Find the Function Rule table[[x,y],[1,1],[2,4],[3,9]]

$\begin{array}{c} x & y \\ 1 & 1 \\ 2 & 4 \\ 3 & 9 \end{array}$

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 $y = a x + b$ .

$1 = a (1) + b 4 = a (2) + b 9 = a (3) + b$

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

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

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

$a + b = 1$

$4 = a (2) + b$

$9 = a (3) + b$

Subtract $b$ from both sides of the equation.

$a = 1 - b$

$4 = a (2) + b$

$9 = a (3) + b$

$a = 1 - b$

$4 = a (2) + b$

$9 = a (3) + b$

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

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

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

$a = 1 - b$

$9 = a (3) + b$

Simplify the right side.

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

Simplify each term.

Apply the distributive property.

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

$a = 1 - b$

$9 = a (3) + b$

Multiply $2$ by $1$ .

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

$a = 1 - b$

$9 = a (3) + b$

Multiply $2$ by $- 1$ .

$4 = 2 - 2 b + b$

$a = 1 - b$

$9 = a (3) + b$

$4 = 2 - 2 b + b$

$a = 1 - b$

$9 = a (3) + b$

Add $- 2 b$ and $b$ .

$4 = 2 - b$

$a = 1 - b$

$9 = a (3) + b$

$4 = 2 - b$

$a = 1 - b$

$9 = a (3) + b$

$4 = 2 - b$

$a = 1 - b$

$9 = a (3) + b$

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

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

$4 = 2 - b$

$a = 1 - b$

Simplify the right side.

Simplify $(1 - b) (3) + b$ .

Simplify each term.

Apply the distributive property.

$9 = 1 \cdot 3 - b \cdot 3 + b$

$4 = 2 - b$

$a = 1 - b$

Multiply $3$ by $1$ .

$9 = 3 - b \cdot 3 + b$

$4 = 2 - b$

$a = 1 - b$

Multiply $3$ by $- 1$ .

$9 = 3 - 3 b + b$

$4 = 2 - b$

$a = 1 - b$

$9 = 3 - 3 b + b$

$4 = 2 - b$

$a = 1 - b$

Add $- 3 b$ and $b$ .

$9 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

$9 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

$9 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

$9 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

Solve for $b$ in $9 = 3 - 2 b$ .

Rewrite the equation as $3 - 2 b = 9$ .

$3 - 2 b = 9$

$4 = 2 - b$

$a = 1 - b$

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

Subtract $3$ from both sides of the equation.

$- 2 b = 9 - 3$

$4 = 2 - b$

$a = 1 - b$

Subtract $3$ from $9$ .

$- 2 b = 6$

$4 = 2 - b$

$a = 1 - b$

$- 2 b = 6$

$4 = 2 - b$

$a = 1 - b$

Divide each term in $- 2 b = 6$ by $- 2$ and simplify.

Divide each term in $- 2 b = 6$ by $- 2$ .

$\frac{- 2 b}{- 2} = \frac{6}{- 2}$

$4 = 2 - b$

$a = 1 - b$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 b}{- 2} = \frac{6}{- 2}$

$4 = 2 - b$

$a = 1 - b$

Divide $b$ by $1$ .

$b = \frac{6}{- 2}$

$4 = 2 - b$

$a = 1 - b$

$b = \frac{6}{- 2}$

$4 = 2 - b$

$a = 1 - b$

$b = \frac{6}{- 2}$

$4 = 2 - b$

$a = 1 - b$

Simplify the right side.

Divide $6$ by $- 2$ .

$b = - 3$

$4 = 2 - b$

$a = 1 - b$

$b = - 3$

$4 = 2 - b$

$a = 1 - b$

$b = - 3$

$4 = 2 - b$

$a = 1 - b$

$b = - 3$

$4 = 2 - b$

$a = 1 - b$

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

Replace all occurrences of $b$ in $4 = 2 - b$ with $- 3$ .

$4 = 2 - (- 3)$

$b = - 3$

$a = 1 - b$

Simplify the right side.

