Find the Function Rule table[[x,y],[0,1],[1,2],[2,9]]

$\begin{array}{c} x & y \\ 0 & 1 \\ 1 & 2 \\ 2 & 9 \end{array}$

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 $y = a x + b$ .

$1 = a (0) + b 2 = a (1) + b 9 = a (2) + b$

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

Rewrite the equation as $b = 1$ .

$b = 1$

$2 = a + b$

$9 = a (2) + b$

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

Replace all occurrences of $b$ in $2 = a + b$ with $1$ .

$2 = a + 1$

$b = 1$

$9 = a (2) + b$

Simplify the left side.

Remove parentheses.

$2 = a + 1$

$b = 1$

$9 = a (2) + b$

$2 = a + 1$

$b = 1$

$9 = a (2) + b$

Replace all occurrences of $b$ in $9 = a (2) + b$ with $1$ .

$9 = a (2) + 1$

$2 = a + 1$

$b = 1$

Simplify $9 = a (2) + 1$ .

Simplify the left side.

Remove parentheses.

$9 = a (2) + 1$

$2 = a + 1$

$b = 1$

$9 = a (2) + 1$

$2 = a + 1$

$b = 1$

Simplify the right side.

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

$9 = 2 a + 1$

$2 = a + 1$

$b = 1$

$9 = 2 a + 1$

$2 = a + 1$

$b = 1$

$9 = 2 a + 1$

$2 = a + 1$

$b = 1$

$9 = 2 a + 1$

$2 = a + 1$

$b = 1$

Solve for $a$ in $9 = 2 a + 1$ .

Rewrite the equation as $2 a + 1 = 9$ .

$2 a + 1 = 9$

$2 = a + 1$

$b = 1$

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

Subtract $1$ from both sides of the equation.

$2 a = 9 - 1$

$2 = a + 1$

$b = 1$

Subtract $1$ from $9$ .

$2 a = 8$

$2 = a + 1$

$b = 1$

$2 a = 8$

$2 = a + 1$

$b = 1$

Divide each term in $2 a = 8$ by $2$ and simplify.

Divide each term in $2 a = 8$ by $2$ .

$\frac{2 a}{2} = \frac{8}{2}$

$2 = a + 1$

$b = 1$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 a}{2} = \frac{8}{2}$

$2 = a + 1$

$b = 1$

Divide $a$ by $1$ .

$a = \frac{8}{2}$

$2 = a + 1$

$b = 1$

$a = \frac{8}{2}$

$2 = a + 1$

$b = 1$

$a = \frac{8}{2}$

$2 = a + 1$

$b = 1$

Simplify the right side.

Divide $8$ by $2$ .

$a = 4$

$2 = a + 1$

$b = 1$

$a = 4$

$2 = a + 1$

$b = 1$

$a = 4$

$2 = a + 1$

$b = 1$

$a = 4$

$2 = a + 1$

$b = 1$

Replace all occurrences of $a$ with $4$ in each equation.

Replace all occurrences of $a$ in $2 = a + 1$ with $4$ .

$2 = (4) + 1$

$a = 4$

$b = 1$

Simplify the right side.

Add $4$ and $1$ .

$2 = 5$

$a = 4$

$b = 1$

$2 = 5$

$a = 4$

$b = 1$

$2 = 5$

$a = 4$

$b = 1$

Since $2 = 5$ is not true, there is no solution.

No solution

No solution

Since $y \neq y$ for the corresponding $x$ values, the function is not linear.

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 $1 = a \cdot 0^{2} + c$ .

Rewrite the equation as $a \cdot 0^{2} + c = 1$ .

$a \cdot 0^{2} + c = 1$

$2 = a + b + c$

$9 = a \cdot 2^{2} + b (2) + c$

Simplify $a \cdot 0^{2} + c$ .

Simplify each term.

Raising $0$ to any positive power yields $0$ .

$a \cdot 0 + c = 1$

$2 = a + b + c$

$9 = a \cdot 2^{2} + b (2) + c$

Multiply $a$ by $0$ .

$0 + c = 1$

$2 = a + b + c$

$9 = a \cdot 2^{2} + b (2) + c$

$0 + c = 1$

$2 = a + b + c$

$9 = a \cdot 2^{2} + b (2) + c$

Add $0$ and $c$ .

