Find the Function Rule table[[x,y],[0,4],[1,8],[2,16]]

$\begin{array}{c} x & y \\ 0 & 4 \\ 1 & 8 \\ 2 & 16 \end{array}$

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 $y = a x + b$ .

$4 = a (0) + b 8 = a (1) + b 16 = a (2) + b$

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

Rewrite the equation as $b = 4$ .

$b = 4$

$8 = a + b$

$16 = a (2) + b$

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

Replace all occurrences of $b$ in $8 = a + b$ with $4$ .

$8 = a + 4$

$b = 4$

$16 = a (2) + b$

Simplify the left side.

Remove parentheses.

$8 = a + 4$

$b = 4$

$16 = a (2) + b$

$8 = a + 4$

$b = 4$

$16 = a (2) + b$

Replace all occurrences of $b$ in $16 = a (2) + b$ with $4$ .

$16 = a (2) + 4$

$8 = a + 4$

$b = 4$

Simplify $16 = a (2) + 4$ .

Simplify the left side.

Remove parentheses.

$16 = a (2) + 4$

$8 = a + 4$

$b = 4$

$16 = a (2) + 4$

$8 = a + 4$

$b = 4$

Simplify the right side.

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

$16 = 2 a + 4$

$8 = a + 4$

$b = 4$

$16 = 2 a + 4$

$8 = a + 4$

$b = 4$

$16 = 2 a + 4$

$8 = a + 4$

$b = 4$

$16 = 2 a + 4$

$8 = a + 4$

$b = 4$

Solve for $a$ in $16 = 2 a + 4$ .

Rewrite the equation as $2 a + 4 = 16$ .

$2 a + 4 = 16$

$8 = a + 4$

$b = 4$

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

Subtract $4$ from both sides of the equation.

$2 a = 16 - 4$

$8 = a + 4$

$b = 4$

Subtract $4$ from $16$ .

$2 a = 12$

$8 = a + 4$

$b = 4$

$2 a = 12$

$8 = a + 4$

$b = 4$

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

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

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

$8 = a + 4$

$b = 4$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

$8 = a + 4$

$b = 4$

Divide $a$ by $1$ .

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

$8 = a + 4$

$b = 4$

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

$8 = a + 4$

$b = 4$

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

$8 = a + 4$

$b = 4$

Simplify the right side.

Divide $12$ by $2$ .

$a = 6$

$8 = a + 4$

$b = 4$

$a = 6$

$8 = a + 4$

$b = 4$

$a = 6$

$8 = a + 4$

$b = 4$

$a = 6$

$8 = a + 4$

$b = 4$

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

Replace all occurrences of $a$ in $8 = a + 4$ with $6$ .

$8 = (6) + 4$

$a = 6$

$b = 4$

Simplify the right side.

Add $6$ and $4$ .

$8 = 10$

$a = 6$

$b = 4$

$8 = 10$

$a = 6$

$b = 4$

$8 = 10$

$a = 6$

$b = 4$

Since $8 = 10$ 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 $4 = a \cdot 0^{2} + c$ .

Rewrite the equation as $a \cdot 0^{2} + c = 4$ .

$a \cdot 0^{2} + c = 4$

$8 = a + b + c$

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

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

Simplify each term.

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

$a \cdot 0 + c = 4$

$8 = a + b + c$

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

Multiply $a$ by $0$ .

$0 + c = 4$

$8 = a + b + c$

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

$0 + c = 4$

$8 = a + b + c$

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

Add $0$ and $c$ .

$c = 4$

$8 = a + b + c$

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

$c = 4$

$8 = a + b + c$

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

$c = 4$

$8 = a + b + c$

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

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

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

$8 = a + b + 4$

$c = 4$

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

Simplify the left side.

Remove parentheses.

$8 = a + b + 4$

$c = 4$

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

$8 = a + b + 4$

$c = 4$

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

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

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

$8 = a + b + 4$

$c = 4$

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

Simplify the left side.

Remove parentheses.

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

$8 = a + b + 4$

$c = 4$

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

$8 = a + b + 4$

$c = 4$

Simplify the right side.

Simplify each term.

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

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

$8 = a + b + 4$

$c = 4$

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

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

$8 = a + b + 4$

$c = 4$

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

$16 = 4 a + 2 b + 4$

$8 = a + b + 4$

$c = 4$

$16 = 4 a + 2 b + 4$

$8 = a + b + 4$

$c = 4$

$16 = 4 a + 2 b + 4$

$8 = a + b + 4$

$c = 4$

$16 = 4 a + 2 b + 4$

$8 = a + b + 4$

$c = 4$

$16 = 4 a + 2 b + 4$

$8 = a + b + 4$

$c = 4$

Solve for $a$ in $8 = a + b + 4$ .

Rewrite the equation as $a + b + 4 = 8$ .

$a + b + 4 = 8$

$16 = 4 a + 2 b + 4$

$c = 4$

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

Subtract $b$ from both sides of the equation.

$a + 4 = 8 - b$

$16 = 4 a + 2 b + 4$

$c = 4$

Subtract $4$ from both sides of the equation.