Simplify $2 - (- 3)$ .

Multiply $- 1$ by $- 3$ .

$4 = 2 + 3$

$b = - 3$

$a = 1 - b$

Add $2$ and $3$ .

$4 = 5$

$b = - 3$

$a = 1 - b$

$4 = 5$

$b = - 3$

$a = 1 - b$

$4 = 5$

$b = - 3$

$a = 1 - b$

Replace all occurrences of $b$ in $a = 1 - b$ with $- 3$ .

$a = 1 - (- 3)$

$4 = 5$

$b = - 3$

Simplify the right side.

Simplify $1 - (- 3)$ .

Multiply $- 1$ by $- 3$ .

$a = 1 + 3$

$4 = 5$

$b = - 3$

Add $1$ and $3$ .

$a = 4$

$4 = 5$

$b = - 3$

$a = 4$

$4 = 5$

$b = - 3$

$a = 4$

$4 = 5$

$b = - 3$

$a = 4$

$4 = 5$

$b = - 3$

Since $4 = 5$ 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 $a$ in $1 = a + b + c$ .

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

$a + b + c = 1$

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

$9 = a \cdot 3^{2} + b (3) + c$

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

Subtract $b$ from both sides of the equation.

$a + c = 1 - b$

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

$9 = a \cdot 3^{2} + b (3) + c$

Subtract $c$ from both sides of the equation.

$a = 1 - b - c$

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

$9 = a \cdot 3^{2} + b (3) + c$

$a = 1 - b - c$

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

$9 = a \cdot 3^{2} + b (3) + c$

$a = 1 - b - c$

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

$9 = a \cdot 3^{2} + b (3) + c$

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

Replace all occurrences of $a$ in $4 = a \cdot 2^{2} + b (2) + c$ with $1 - b - c$ .

$4 = (1 - b - c) \cdot 2^{2} + b (2) + c$

$a = 1 - b - c$

$9 = a \cdot 3^{2} + b (3) + c$

Simplify the right side.

Simplify $(1 - b - c) \cdot 2^{2} + b (2) + c$ .

Simplify each term.

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

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

$a = 1 - b - c$

$9 = a \cdot 3^{2} + b (3) + c$

Apply the distributive property.

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

$a = 1 - b - c$

$9 = a \cdot 3^{2} + b (3) + c$

Simplify.

Multiply $4$ by $1$ .

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

$a = 1 - b - c$

$9 = a \cdot 3^{2} + b (3) + c$

Multiply $4$ by $- 1$ .

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

$a = 1 - b - c$

$9 = a \cdot 3^{2} + b (3) + c$

Multiply $4$ by $- 1$ .

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

$a = 1 - b - c$

$9 = a \cdot 3^{2} + b (3) + c$

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

$a = 1 - b - c$

$9 = a \cdot 3^{2} + b (3) + c$

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

$4 = 4 - 4 b - 4 c + 2 b + c$

$a = 1 - b - c$

$9 = a \cdot 3^{2} + b (3) + c$

$4 = 4 - 4 b - 4 c + 2 b + c$

$a = 1 - b - c$

$9 = a \cdot 3^{2} + b (3) + c$

Simplify by adding terms.

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

$4 = 4 - 2 b - 4 c + c$

$a = 1 - b - c$

$9 = a \cdot 3^{2} + b (3) + c$

Add $- 4 c$ and $c$ .

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$9 = a \cdot 3^{2} + b (3) + c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$9 = a \cdot 3^{2} + b (3) + c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$9 = a \cdot 3^{2} + b (3) + c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$9 = a \cdot 3^{2} + b (3) + c$

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

$9 = (1 - b - c) \cdot 3^{2} + b (3) + c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Simplify the right side.

Simplify $(1 - b - c) \cdot 3^{2} + b (3) + c$ .

Simplify each term.

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

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

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Apply the distributive property.

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

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Simplify.

Multiply $9$ by $1$ .

$9 = 9 - b \cdot 9 - c \cdot 9 + b (3) + c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Multiply $9$ by $- 1$ .

