Find the Function Rule table[[x,y],[0,3],[1,6],[2,12]]

$\begin{array}{c} x & y \\ 0 & 3 \\ 1 & 6 \\ 2 & 12 \end{array}$

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 $y = a x + b$ .

$3 = a (0) + b 6 = a (1) + b 12 = a (2) + b$

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

Rewrite the equation as $b = 3$ .

$b = 3$

$6 = a + b$

$12 = a (2) + b$

Replace all occurrences of $b$ with $3$ in each equation.

Replace all occurrences of $b$ in $6 = a + b$ with $3$ .

$6 = a + 3$

$b = 3$

$12 = a (2) + b$

Simplify the left side.

Remove parentheses.

$6 = a + 3$

$b = 3$

$12 = a (2) + b$

$6 = a + 3$

$b = 3$

$12 = a (2) + b$

Replace all occurrences of $b$ in $12 = a (2) + b$ with $3$ .

$12 = a (2) + 3$

$6 = a + 3$

$b = 3$

Simplify $12 = a (2) + 3$ .

Simplify the left side.

Remove parentheses.

$12 = a (2) + 3$

$6 = a + 3$

$b = 3$

$12 = a (2) + 3$

$6 = a + 3$

$b = 3$

Simplify the right side.

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

$12 = 2 a + 3$

$6 = a + 3$

$b = 3$

$12 = 2 a + 3$

$6 = a + 3$

$b = 3$

$12 = 2 a + 3$

$6 = a + 3$

$b = 3$

$12 = 2 a + 3$

$6 = a + 3$

$b = 3$

Solve for $a$ in $12 = 2 a + 3$ .

Rewrite the equation as $2 a + 3 = 12$ .

$2 a + 3 = 12$

$6 = a + 3$

$b = 3$

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

Subtract $3$ from both sides of the equation.

$2 a = 12 - 3$

$6 = a + 3$

$b = 3$

Subtract $3$ from $12$ .

$2 a = 9$

$6 = a + 3$

$b = 3$

$2 a = 9$

$6 = a + 3$

$b = 3$

Divide each term in $2 a = 9$ by $2$ and simplify.

Divide each term in $2 a = 9$ by $2$ .

$\frac{2 a}{2} = \frac{9}{2}$

$6 = a + 3$

$b = 3$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 a}{2} = \frac{9}{2}$

$6 = a + 3$

$b = 3$

Divide $a$ by $1$ .

$a = \frac{9}{2}$

$6 = a + 3$

$b = 3$

$a = \frac{9}{2}$

$6 = a + 3$

$b = 3$

$a = \frac{9}{2}$

$6 = a + 3$

$b = 3$

$a = \frac{9}{2}$

$6 = a + 3$

$b = 3$

$a = \frac{9}{2}$

$6 = a + 3$

$b = 3$

Replace all occurrences of $a$ with $\frac{9}{2}$ in each equation.

Replace all occurrences of $a$ in $6 = a + 3$ with $\frac{9}{2}$ .

$6 = (\frac{9}{2}) + 3$

$a = \frac{9}{2}$

$b = 3$

Simplify the right side.

Simplify $(\frac{9}{2}) + 3$ .

To write $3$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$6 = \frac{9}{2} + 3 \cdot \frac{2}{2}$

$a = \frac{9}{2}$

$b = 3$

Combine $3$ and $\frac{2}{2}$ .

$6 = \frac{9}{2} + \frac{3 \cdot 2}{2}$

$a = \frac{9}{2}$

$b = 3$

Combine the numerators over the common denominator.

$6 = \frac{9 + 3 \cdot 2}{2}$

$a = \frac{9}{2}$

$b = 3$

Simplify the numerator.

Multiply $3$ by $2$ .

$6 = \frac{9 + 6}{2}$

$a = \frac{9}{2}$

$b = 3$

Add $9$ and $6$ .

$6 = \frac{15}{2}$

$a = \frac{9}{2}$

$b = 3$

$6 = \frac{15}{2}$

$a = \frac{9}{2}$

$b = 3$

$6 = \frac{15}{2}$

$a = \frac{9}{2}$

$b = 3$

$6 = \frac{15}{2}$

$a = \frac{9}{2}$

$b = 3$

$6 = \frac{15}{2}$

$a = \frac{9}{2}$

$b = 3$

Since $6 = \frac{15}{2}$ 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 $3 = a \cdot 0^{2} + c$ .

Rewrite the equation as $a \cdot 0^{2} + c = 3$ .

$a \cdot 0^{2} + c = 3$

$6 = a + b + c$