Find the Function Rule table[[x,y],[2,8],[3,13],[4,20]]

$\begin{array}{c} x & y \\ 2 & 8 \\ 3 & 13 \\ 4 & 20 \end{array}$

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 $y = a x + b$ .

$8 = a (2) + b 13 = a (3) + b 20 = a (4) + b$

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

Solve for $b$ in $8 = a (2) + b$ .

Rewrite the equation as $a (2) + b = 8$ .

$a (2) + b = 8$

$13 = a (3) + b$

$20 = a (4) + b$

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

$2 a + b = 8$

$13 = a (3) + b$

$20 = a (4) + b$

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

$b = 8 - 2 a$

$13 = a (3) + b$

$20 = a (4) + b$

$b = 8 - 2 a$

$13 = a (3) + b$

$20 = a (4) + b$

Replace all occurrences of $b$ with $8 - 2 a$ in each equation.

Replace all occurrences of $b$ in $13 = a (3) + b$ with $8 - 2 a$ .

$13 = a (3) + 8 - 2 a$

$b = 8 - 2 a$

$20 = a (4) + b$

Simplify $13 = a (3) + 8 - 2 a$ .

Simplify the left side.

Remove parentheses.

$13 = a (3) + 8 - 2 a$

$b = 8 - 2 a$

$20 = a (4) + b$

$13 = a (3) + 8 - 2 a$

$b = 8 - 2 a$

$20 = a (4) + b$

Simplify the right side.

Simplify $a (3) + 8 - 2 a$ .

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

$13 = 3 a + 8 - 2 a$

$b = 8 - 2 a$

$20 = a (4) + b$

Subtract $2 a$ from $3 a$ .

$13 = a + 8$

$b = 8 - 2 a$

$20 = a (4) + b$

$13 = a + 8$

$b = 8 - 2 a$

$20 = a (4) + b$

$13 = a + 8$

$b = 8 - 2 a$

$20 = a (4) + b$

$13 = a + 8$

$b = 8 - 2 a$

$20 = a (4) + b$

Replace all occurrences of $b$ in $20 = a (4) + b$ with $8 - 2 a$ .

$20 = a (4) + 8 - 2 a$

$13 = a + 8$

$b = 8 - 2 a$

Simplify $20 = a (4) + 8 - 2 a$ .

Simplify the left side.

Remove parentheses.

$20 = a (4) + 8 - 2 a$

$13 = a + 8$

$b = 8 - 2 a$

$20 = a (4) + 8 - 2 a$

$13 = a + 8$

$b = 8 - 2 a$

Simplify the right side.

Simplify $a (4) + 8 - 2 a$ .

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

$20 = 4 a + 8 - 2 a$

$13 = a + 8$

$b = 8 - 2 a$

Subtract $2 a$ from $4 a$ .

$20 = 2 a + 8$

$13 = a + 8$

$b = 8 - 2 a$

$20 = 2 a + 8$

$13 = a + 8$

$b = 8 - 2 a$

$20 = 2 a + 8$

$13 = a + 8$

$b = 8 - 2 a$

$20 = 2 a + 8$

$13 = a + 8$

$b = 8 - 2 a$

$20 = 2 a + 8$

$13 = a + 8$

$b = 8 - 2 a$

Solve for $a$ in $20 = 2 a + 8$ .

Rewrite the equation as $2 a + 8 = 20$ .

$2 a + 8 = 20$

$13 = a + 8$

$b = 8 - 2 a$

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

Subtract $8$ from both sides of the equation.

$2 a = 20 - 8$

$13 = a + 8$

$b = 8 - 2 a$

Subtract $8$ from $20$ .

$2 a = 12$

$13 = a + 8$

$b = 8 - 2 a$

$2 a = 12$

$13 = a + 8$

$b = 8 - 2 a$

Divide each term in $2 a = 12$ by $2$ and simplify.

Divide each term in $2 a = 12$ by $2$ .

$\frac{2 a}{2} = \frac{12}{2}$

$13 = a + 8$

$b = 8 - 2 a$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 a}{2} = \frac{12}{2}$

$13 = a + 8$

$b = 8 - 2 a$

Divide $a$ by $1$ .

$a = \frac{12}{2}$

$13 = a + 8$

$b = 8 - 2 a$

$a = \frac{12}{2}$

$13 = a + 8$

$b = 8 - 2 a$

$a = \frac{12}{2}$

$13 = a + 8$

$b = 8 - 2 a$

Simplify the right side.