$9 = 9 - 9 b - c \cdot 9 + b (3) + c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Multiply $9$ by $- 1$ .

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

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

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

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

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

$9 = 9 - 9 b - 9 c + 3 b + c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$9 = 9 - 9 b - 9 c + 3 b + c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Simplify by adding terms.

Add $- 9 b$ and $3 b$ .

$9 = 9 - 6 b - 9 c + c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Add $- 9 c$ and $c$ .

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Solve for $b$ in $9 = 9 - 6 b - 8 c$ .

Rewrite the equation as $9 - 6 b - 8 c = 9$ .

$9 - 6 b - 8 c = 9$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

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

Subtract $9$ from both sides of the equation.

$- 6 b - 8 c = 9 - 9$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Add $8 c$ to both sides of the equation.

$- 6 b = 9 - 9 + 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Subtract $9$ from $9$ .

$- 6 b = 0 + 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Add $0$ and $8 c$ .

$- 6 b = 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$- 6 b = 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Divide each term in $- 6 b = 8 c$ by $- 6$ and simplify.

Divide each term in $- 6 b = 8 c$ by $- 6$ .

$\frac{- 6 b}{- 6} = \frac{8 c}{- 6}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Simplify the left side.

Cancel the common factor of $- 6$ .

Cancel the common factor.

$\frac{- 6 b}{- 6} = \frac{8 c}{- 6}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Divide $b$ by $1$ .

$b = \frac{8 c}{- 6}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$b = \frac{8 c}{- 6}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$b = \frac{8 c}{- 6}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Simplify the right side.

Cancel the common factor of $8$ and $- 6$ .

Factor $2$ out of $8 c$ .

$b = \frac{2 (4 c)}{- 6}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Cancel the common factors.

Factor $2$ out of $- 6$ .

$b = \frac{2 (4 c)}{2 (- 3)}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Cancel the common factor.

$b = \frac{2 (4 c)}{2 \cdot - 3}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Rewrite the expression.

$b = \frac{4 c}{- 3}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$b = \frac{4 c}{- 3}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$b = \frac{4 c}{- 3}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Move the negative in front of the fraction.

$b = - \frac{4 c}{3}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$b = - \frac{4 c}{3}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$b = - \frac{4 c}{3}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$b = - \frac{4 c}{3}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Replace all occurrences of $b$ with $- \frac{4 c}{3}$ in each equation.

Replace all occurrences of $b$ in $4 = 4 - 2 b - 3 c$ with $- \frac{4 c}{3}$ .

$4 = 4 - 2 (- \frac{4 c}{3}) - 3 c$

$b = - \frac{4 c}{3}$

$a = 1 - b - c$

Simplify the right side.

Simplify $4 - 2 (- \frac{4 c}{3}) - 3 c$ .

Multiply $- 2 (- \frac{4 c}{3})$ .

Multiply $- 1$ by $- 2$ .

$4 = 4 + 2 (\frac{4 c}{3}) - 3 c$

$b = - \frac{4 c}{3}$

$a = 1 - b - c$

Combine $2$ and $\frac{4 c}{3}$ .

$4 = 4 + \frac{2 (4 c)}{3} - 3 c$

$b = - \frac{4 c}{3}$

$a = 1 - b - c$

Multiply $4$ by $2$ .

$4 = 4 + \frac{8 c}{3} - 3 c$

$b = - \frac{4 c}{3}$

$a = 1 - b - c$

$4 = 4 + \frac{8 c}{3} - 3 c$

$b = - \frac{4 c}{3}$

$a = 1 - b - c$

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

$4 = 4 + \frac{8 c}{3} - 3 c \cdot \frac{3}{3}$

$b = - \frac{4 c}{3}$

$a = 1 - b - c$

Simplify terms.

Combine $- 3 c$ and $\frac{3}{3}$ .

$4 = 4 + \frac{8 c}{3} + \frac{- 3 c \cdot 3}{3}$

$b = - \frac{4 c}{3}$

$a = 1 - b - c$

Combine the numerators over the common denominator.