$a = 8 - b - 4$

$16 = 4 a + 2 b + 4$

$c = 4$

Subtract $4$ from $8$ .

$a = - b + 4$

$16 = 4 a + 2 b + 4$

$c = 4$

$a = - b + 4$

$16 = 4 a + 2 b + 4$

$c = 4$

$a = - b + 4$

$16 = 4 a + 2 b + 4$

$c = 4$

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

Replace all occurrences of $a$ in $16 = 4 a + 2 b + 4$ with $- b + 4$ .

$16 = 4 (- b + 4) + 2 b + 4$

$a = - b + 4$

$c = 4$

Simplify the right side.

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

Simplify each term.

Apply the distributive property.

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

$a = - b + 4$

$c = 4$

Multiply $- 1$ by $4$ .

$16 = - 4 b + 4 \cdot 4 + 2 b + 4$

$a = - b + 4$

$c = 4$

Multiply $4$ by $4$ .

$16 = - 4 b + 16 + 2 b + 4$

$a = - b + 4$

$c = 4$

$16 = - 4 b + 16 + 2 b + 4$

$a = - b + 4$

$c = 4$

Simplify by adding terms.

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

$16 = - 2 b + 16 + 4$

$a = - b + 4$

$c = 4$

Add $16$ and $4$ .

$16 = - 2 b + 20$

$a = - b + 4$

$c = 4$

$16 = - 2 b + 20$

$a = - b + 4$

$c = 4$

$16 = - 2 b + 20$

$a = - b + 4$

$c = 4$

$16 = - 2 b + 20$

$a = - b + 4$

$c = 4$

$16 = - 2 b + 20$

$a = - b + 4$

$c = 4$

Solve for $b$ in $16 = - 2 b + 20$ .

Rewrite the equation as $- 2 b + 20 = 16$ .

$- 2 b + 20 = 16$

$a = - b + 4$

$c = 4$

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

Subtract $20$ from both sides of the equation.

$- 2 b = 16 - 20$

$a = - b + 4$

$c = 4$

Subtract $20$ from $16$ .

$- 2 b = - 4$

$a = - b + 4$

$c = 4$

$- 2 b = - 4$

$a = - b + 4$

$c = 4$

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

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

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

$a = - b + 4$

$c = 4$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

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

$a = - b + 4$

$c = 4$

Divide $b$ by $1$ .

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

$a = - b + 4$

$c = 4$

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

$a = - b + 4$

$c = 4$

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

$a = - b + 4$

$c = 4$

Simplify the right side.

Divide $- 4$ by $- 2$ .

$b = 2$

$a = - b + 4$

$c = 4$

$b = 2$

$a = - b + 4$

$c = 4$

$b = 2$

$a = - b + 4$

$c = 4$

$b = 2$

$a = - b + 4$

$c = 4$

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

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

$a = - (2) + 4$

$b = 2$

$c = 4$

Simplify the right side.

Simplify $- (2) + 4$ .

Multiply $- 1$ by $2$ .

$a = - 2 + 4$

$b = 2$

$c = 4$

Add $- 2$ and $4$ .

$a = 2$

$b = 2$

$c = 4$

$a = 2$

$b = 2$

$c = 4$

$a = 2$

$b = 2$

$c = 4$

$a = 2$

$b = 2$

$c = 4$

List all of the solutions.

$a = 2, b = 2, c = 4$

$a = 2, b = 2, c = 4$

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

Simplify each term.

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

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

Multiply $2$ by $0$ .

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

Multiply $2$ by $0$ .

$y = 0 + 0 + 4$

$y = 0 + 0 + 4$

Simplify by adding numbers.

Add $0$ and $0$ .

$y = 0 + 4$

Add $0$ and $4$ .

$y = 4$

$y = 4$

$y = 4$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 0$ . This check passes since $y = 4$ and $y = 4$ .

$4 = 4$

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

Simplify each term.

One to any power is one.

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

Multiply $2$ by $1$ .

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

Multiply $2$ by $1$ .

$y = 2 + 2 + 4$

$y = 2 + 2 + 4$

Simplify by adding numbers.

Add $2$ and $2$ .

$y = 4 + 4$

Add $4$ and $4$ .

$y = 8$

$y = 8$

$y = 8$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 1$ . This check passes since $y = 8$ and $y = 8$ .

$8 = 8$

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

Simplify each term.

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

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

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

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

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

$y = 2^{1 + 2} + (2) \cdot (2) + 4$

$y = 2^{1 + 2} + (2) \cdot (2) + 4$

Add $1$ and $2$ .

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

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

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

$y = 8 + (2) \cdot (2) + 4$

Multiply $2$ by $2$ .

$y = 8 + 4 + 4$

$y = 8 + 4 + 4$

Simplify by adding numbers.

Add $8$ and $4$ .

$y = 12 + 4$

Add $12$ and $4$ .

$y = 16$

$y = 16$

$y = 16$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 2$ . This check passes since $y = 16$ and $y = 16$ .

$16 = 16$

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

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

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