Divide $12$ by $2$ .

$a = 6$

$13 = a + 8$

$b = 8 - 2 a$

$a = 6$

$13 = a + 8$

$b = 8 - 2 a$

$a = 6$

$13 = a + 8$

$b = 8 - 2 a$

$a = 6$

$13 = a + 8$

$b = 8 - 2 a$

Replace all occurrences of $a$ with $6$ in each equation.

Replace all occurrences of $a$ in $13 = a + 8$ with $6$ .

$13 = (6) + 8$

$a = 6$

$b = 8 - 2 a$

Simplify the right side.

Add $6$ and $8$ .

$13 = 14$

$a = 6$

$b = 8 - 2 a$

$13 = 14$

$a = 6$

$b = 8 - 2 a$

Replace all occurrences of $a$ in $b = 8 - 2 a$ with $6$ .

$b = 8 - 2 \cdot 6$

$13 = 14$

$a = 6$

Simplify the right side.

Simplify $8 - 2 \cdot 6$ .

Multiply $- 2$ by $6$ .

$b = 8 - 12$

$13 = 14$

$a = 6$

Subtract $12$ from $8$ .

$b = - 4$

$13 = 14$

$a = 6$

$b = - 4$

$13 = 14$

$a = 6$

$b = - 4$

$13 = 14$

$a = 6$

$b = - 4$

$13 = 14$

$a = 6$

Since $13 = 14$ 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 $8 = a \cdot 2^{2} + b (2) + c$ .

Rewrite the equation as $a \cdot 2^{2} + b (2) + c = 8$ .

$a \cdot 2^{2} + b (2) + c = 8$

$13 = a \cdot 3^{2} + b (3) + c$

$20 = a \cdot 4^{2} + b (4) + c$

Simplify each term.

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

$a \cdot 4 + b (2) + c = 8$

$13 = a \cdot 3^{2} + b (3) + c$

$20 = a \cdot 4^{2} + b (4) + c$

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

$4 \cdot a + b (2) + c = 8$

$13 = a \cdot 3^{2} + b (3) + c$

$20 = a \cdot 4^{2} + b (4) + c$

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

$4 a + 2 b + c = 8$

$13 = a \cdot 3^{2} + b (3) + c$

$20 = a \cdot 4^{2} + b (4) + c$

$4 a + 2 b + c = 8$

$13 = a \cdot 3^{2} + b (3) + c$

$20 = a \cdot 4^{2} + b (4) + c$

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

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

$2 b + c = 8 - 4 a$

$13 = a \cdot 3^{2} + b (3) + c$

$20 = a \cdot 4^{2} + b (4) + c$

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

$c = 8 - 4 a - 2 b$

$13 = a \cdot 3^{2} + b (3) + c$

$20 = a \cdot 4^{2} + b (4) + c$

$c = 8 - 4 a - 2 b$

$13 = a \cdot 3^{2} + b (3) + c$

$20 = a \cdot 4^{2} + b (4) + c$

$c = 8 - 4 a - 2 b$

$13 = a \cdot 3^{2} + b (3) + c$

$20 = a \cdot 4^{2} + b (4) + c$

Replace all occurrences of $c$ with $8 - 4 a - 2 b$ in each equation.

Replace all occurrences of $c$ in $13 = a \cdot 3^{2} + b (3) + c$ with $8 - 4 a - 2 b$ .

$13 = a \cdot 3^{2} + b (3) + 8 - 4 a - 2 b$

$c = 8 - 4 a - 2 b$

$20 = a \cdot 4^{2} + b (4) + c$

Simplify $13 = a \cdot 3^{2} + b (3) + 8 - 4 a - 2 b$ .

Simplify the left side.

Remove parentheses.

$13 = a \cdot 3^{2} + b (3) + 8 - 4 a - 2 b$

$c = 8 - 4 a - 2 b$

$20 = a \cdot 4^{2} + b (4) + c$

$13 = a \cdot 3^{2} + b (3) + 8 - 4 a - 2 b$

$c = 8 - 4 a - 2 b$

$20 = a \cdot 4^{2} + b (4) + c$

Simplify the right side.

Simplify $a \cdot 3^{2} + b (3) + 8 - 4 a - 2 b$ .

Simplify each term.