$c = 1$

$2 = a + b + c$

$9 = a \cdot 2^{2} + b (2) + c$

$c = 1$

$2 = a + b + c$

$9 = a \cdot 2^{2} + b (2) + c$

$c = 1$

$2 = a + b + c$

$9 = a \cdot 2^{2} + b (2) + c$

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

Replace all occurrences of $c$ in $2 = a + b + c$ with $1$ .

$2 = a + b + 1$

$c = 1$

$9 = a \cdot 2^{2} + b (2) + c$

Simplify the left side.

Remove parentheses.

$2 = a + b + 1$

$c = 1$

$9 = a \cdot 2^{2} + b (2) + c$

$2 = a + b + 1$

$c = 1$

$9 = a \cdot 2^{2} + b (2) + c$

Replace all occurrences of $c$ in $9 = a \cdot 2^{2} + b (2) + c$ with $1$ .

$9 = a \cdot 2^{2} + b (2) + 1$

$2 = a + b + 1$

$c = 1$

Simplify $9 = a \cdot 2^{2} + b (2) + 1$ .

Simplify the left side.

Remove parentheses.

$9 = a \cdot 2^{2} + b (2) + 1$

$2 = a + b + 1$

$c = 1$

$9 = a \cdot 2^{2} + b (2) + 1$

$2 = a + b + 1$

$c = 1$

Simplify the right side.

Simplify each term.

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

$9 = a \cdot 4 + b (2) + 1$

$2 = a + b + 1$

$c = 1$

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

$9 = 4 \cdot a + b (2) + 1$

$2 = a + b + 1$

$c = 1$

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

$9 = 4 a + 2 b + 1$

$2 = a + b + 1$

$c = 1$

$9 = 4 a + 2 b + 1$

$2 = a + b + 1$

$c = 1$

$9 = 4 a + 2 b + 1$

$2 = a + b + 1$

$c = 1$

$9 = 4 a + 2 b + 1$

$2 = a + b + 1$

$c = 1$

$9 = 4 a + 2 b + 1$

$2 = a + b + 1$

$c = 1$

Solve for $a$ in $2 = a + b + 1$ .

Rewrite the equation as $a + b + 1 = 2$ .

$a + b + 1 = 2$

$9 = 4 a + 2 b + 1$

$c = 1$

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

Subtract $b$ from both sides of the equation.

$a + 1 = 2 - b$

$9 = 4 a + 2 b + 1$

$c = 1$

Subtract $1$ from both sides of the equation.

$a = 2 - b - 1$

$9 = 4 a + 2 b + 1$

$c = 1$

Subtract $1$ from $2$ .

$a = - b + 1$

$9 = 4 a + 2 b + 1$

$c = 1$

$a = - b + 1$

$9 = 4 a + 2 b + 1$

$c = 1$

$a = - b + 1$

$9 = 4 a + 2 b + 1$

$c = 1$

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

Replace all occurrences of $a$ in $9 = 4 a + 2 b + 1$ with $- b + 1$ .

$9 = 4 (- b + 1) + 2 b + 1$

$a = - b + 1$

$c = 1$

Simplify the right side.

Simplify $4 (- b + 1) + 2 b + 1$ .

Simplify each term.

Apply the distributive property.

$9 = 4 (- b) + 4 \cdot 1 + 2 b + 1$

$a = - b + 1$

$c = 1$

Multiply $- 1$ by $4$ .

$9 = - 4 b + 4 \cdot 1 + 2 b + 1$

$a = - b + 1$

$c = 1$

Multiply $4$ by $1$ .

$9 = - 4 b + 4 + 2 b + 1$

$a = - b + 1$

$c = 1$

$9 = - 4 b + 4 + 2 b + 1$

$a = - b + 1$

$c = 1$

Simplify by adding terms.

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

$9 = - 2 b + 4 + 1$

$a = - b + 1$

$c = 1$

Add $4$ and $1$ .

$9 = - 2 b + 5$

$a = - b + 1$

$c = 1$

$9 = - 2 b + 5$

$a = - b + 1$

$c = 1$

$9 = - 2 b + 5$

$a = - b + 1$

$c = 1$

$9 = - 2 b + 5$

$a = - b + 1$

$c = 1$

$9 = - 2 b + 5$

$a = - b + 1$

$c = 1$

Solve for $b$ in $9 = - 2 b + 5$ .

