 4 a - 2 b$

$144 = a \cdot 4^{2} + b (4) + c$

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

$81 = 9 \cdot a + b (3) + 36 - 4 a - 2 b$

$c = 36 - 4 a - 2 b$

$144 = a \cdot 4^{2} + b (4) + c$

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

$81 = 9 a + 3 b + 36 - 4 a - 2 b$

$c = 36 - 4 a - 2 b$

$144 = a \cdot 4^{2} + b (4) + c$

$81 = 9 a + 3 b + 36 - 4 a - 2 b$

$c = 36 - 4 a - 2 b$

$144 = a \cdot 4^{2} + b (4) + c$

Simplify by adding terms.

Subtract $4 a$ from $9 a$ .

$81 = 5 a + 3 b + 36 - 2 b$

$c = 36 - 4 a - 2 b$

$144 = a \cdot 4^{2} + b (4) + c$

Subtract $2 b$ from $3 b$ .

$81 = 5 a + b + 36$

$c = 36 - 4 a - 2 b$

$144 = a \cdot 4^{2} + b (4) + c$

$81 = 5 a + b + 36$

$c = 36 - 4 a - 2 b$

$144 = a \cdot 4^{2} + b (4) + c$

$81 = 5 a + b + 36$

$c = 36 - 4 a - 2 b$

$144 = a \cdot 4^{2} + b (4) + c$

$81 = 5 a + b + 36$

$c = 36 - 4 a - 2 b$

$144 = a \cdot 4^{2} + b (4) + c$

$81 = 5 a + b + 36$

$c = 36 - 4 a - 2 b$

$144 = a \cdot 4^{2} + b (4) + c$

Replace all occurrences of $c$ in $144 = a \cdot 4^{2} + b (4) + c$ with $36 - 4 a - 2 b$ .

$144 = a \cdot 4^{2} + b (4) + 36 - 4 a - 2 b$

$81 = 5 a + b + 36$

$c = 36 - 4 a - 2 b$

Simplify $144 = a \cdot 4^{2} + b (4) + 36 - 4 a - 2 b$ .

Simplify the left side.

Remove parentheses.

$144 = a \cdot 4^{2} + b (4) + 36 - 4 a - 2 b$

$81 = 5 a + b + 36$

$c = 36 - 4 a - 2 b$

$144 = a \cdot 4^{2} + b (4) + 36 - 4 a - 2 b$

$81 = 5 a + b + 36$

$c = 36 - 4 a - 2 b$

Simplify the right side.

Simplify $a \cdot 4^{2} + b (4) + 36 - 4 a - 2 b$ .

Simplify each term.

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

$144 = a \cdot 16 + b (4) + 36 - 4 a - 2 b$

$81 = 5 a + b + 36$

$c = 36 - 4 a - 2 b$

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

$144 = 16 \cdot a + b (4) + 36 - 4 a - 2 b$

$81 = 5 a + b + 36$

$c = 36 - 4 a - 2 b$

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

$144 = 16 a + 4 b + 36 - 4 a - 2 b$

$81 = 5 a + b + 36$

$c = 36 - 4 a - 2 b$

$144 = 16 a + 4 b + 36 - 4 a - 2 b$

$81 = 5 a + b + 36$

$c = 36 - 4 a - 2 b$

Simplify by adding terms.

Subtract $4 a$ from $16 a$ .

$144 = 12 a + 4 b + 36 - 2 b$

$81 = 5 a + b + 36$

$c = 36 - 4 a - 2 b$

Subtract $2 b$ from $4 b$ .

$144 = 12 a + 2 b + 36$

$81 = 5 a + b + 36$

$c = 36 - 4 a - 2 b$

$144 = 12 a + 2 b + 36$

$81 = 5 a + b + 36$

$c = 36 - 4 a - 2 b$

$144 = 12 a + 2 b + 36$

$81 = 5 a + b + 36$

$c = 36 - 4 a - 2 b$

$144 = 12 a + 2 b + 36$

$81 = 5 a + b + 36$

$c = 36 - 4 a - 2 b$

$144 = 12 a + 2 b + 36$

$81 = 5 a + b + 36$

$c = 36 - 4 a - 2 b$

$144 = 12 a + 2 b + 36$

$81 = 5 a + b + 36$

$c = 36 - 4 a - 2 b$

Solve for $b$ in $81 = 5 a + b + 36$ .

Rewrite the equation as $5 a + b + 36 = 81$ .