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

$13 = a \cdot 9 + b (3) + 8 - 4 a - 2 b$

$c = 8 - 4 a - 2 b$

$20 = a \cdot 4^{2} + b (4) + c$

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

$13 = 9 \cdot a + b (3) + 8 - 4 a - 2 b$

$c = 8 - 4 a - 2 b$

$20 = a \cdot 4^{2} + b (4) + c$

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

$13 = 9 a + 3 b + 8 - 4 a - 2 b$

$c = 8 - 4 a - 2 b$

$20 = a \cdot 4^{2} + b (4) + c$

$13 = 9 a + 3 b + 8 - 4 a - 2 b$

$c = 8 - 4 a - 2 b$

$20 = a \cdot 4^{2} + b (4) + c$

Simplify by adding terms.

Subtract $4 a$ from $9 a$ .

$13 = 5 a + 3 b + 8 - 2 b$

$c = 8 - 4 a - 2 b$

$20 = a \cdot 4^{2} + b (4) + c$

Subtract $2 b$ from $3 b$ .

$13 = 5 a + b + 8$

$c = 8 - 4 a - 2 b$

$20 = a \cdot 4^{2} + b (4) + c$

$13 = 5 a + b + 8$

$c = 8 - 4 a - 2 b$

$20 = a \cdot 4^{2} + b (4) + c$

$13 = 5 a + b + 8$

$c = 8 - 4 a - 2 b$

$20 = a \cdot 4^{2} + b (4) + c$

$13 = 5 a + b + 8$

$c = 8 - 4 a - 2 b$

$20 = a \cdot 4^{2} + b (4) + c$

$13 = 5 a + b + 8$

$c = 8 - 4 a - 2 b$

$20 = a \cdot 4^{2} + b (4) + c$

Replace all occurrences of $c$ in $20 = a \cdot 4^{2} + b (4) + c$ with $8 - 4 a - 2 b$ .

$20 = a \cdot 4^{2} + b (4) + 8 - 4 a - 2 b$

$13 = 5 a + b + 8$

$c = 8 - 4 a - 2 b$

Simplify $20 = a \cdot 4^{2} + b (4) + 8 - 4 a - 2 b$ .

Simplify the left side.

Remove parentheses.

$20 = a \cdot 4^{2} + b (4) + 8 - 4 a - 2 b$

$13 = 5 a + b + 8$

$c = 8 - 4 a - 2 b$

$20 = a \cdot 4^{2} + b (4) + 8 - 4 a - 2 b$

$13 = 5 a + b + 8$

$c = 8 - 4 a - 2 b$

Simplify the right side.

Simplify $a \cdot 4^{2} + b (4) + 8 - 4 a - 2 b$ .

Simplify each term.

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

$20 = a \cdot 16 + b (4) + 8 - 4 a - 2 b$

$13 = 5 a + b + 8$

$c = 8 - 4 a - 2 b$

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

$20 = 16 \cdot a + b (4) + 8 - 4 a - 2 b$

$13 = 5 a + b + 8$

$c = 8 - 4 a - 2 b$

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

$20 = 16 a + 4 b + 8 - 4 a - 2 b$

$13 = 5 a + b + 8$

$c = 8 - 4 a - 2 b$

$20 = 16 a + 4 b + 8 - 4 a - 2 b$

$13 = 5 a + b + 8$

$c = 8 - 4 a - 2 b$

Simplify by adding terms.

Subtract $4 a$ from $16 a$ .

$20 = 12 a + 4 b + 8 - 2 b$

$13 = 5 a + b + 8$

$c = 8 - 4 a - 2 b$

Subtract $2 b$ from $4 b$ .

$20 = 12 a + 2 b + 8$

$13 = 5 a + b + 8$

$c = 8 - 4 a - 2 b$

$20 = 12 a + 2 b + 8$

$13 = 5 a + b + 8$

$c = 8 - 4 a - 2 b$

$20 = 12 a + 2 b + 8$

$13 = 5 a + b + 8$

$c = 8 - 4 a - 2 b$

$20 = 12 a + 2 b + 8$

$13 = 5 a + b + 8$

$c = 8 - 4 a - 2 b$

$20 = 12 a + 2 b + 8$

$13 = 5 a + b + 8$

$c = 8 - 4 a - 2 b$

$20 = 12 a + 2 b + 8$

$13 = 5 a + b + 8$

$c = 8 - 4 a - 2 b$

Solve for $b$ in $13 = 5 a + b + 8$ .

Rewrite the equation as $5 a + b + 8 = 13$ .