$4 = 4 + \frac{8 c - 3 c \cdot 3}{3}$

$b = - \frac{4 c}{3}$

$a = 1 - b - c$

$4 = 4 + \frac{8 c - 3 c \cdot 3}{3}$

$b = - \frac{4 c}{3}$

$a = 1 - b - c$

Simplify each term.

Simplify the numerator.

Factor $c$ out of $8 c - 3 c \cdot 3$ .

Factor $c$ out of $8 c$ .

$4 = 4 + \frac{c \cdot 8 - 3 c \cdot 3}{3}$

$b = - \frac{4 c}{3}$

$a = 1 - b - c$

Factor $c$ out of $- 3 c \cdot 3$ .

$4 = 4 + \frac{c \cdot 8 + c (- 3 \cdot 3)}{3}$

$b = - \frac{4 c}{3}$

$a = 1 - b - c$

Factor $c$ out of $c \cdot 8 + c (- 3 \cdot 3)$ .

$4 = 4 + \frac{c (8 - 3 \cdot 3)}{3}$

$b = - \frac{4 c}{3}$

$a = 1 - b - c$

$4 = 4 + \frac{c (8 - 3 \cdot 3)}{3}$

$b = - \frac{4 c}{3}$

$a = 1 - b - c$

Multiply $- 3$ by $3$ .

$4 = 4 + \frac{c (8 - 9)}{3}$

$b = - \frac{4 c}{3}$

$a = 1 - b - c$

Subtract $9$ from $8$ .

$4 = 4 + \frac{c \cdot - 1}{3}$

$b = - \frac{4 c}{3}$

$a = 1 - b - c$

$4 = 4 + \frac{c \cdot - 1}{3}$

$b = - \frac{4 c}{3}$

$a = 1 - b - c$

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

$4 = 4 + \frac{- 1 \cdot c}{3}$

$b = - \frac{4 c}{3}$

$a = 1 - b - c$

Move the negative in front of the fraction.

$4 = 4 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

$a = 1 - b - c$

$4 = 4 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

$a = 1 - b - c$

$4 = 4 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

$a = 1 - b - c$

$4 = 4 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

$a = 1 - b - c$

Replace all occurrences of $b$ in $a = 1 - b - c$ with $- \frac{4 c}{3}$ .

$a = 1 - (- \frac{4 c}{3}) - c$

$4 = 4 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

Simplify the right side.

Simplify $1 - (- \frac{4 c}{3}) - c$ .

Multiply $- (- \frac{4 c}{3})$ .

Multiply $- 1$ by $- 1$ .

$a = 1 + 1 (\frac{4 c}{3}) - c$

$4 = 4 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

Multiply $\frac{4 c}{3}$ by $1$ .

$a = 1 + \frac{4 c}{3} - c$

$4 = 4 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

$a = 1 + \frac{4 c}{3} - c$

$4 = 4 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

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

$a = 1 + \frac{4 c}{3} - c \cdot \frac{3}{3}$

$4 = 4 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

Simplify terms.

Combine $- c$ and $\frac{3}{3}$ .

$a = 1 + \frac{4 c}{3} + \frac{- c \cdot 3}{3}$

$4 = 4 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

Combine the numerators over the common denominator.

$a = 1 + \frac{4 c - c \cdot 3}{3}$

$4 = 4 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

$a = 1 + \frac{4 c - c \cdot 3}{3}$

$4 = 4 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

Simplify each term.

Simplify the numerator.

Factor $c$ out of $4 c - c \cdot 3$ .

Factor $c$ out of $4 c$ .

$a = 1 + \frac{c \cdot 4 - c \cdot 3}{3}$

$4 = 4 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

Factor $c$ out of $- c \cdot 3$ .

$a = 1 + \frac{c \cdot 4 + c (- 1 \cdot 3)}{3}$

$4 = 4 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

Factor $c$ out of $c \cdot 4 + c (- 1 \cdot 3)$ .

