Find the Function Rule table[[x,y],[-1,3],[0,9],[1,27]]

$\begin{array}{c} x & y \\ - 1 & 3 \\ 0 & 9 \\ 1 & 27 \end{array}$

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 $y = a x + b$ .

$3 = a (- 1) + b 9 = a (0) + b 27 = a (1) + b$

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

Rewrite the equation as $b = 9$ .

$b = 9$

$3 = a (- 1) + b$

$27 = a + b$

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

Replace all occurrences of $b$ in $3 = a (- 1) + b$ with $9$ .

$3 = a (- 1) + 9$

$b = 9$

$27 = a + b$

Simplify $3 = a (- 1) + 9$ .

Simplify the left side.

Remove parentheses.

$3 = a (- 1) + 9$

$b = 9$

$27 = a + b$

$3 = a (- 1) + 9$

$b = 9$

$27 = a + b$

Simplify the right side.

Simplify each term.

Move $- 1$ to the left of $a$ .

$3 = - 1 a + 9$

$b = 9$

$27 = a + b$

Rewrite $- 1 a$ as $- a$ .

$3 = - a + 9$

$b = 9$

$27 = a + b$

$3 = - a + 9$

$b = 9$

$27 = a + b$

$3 = - a + 9$

$b = 9$

$27 = a + b$

$3 = - a + 9$

$b = 9$

$27 = a + b$

Replace all occurrences of $b$ in $27 = a + b$ with $9$ .

$27 = a + 9$

$3 = - a + 9$

$b = 9$

Simplify the left side.

Remove parentheses.

$27 = a + 9$

$3 = - a + 9$

$b = 9$

$27 = a + 9$

$3 = - a + 9$

$b = 9$

$27 = a + 9$

$3 = - a + 9$

$b = 9$

Solve for $a$ in $27 = a + 9$ .

Rewrite the equation as $a + 9 = 27$ .

$a + 9 = 27$

$3 = - a + 9$

$b = 9$

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

Subtract $9$ from both sides of the equation.

$a = 27 - 9$

$3 = - a + 9$

$b = 9$

Subtract $9$ from $27$ .

$a = 18$

$3 = - a + 9$

$b = 9$

$a = 18$

$3 = - a + 9$

$b = 9$

$a = 18$

$3 = - a + 9$

$b = 9$

Replace all occurrences of $a$ with $18$ in each equation.

Replace all occurrences of $a$ in $3 = - a + 9$ with $18$ .

$3 = - (18) + 9$

$a = 18$

$b = 9$

Simplify the right side.

Simplify $- (18) + 9$ .

Multiply $- 1$ by $18$ .

$3 = - 18 + 9$

$a = 18$

$b = 9$

Add $- 18$ and $9$ .

$3 = - 9$

$a = 18$

$b = 9$

$3 = - 9$

$a = 18$

$b = 9$

$3 = - 9$

$a = 18$

$b = 9$

$3 = - 9$

$a = 18$

$b = 9$

Since $3 = - 9$ 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 $9 = a \cdot 0^{2} + c$ .

Rewrite the equation as $a \cdot 0^{2} + c = 9$ .

$a \cdot 0^{2} + c = 9$

$3 = a {(- 1)}^{2} + b (- 1) + c$

$27 = a + b + c$

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

Simplify each term.

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

$a \cdot 0 + c = 9$

$3 = a {(- 1)}^{2} + b (- 1) + c$

$27 = a + b + c$

Multiply $a$ by $0$ .

$0 + c = 9$

$3 = a {(- 1)}^{2} + b (- 1) + c$

$27 = a + b + c$

$0 + c = 9$

$3 = a {(- 1)}^{2} + b (- 1) + c$

$27 = a + b + c$

Add $0$ and $c$ .