$5 a + b + 8 = 13$

$20 = 12 a + 2 b + 8$

$c = 8 - 4 a - 2 b$

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

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

$b + 8 = 13 - 5 a$

$20 = 12 a + 2 b + 8$

$c = 8 - 4 a - 2 b$

Subtract $8$ from both sides of the equation.

$b = 13 - 5 a - 8$

$20 = 12 a + 2 b + 8$

$c = 8 - 4 a - 2 b$

Subtract $8$ from $13$ .

$b = - 5 a + 5$

$20 = 12 a + 2 b + 8$

$c = 8 - 4 a - 2 b$

$b = - 5 a + 5$

$20 = 12 a + 2 b + 8$

$c = 8 - 4 a - 2 b$

$b = - 5 a + 5$

$20 = 12 a + 2 b + 8$

$c = 8 - 4 a - 2 b$

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

Replace all occurrences of $b$ in $20 = 12 a + 2 b + 8$ with $- 5 a + 5$ .

$20 = 12 a + 2 (- 5 a + 5) + 8$

$b = - 5 a + 5$

$c = 8 - 4 a - 2 b$

Simplify the right side.

Simplify $12 a + 2 (- 5 a + 5) + 8$ .

Simplify each term.

Apply the distributive property.

$20 = 12 a + 2 (- 5 a) + 2 \cdot 5 + 8$

$b = - 5 a + 5$

$c = 8 - 4 a - 2 b$

Multiply $- 5$ by $2$ .

$20 = 12 a - 10 a + 2 \cdot 5 + 8$

$b = - 5 a + 5$

$c = 8 - 4 a - 2 b$

Multiply $2$ by $5$ .

$20 = 12 a - 10 a + 10 + 8$

$b = - 5 a + 5$

$c = 8 - 4 a - 2 b$

$20 = 12 a - 10 a + 10 + 8$

$b = - 5 a + 5$

$c = 8 - 4 a - 2 b$

Simplify by adding terms.

Subtract $10 a$ from $12 a$ .

$20 = 2 a + 10 + 8$

$b = - 5 a + 5$

$c = 8 - 4 a - 2 b$

Add $10$ and $8$ .

$20 = 2 a + 18$

$b = - 5 a + 5$

$c = 8 - 4 a - 2 b$

$20 = 2 a + 18$

$b = - 5 a + 5$

$c = 8 - 4 a - 2 b$

$20 = 2 a + 18$

$b = - 5 a + 5$

$c = 8 - 4 a - 2 b$

$20 = 2 a + 18$

$b = - 5 a + 5$

$c = 8 - 4 a - 2 b$

Replace all occurrences of $b$ in $c = 8 - 4 a - 2 b$ with $- 5 a + 5$ .

$c = 8 - 4 a - 2 (- 5 a + 5)$

$20 = 2 a + 18$

$b = - 5 a + 5$

Simplify the right side.

Simplify $8 - 4 a - 2 (- 5 a + 5)$ .

Simplify each term.

Apply the distributive property.

$c = 8 - 4 a - 2 (- 5 a) - 2 \cdot 5$

$20 = 2 a + 18$

$b = - 5 a + 5$

Multiply $- 5$ by $- 2$ .

$c = 8 - 4 a + 10 a - 2 \cdot 5$

$20 = 2 a + 18$

$b = - 5 a + 5$

Multiply $- 2$ by $5$ .

$c = 8 - 4 a + 10 a - 10$

$20 = 2 a + 18$

$b = - 5 a + 5$

$c = 8 - 4 a + 10 a - 10$

$20 = 2 a + 18$

$b = - 5 a + 5$

Simplify by adding terms.

Subtract $10$ from $8$ .

$c = - 4 a + 10 a - 2$

$20 = 2 a + 18$

$b = - 5 a + 5$

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

$c = 6 a - 2$

$20 = 2 a + 18$

$b = - 5 a + 5$

$c = 6 a - 2$

$20 = 2 a + 18$

$b = - 5 a + 5$

$c = 6 a - 2$

$20 = 2 a + 18$

$b = - 5 a + 5$

$c = 6 a - 2$

$20 = 2 a + 18$

$b = - 5 a + 5$

$c = 6 a - 2$

$20 = 2 a + 18$

$b = - 5 a + 5$

Solve for $a$ in $20 = 2 a + 18$ .

Rewrite the equation as $2 a + 18 = 20$ .