$5 a + b + 36 = 81$

$144 = 12 a + 2 b + 36$

$c = 36 - 4 a - 2 b$

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

Subtract $5 a$ from both sides of the equation.

$b + 36 = 81 - 5 a$

$144 = 12 a + 2 b + 36$

$c = 36 - 4 a - 2 b$

Subtract $36$ from both sides of the equation.

$b = 81 - 5 a - 36$

$144 = 12 a + 2 b + 36$

$c = 36 - 4 a - 2 b$

Subtract $36$ from $81$ .

$b = - 5 a + 45$

$144 = 12 a + 2 b + 36$

$c = 36 - 4 a - 2 b$

$b = - 5 a + 45$

$144 = 12 a + 2 b + 36$

$c = 36 - 4 a - 2 b$

$b = - 5 a + 45$

$144 = 12 a + 2 b + 36$

$c = 36 - 4 a - 2 b$

Replace all occurrences of $b$ with $- 5 a + 45$ in each equation.

Replace all occurrences of $b$ in $144 = 12 a + 2 b + 36$ with $- 5 a + 45$ .

$144 = 12 a + 2 (- 5 a + 45) + 36$

$b = - 5 a + 45$

$c = 36 - 4 a - 2 b$

Simplify the right side.

Simplify $12 a + 2 (- 5 a + 45) + 36$ .

Simplify each term.

Apply the distributive property.

$144 = 12 a + 2 (- 5 a) + 2 \cdot 45 + 36$

$b = - 5 a + 45$

$c = 36 - 4 a - 2 b$

Multiply $- 5$ by $2$ .

$144 = 12 a - 10 a + 2 \cdot 45 + 36$

$b = - 5 a + 45$

$c = 36 - 4 a - 2 b$

Multiply $2$ by $45$ .

$144 = 12 a - 10 a + 90 + 36$

$b = - 5 a + 45$

$c = 36 - 4 a - 2 b$

$144 = 12 a - 10 a + 90 + 36$

$b = - 5 a + 45$

$c = 36 - 4 a - 2 b$

Simplify by adding terms.

Subtract $10 a$ from $12 a$ .

$144 = 2 a + 90 + 36$

$b = - 5 a + 45$

$c = 36 - 4 a - 2 b$

Add $90$ and $36$ .

$144 = 2 a + 126$

$b = - 5 a + 45$

$c = 36 - 4 a - 2 b$

$144 = 2 a + 126$

$b = - 5 a + 45$

$c = 36 - 4 a - 2 b$

$144 = 2 a + 126$

$b = - 5 a + 45$

$c = 36 - 4 a - 2 b$

$144 = 2 a + 126$

$b = - 5 a + 45$

$c = 36 - 4 a - 2 b$

Replace all occurrences of $b$ in $c = 36 - 4 a - 2 b$ with $- 5 a + 45$ .

$c = 36 - 4 a - 2 (- 5 a + 45)$

$144 = 2 a + 126$

$b = - 5 a + 45$

Simplify the right side.

Simplify $36 - 4 a - 2 (- 5 a + 45)$ .

Simplify each term.

Apply the distributive property.

$c = 36 - 4 a - 2 (- 5 a) - 2 \cdot 45$

$144 = 2 a + 126$

$b = - 5 a + 45$

Multiply $- 5$ by $- 2$ .

$c = 36 - 4 a + 10 a - 2 \cdot 45$

$144 = 2 a + 126$

$b = - 5 a + 45$

Multiply $- 2$ by $45$ .

$c = 36 - 4 a + 10 a - 90$

$144 = 2 a + 126$

$b = - 5 a + 45$

$c = 36 - 4 a + 10 a - 90$

$144 = 2 a + 126$

$b = - 5 a + 45$

Simplify by adding terms.

Subtract $90$ from $36$ .

$c = - 4 a + 10 a - 54$

$144 = 2 a + 126$

$b = - 5 a + 45$

Add $- 4 a$ and $10 a$ .

$c = 6 a - 54$

$144 = 2 a + 126$

$b = - 5 a + 45$

$c = 6 a - 54$

$144 = 2 a + 126$

$b = - 5 a + 45$

$c = 6 a - 54$

$144 = 2 a + 126$

$b = - 5 a + 45$

$c = 6 a - 54$

$144 = 2 a + 126$

$b = - 5 a + 45$

$c = 6 a - 54$

$144 = 2 a + 126$

$b = - 5 a + 45$

Solve for $a$ in $144 = 2 a + 126$ .

Rewrite the equation as $2 a + 126 = 144$ .