$c = 9$

$3 = a {(- 1)}^{2} + b (- 1) + c$

$27 = a + b + c$

$c = 9$

$3 = a {(- 1)}^{2} + b (- 1) + c$

$27 = a + b + c$

$c = 9$

$3 = a {(- 1)}^{2} + b (- 1) + c$

$27 = a + b + c$

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

Replace all occurrences of $c$ in $3 = a {(- 1)}^{2} + b (- 1) + c$ with $9$ .

$3 = a {(- 1)}^{2} + b (- 1) + 9$

$c = 9$

$27 = a + b + c$

Simplify $3 = a {(- 1)}^{2} + b (- 1) + 9$ .

Simplify the left side.

Remove parentheses.

$3 = a {(- 1)}^{2} + b (- 1) + 9$

$c = 9$

$27 = a + b + c$

$3 = a {(- 1)}^{2} + b (- 1) + 9$

$c = 9$

$27 = a + b + c$

Simplify the right side.

Simplify each term.

Raise $- 1$ to the power of $2$ .

$3 = a \cdot 1 + b (- 1) + 9$

$c = 9$

$27 = a + b + c$

Multiply $a$ by $1$ .

$3 = a + b (- 1) + 9$

$c = 9$

$27 = a + b + c$

Move $- 1$ to the left of $b$ .

$3 = a - 1 b + 9$

$c = 9$

$27 = a + b + c$

Rewrite $- 1 b$ as $- b$ .

$3 = a - b + 9$

$c = 9$

$27 = a + b + c$

$3 = a - b + 9$

$c = 9$

$27 = a + b + c$

$3 = a - b + 9$

$c = 9$

$27 = a + b + c$

$3 = a - b + 9$

$c = 9$

$27 = a + b + c$

Replace all occurrences of $c$ in $27 = a + b + c$ with $9$ .

$27 = a + b + 9$

$3 = a - b + 9$

$c = 9$

Simplify the left side.

Remove parentheses.

$27 = a + b + 9$

$3 = a - b + 9$

$c = 9$

$27 = a + b + 9$

$3 = a - b + 9$

$c = 9$

$27 = a + b + 9$

$3 = a - b + 9$

$c = 9$

Solve for $a$ in $27 = a + b + 9$ .

Rewrite the equation as $a + b + 9 = 27$ .

$a + b + 9 = 27$

$3 = a - b + 9$

$c = 9$

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

Subtract $b$ from both sides of the equation.

$a + 9 = 27 - b$

$3 = a - b + 9$

$c = 9$

Subtract $9$ from both sides of the equation.

$a = 27 - b - 9$

$3 = a - b + 9$

$c = 9$

Subtract $9$ from $27$ .

$a = - b + 18$

$3 = a - b + 9$

$c = 9$

$a = - b + 18$

$3 = a - b + 9$

$c = 9$

$a = - b + 18$

$3 = a - b + 9$

$c = 9$

Replace all occurrences of $a$ with $- b + 18$ in each equation.

Replace all occurrences of $a$ in $3 = a - b + 9$ with $- b + 18$ .

$3 = (- b + 18) - b + 9$

$a = - b + 18$

$c = 9$

Simplify the right side.

Simplify $(- b + 18) - b + 9$ .

Subtract $b$ from $- b$ .

$3 = - 2 b + 18 + 9$

$a = - b + 18$

$c = 9$

Add $18$ and $9$ .

$3 = - 2 b + 27$

$a = - b + 18$

$c = 9$

$3 = - 2 b + 27$

$a = - b + 18$

$c = 9$

$3 = - 2 b + 27$

$a = - b + 18$

$c = 9$

$3 = - 2 b + 27$

$a = - b + 18$

$c = 9$

Solve for $b$ in $3 = - 2 b + 27$ .

Rewrite the equation as $- 2 b + 27 = 3$ .

$- 2 b + 27 = 3$

$a = - b + 18$

$c = 9$

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

Subtract $27$ from both sides of the equation.

$- 2 b = 3 - 27$

$a = - b + 18$

$c = 9$

Subtract $27$ from $3$ .