$2 a + 18 = 20$

$c = 6 a - 2$

$b = - 5 a + 5$

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

Subtract $18$ from both sides of the equation.

$2 a = 20 - 18$

$c = 6 a - 2$

$b = - 5 a + 5$

Subtract $18$ from $20$ .

$2 a = 2$

$c = 6 a - 2$

$b = - 5 a + 5$

$2 a = 2$

$c = 6 a - 2$

$b = - 5 a + 5$

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

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

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

$c = 6 a - 2$

$b = - 5 a + 5$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

$c = 6 a - 2$

$b = - 5 a + 5$

Divide $a$ by $1$ .

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

$c = 6 a - 2$

$b = - 5 a + 5$

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

$c = 6 a - 2$

$b = - 5 a + 5$

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

$c = 6 a - 2$

$b = - 5 a + 5$

Simplify the right side.

Divide $2$ by $2$ .

$a = 1$

$c = 6 a - 2$

$b = - 5 a + 5$

$a = 1$

$c = 6 a - 2$

$b = - 5 a + 5$

$a = 1$

$c = 6 a - 2$

$b = - 5 a + 5$

$a = 1$

$c = 6 a - 2$

$b = - 5 a + 5$

Replace all occurrences of $a$ with $1$ in each equation.

Replace all occurrences of $a$ in $c = 6 a - 2$ with $1$ .

$c = 6 (1) - 2$

$a = 1$

$b = - 5 a + 5$

Simplify the right side.

Simplify $6 (1) - 2$ .

Multiply $6$ by $1$ .

$c = 6 - 2$

$a = 1$

$b = - 5 a + 5$

Subtract $2$ from $6$ .

$c = 4$

$a = 1$

$b = - 5 a + 5$

$c = 4$

$a = 1$

$b = - 5 a + 5$

$c = 4$

$a = 1$

$b = - 5 a + 5$

Replace all occurrences of $a$ in $b = - 5 a + 5$ with $1$ .

$b = - 5 \cdot 1 + 5$

$c = 4$

$a = 1$

Simplify the right side.

Simplify $- 5 \cdot 1 + 5$ .

Multiply $- 5$ by $1$ .

$b = - 5 + 5$

$c = 4$

$a = 1$

Add $- 5$ and $5$ .

$b = 0$

$c = 4$

$a = 1$

$b = 0$

$c = 4$

$a = 1$

$b = 0$

$c = 4$

$a = 1$

$b = 0$

$c = 4$

$a = 1$

List all of the solutions.

$b = 0, c = 4, a = 1$

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 $a = 1$ , $b = 0$ , $c = 4$ , and $x = 2$ .

Simplify each term.

Multiply ${(2)}^{2}$ by $1$ .

$y = {(2)}^{2} + (0) \cdot (2) + 4$

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

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

Multiply $0$ by $2$ .

$y = 4 + 0 + 4$

Simplify by adding numbers.

Add $4$ and $0$ .

$y = 4 + 4$

Add $4$ and $4$ .

$y = 8$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 2$ . This check passes since $y = 8$ and $y = 8$ .

$8 = 8$

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = 1$ , $b = 0$ , $c = 4$ , and $x = 3$ .

Simplify each term.

Multiply ${(3)}^{2}$ by $1$ .

$y = {(3)}^{2} + (0) \cdot (3) + 4$

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

$y = 9 + (0) \cdot (3) + 4$

Multiply $0$ by $3$ .

$y = 9 + 0 + 4$

Simplify by adding numbers.

Add $9$ and $0$ .

$y = 9 + 4$

Add $9$ and $4$ .

$y = 13$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 3$ . This check passes since $y = 13$ and $y = 13$ .

$13 = 13$

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = 1$ , $b = 0$ , $c = 4$ , and $x = 4$ .

Simplify each term.

Multiply ${(4)}^{2}$ by $1$ .

$y = {(4)}^{2} + (0) \cdot (4) + 4$

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

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

Multiply $0$ by $4$ .

$y = 16 + 0 + 4$

Simplify by adding numbers.

Add $16$ and $0$ .

$y = 16 + 4$

Add $16$ and $4$ .

$y = 20$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 4$ . This check passes since $y = 20$ and $y = 20$ .

$20 = 20$

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 $y = x^{2} + 4$ .

$y = x^{2} + 4$

Find the Function Rule table[[x,y],[2,8],[3,13],[4,20]]

Related Questions

Support LCMGCF.com by disabling your adblocker.