$2 a + 126 = 144$

$c = 6 a - 54$

$b = - 5 a + 45$

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

Subtract $126$ from both sides of the equation.

$2 a = 144 - 126$

$c = 6 a - 54$

$b = - 5 a + 45$

Subtract $126$ from $144$ .

$2 a = 18$

$c = 6 a - 54$

$b = - 5 a + 45$

$2 a = 18$

$c = 6 a - 54$

$b = - 5 a + 45$

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

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

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

$c = 6 a - 54$

$b = - 5 a + 45$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

$c = 6 a - 54$

$b = - 5 a + 45$

Divide $a$ by $1$ .

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

$c = 6 a - 54$

$b = - 5 a + 45$

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

$c = 6 a - 54$

$b = - 5 a + 45$

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

$c = 6 a - 54$

$b = - 5 a + 45$

Simplify the right side.

Divide $18$ by $2$ .

$a = 9$

$c = 6 a - 54$

$b = - 5 a + 45$

$a = 9$

$c = 6 a - 54$

$b = - 5 a + 45$

$a = 9$

$c = 6 a - 54$

$b = - 5 a + 45$

$a = 9$

$c = 6 a - 54$

$b = - 5 a + 45$

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

Replace all occurrences of $a$ in $c = 6 a - 54$ with $9$ .

$c = 6 (9) - 54$

$a = 9$

$b = - 5 a + 45$

Simplify the right side.

Simplify $6 (9) - 54$ .

Multiply $6$ by $9$ .

$c = 54 - 54$

$a = 9$

$b = - 5 a + 45$

Subtract $54$ from $54$ .

$c = 0$

$a = 9$

$b = - 5 a + 45$

$c = 0$

$a = 9$

$b = - 5 a + 45$

$c = 0$

$a = 9$

$b = - 5 a + 45$

Replace all occurrences of $a$ in $b = - 5 a + 45$ with $9$ .

$b = - 5 \cdot 9 + 45$

$c = 0$

$a = 9$

Simplify the right side.

Simplify $- 5 \cdot 9 + 45$ .

Multiply $- 5$ by $9$ .

$b = - 45 + 45$

$c = 0$

$a = 9$

Add $- 45$ and $45$ .

$b = 0$

$c = 0$

$a = 9$

$b = 0$

$c = 0$

$a = 9$

$b = 0$

$c = 0$

$a = 9$

$b = 0$

$c = 0$

$a = 9$

List all of the solutions.

$b = 0, c = 0, a = 9$

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 $a = 9$ , $b = 0$ , $c = 0$ , and $x = 2$ .

Simplify each term.

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

$y = 9 \cdot 4 + (0) \cdot (2) + 0$

Multiply $9$ by $4$ .

$y = 36 + (0) \cdot (2) + 0$

Multiply $0$ by $2$ .

$y = 36 + 0 + 0$

Simplify by adding numbers.

Add $36$ and $0$ .

$y = 36 + 0$

Add $36$ and $0$ .

$y = 36$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 2$ . This check passes since $y = 36$ and $y = 36$ .

$36 = 36$

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = 9$ , $b = 0$ , $c = 0$ , and $x = 3$ .

Simplify each term.

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

$y = 9 \cdot 9 + (0) \cdot (3) + 0$

Multiply $9$ by $9$ .

$y = 81 + (0) \cdot (3) + 0$

Multiply $0$ by $3$ .

$y = 81 + 0 + 0$

Simplify by adding numbers.

Add $81$ and $0$ .

$y = 81 + 0$

Add $81$ and $0$ .

$y = 81$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 3$ . This check passes since $y = 81$ and $y = 81$ .

$81 = 81$

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = 9$ , $b = 0$ , $c = 0$ , and $x = 4$ .

Simplify each term.

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

$y = 9 \cdot 16 + (0) \cdot (4) + 0$

Multiply $9$ by $16$ .

$y = 144 + (0) \cdot (4) + 0$

Multiply $0$ by $4$ .

$y = 144 + 0 + 0$

Simplify by adding numbers.

Add $144$ and $0$ .

$y = 144 + 0$

Add $144$ and $0$ .

$y = 144$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 4$ . This check passes since $y = 144$ and $y = 144$ .

$144 = 144$

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 $y = 9 x^{2}$ .

$y = 9 x^{2}$

Find the Function Rule table[[x,y],[2,36],[3,81],[4,144]]

Related Questions

Support LCMGCF.com by disabling your adblocker.