$- 2 b = - 24$

$a = - b + 18$

$c = 9$

$- 2 b = - 24$

$a = - b + 18$

$c = 9$

Divide each term in $- 2 b = - 24$ by $- 2$ and simplify.

Divide each term in $- 2 b = - 24$ by $- 2$ .

$\frac{- 2 b}{- 2} = \frac{- 24}{- 2}$

$a = - b + 18$

$c = 9$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 b}{- 2} = \frac{- 24}{- 2}$

$a = - b + 18$

$c = 9$

Divide $b$ by $1$ .

$b = \frac{- 24}{- 2}$

$a = - b + 18$

$c = 9$

$b = \frac{- 24}{- 2}$

$a = - b + 18$

$c = 9$

$b = \frac{- 24}{- 2}$

$a = - b + 18$

$c = 9$

Simplify the right side.

Divide $- 24$ by $- 2$ .

$b = 12$

$a = - b + 18$

$c = 9$

$b = 12$

$a = - b + 18$

$c = 9$

$b = 12$

$a = - b + 18$

$c = 9$

$b = 12$

$a = - b + 18$

$c = 9$

Replace all occurrences of $b$ with $12$ in each equation.

Replace all occurrences of $b$ in $a = - b + 18$ with $12$ .

$a = - (12) + 18$

$b = 12$

$c = 9$

Simplify the right side.

Simplify $- (12) + 18$ .

Multiply $- 1$ by $12$ .

$a = - 12 + 18$

$b = 12$

$c = 9$

Add $- 12$ and $18$ .

$a = 6$

$b = 12$

$c = 9$

$a = 6$

$b = 12$

$c = 9$

$a = 6$

$b = 12$

$c = 9$

$a = 6$

$b = 12$

$c = 9$

List all of the solutions.

$a = 6, b = 12, c = 9$

$a = 6, b = 12, c = 9$

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 $a = 6$ , $b = 12$ , $c = 9$ , and $x = - 1$ .

Simplify each term.

Raise $- 1$ to the power of $2$ .

$y = 6 \cdot 1 + (12) \cdot (- 1) + 9$

Multiply $6$ by $1$ .

$y = 6 + (12) \cdot (- 1) + 9$

Multiply $12$ by $- 1$ .

$y = 6 - 12 + 9$

$y = 6 - 12 + 9$

Simplify by adding and subtracting.

Subtract $12$ from $6$ .

$y = - 6 + 9$

Add $- 6$ and $9$ .

$y = 3$

$y = 3$

$y = 3$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = - 1$ . This check passes since $y = 3$ and $y = 3$ .

$3 = 3$

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = 6$ , $b = 12$ , $c = 9$ , and $x = 0$ .

Simplify each term.

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

$y = 6 \cdot 0 + (12) \cdot (0) + 9$

Multiply $6$ by $0$ .

$y = 0 + (12) \cdot (0) + 9$

Multiply $12$ by $0$ .

$y = 0 + 0 + 9$

$y = 0 + 0 + 9$

Simplify by adding numbers.

Add $0$ and $0$ .

$y = 0 + 9$

Add $0$ and $9$ .

$y = 9$

$y = 9$

$y = 9$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 0$ . This check passes since $y = 9$ and $y = 9$ .

$9 = 9$

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = 6$ , $b = 12$ , $c = 9$ , and $x = 1$ .

Simplify each term.

One to any power is one.

$y = 6 \cdot 1 + (12) \cdot (1) + 9$

Multiply $6$ by $1$ .

$y = 6 + (12) \cdot (1) + 9$

Multiply $12$ by $1$ .

$y = 6 + 12 + 9$

$y = 6 + 12 + 9$

Simplify by adding numbers.

Add $6$ and $12$ .

$y = 18 + 9$

Add $18$ and $9$ .

$y = 27$

$y = 27$

$y = 27$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 1$ . This check passes since $y = 27$ and $y = 27$ .

$27 = 27$

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 $y = 6 x^{2} + 12 x + 9$ .

$y = 6 x^{2} + 12 x + 9$

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