Rewrite the equation as $- 2 b + 5 = 9$ .

$- 2 b + 5 = 9$

$a = - b + 1$

$c = 1$

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

Subtract $5$ from both sides of the equation.

$- 2 b = 9 - 5$

$a = - b + 1$

$c = 1$

Subtract $5$ from $9$ .

$- 2 b = 4$

$a = - b + 1$

$c = 1$

$- 2 b = 4$

$a = - b + 1$

$c = 1$

Divide each term in $- 2 b = 4$ by $- 2$ and simplify.

Divide each term in $- 2 b = 4$ by $- 2$ .

$\frac{- 2 b}{- 2} = \frac{4}{- 2}$

$a = - b + 1$

$c = 1$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 b}{- 2} = \frac{4}{- 2}$

$a = - b + 1$

$c = 1$

Divide $b$ by $1$ .

$b = \frac{4}{- 2}$

$a = - b + 1$

$c = 1$

$b = \frac{4}{- 2}$

$a = - b + 1$

$c = 1$

$b = \frac{4}{- 2}$

$a = - b + 1$

$c = 1$

Simplify the right side.

Divide $4$ by $- 2$ .

$b = - 2$

$a = - b + 1$

$c = 1$

$b = - 2$

$a = - b + 1$

$c = 1$

$b = - 2$

$a = - b + 1$

$c = 1$

$b = - 2$

$a = - b + 1$

$c = 1$

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

Replace all occurrences of $b$ in $a = - b + 1$ with $- 2$ .

$a = - (- 2) + 1$

$b = - 2$

$c = 1$

Simplify the right side.

Simplify $- (- 2) + 1$ .

Multiply $- 1$ by $- 2$ .

$a = 2 + 1$

$b = - 2$

$c = 1$

Add $2$ and $1$ .

$a = 3$

$b = - 2$

$c = 1$

$a = 3$

$b = - 2$

$c = 1$

$a = 3$

$b = - 2$

$c = 1$

$a = 3$

$b = - 2$

$c = 1$

List all of the solutions.

$a = 3, b = - 2, c = 1$

$a = 3, b = - 2, c = 1$

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 $a = 3$ , $b = - 2$ , $c = 1$ , and $x = 0$ .

Simplify each term.

Raising $0$ to any positive power yields $0$ .

$y = 3 \cdot 0 + (- 2) \cdot (0) + 1$

Multiply $3$ by $0$ .

$y = 0 + (- 2) \cdot (0) + 1$

Multiply $- 2$ by $0$ .

$y = 0 + 0 + 1$

$y = 0 + 0 + 1$

Simplify by adding numbers.

Add $0$ and $0$ .

$y = 0 + 1$

Add $0$ and $1$ .

$y = 1$

$y = 1$

$y = 1$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 0$ . This check passes since $y = 1$ and $y = 1$ .

$1 = 1$

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

Simplify each term.

One to any power is one.

$y = 3 \cdot 1 + (- 2) \cdot (1) + 1$

Multiply $3$ by $1$ .

$y = 3 + (- 2) \cdot (1) + 1$

Multiply $- 2$ by $1$ .

$y = 3 - 2 + 1$

$y = 3 - 2 + 1$

Simplify by adding and subtracting.

Subtract $2$ from $3$ .

$y = 1 + 1$

Add $1$ and $1$ .

$y = 2$

$y = 2$

$y = 2$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 1$ . This check passes since $y = 2$ and $y = 2$ .

$2 = 2$

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

Simplify each term.

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

$y = 3 \cdot 4 + (- 2) \cdot (2) + 1$

Multiply $3$ by $4$ .

$y = 12 + (- 2) \cdot (2) + 1$

Multiply $- 2$ by $2$ .

$y = 12 - 4 + 1$

$y = 12 - 4 + 1$

Simplify by adding and subtracting.

Subtract $4$ from $12$ .

$y = 8 + 1$

Add $8$ and $1$ .

$y = 9$

$y = 9$

$y = 9$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 2$ . This check passes since $y = 9$ and $y = 9$ .

$9 = 9$

Since $y = y$ for the corresponding $x$ values, the function is quadratic.

The function is quadratic

The function is quadratic

The function is quadratic

Step 3

Since all $y = y$ , the function is quadratic and follows the form $y = 3 x^{2} - 2 x + 1$ .

$y = 3 x^{2} - 2 x + 1$

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