$a = 1 + \frac{c (4 - 1 \cdot 3)}{3}$

$4 = 4 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

$a = 1 + \frac{c (4 - 1 \cdot 3)}{3}$

$4 = 4 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

Multiply $- 1$ by $3$ .

$a = 1 + \frac{c (4 - 3)}{3}$

$4 = 4 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

Subtract $3$ from $4$ .

$a = 1 + \frac{c \cdot 1}{3}$

$4 = 4 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

$a = 1 + \frac{c \cdot 1}{3}$

$4 = 4 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

Multiply $c$ by $1$ .

$a = 1 + \frac{c}{3}$

$4 = 4 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

$a = 1 + \frac{c}{3}$

$4 = 4 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

$a = 1 + \frac{c}{3}$

$4 = 4 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

$a = 1 + \frac{c}{3}$

$4 = 4 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

$a = 1 + \frac{c}{3}$

$4 = 4 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

Solve for $c$ in $4 = 4 - \frac{c}{3}$ .

Rewrite the equation as $4 - \frac{c}{3} = 4$ .

$4 - \frac{c}{3} = 4$

$a = 1 + \frac{c}{3}$

$b = - \frac{4 c}{3}$

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

Subtract $4$ from both sides of the equation.

$- \frac{c}{3} = 4 - 4$

$a = 1 + \frac{c}{3}$

$b = - \frac{4 c}{3}$

Subtract $4$ from $4$ .

$- \frac{c}{3} = 0$

$a = 1 + \frac{c}{3}$

$b = - \frac{4 c}{3}$

$- \frac{c}{3} = 0$

$a = 1 + \frac{c}{3}$

$b = - \frac{4 c}{3}$

Set the numerator equal to zero.

$c = 0$

$a = 1 + \frac{c}{3}$

$b = - \frac{4 c}{3}$

$c = 0$

$a = 1 + \frac{c}{3}$

$b = - \frac{4 c}{3}$

Replace all occurrences of $c$ with $0$ in each equation.

Replace all occurrences of $c$ in $a = 1 + \frac{c}{3}$ with $0$ .

$a = 1 + \frac{0}{3}$

$c = 0$

$b = - \frac{4 c}{3}$

Simplify the right side.

Simplify $1 + \frac{0}{3}$ .

Divide $0$ by $3$ .

$a = 1 + 0$

$c = 0$

$b = - \frac{4 c}{3}$

Add $1$ and $0$ .

$a = 1$

$c = 0$

$b = - \frac{4 c}{3}$

$a = 1$

$c = 0$

$b = - \frac{4 c}{3}$

$a = 1$

$c = 0$

$b = - \frac{4 c}{3}$

Replace all occurrences of $c$ in $b = - \frac{4 c}{3}$ with $0$ .

$b = - \frac{4 (0)}{3}$

$a = 1$

$c = 0$

Simplify the right side.

Simplify $- \frac{4 (0)}{3}$ .

Cancel the common factor of $0$ and $3$ .

Factor $3$ out of $4 (0)$ .

$b = - \frac{3 (4 \cdot (0))}{3}$

$a = 1$

$c = 0$

Cancel the common factors.

Factor $3$ out of $3$ .

$b = - \frac{3 (4 \cdot (0))}{3 (1)}$

$a = 1$

$c = 0$

Cancel the common factor.

$b = - \frac{3 (4 \cdot (0))}{3 \cdot 1}$

$a = 1$

$c = 0$

Rewrite the expression.

$b = - \frac{4 \cdot (0)}{1}$

$a = 1$

$c = 0$

Divide $4 \cdot (0)$ by $1$ .

$b = - (4 \cdot (0))$

$a = 1$

$c = 0$

$b = - (4 \cdot (0))$

$a = 1$

$c = 0$

$b = - (4 \cdot (0))$

$a = 1$

$c = 0$

Multiply $- (4 \cdot (0))$ .

Multiply $4$ by $0$ .

$b = - 0$

$a = 1$

$c = 0$

Multiply $- 1$ by $0$ .

$b = 0$

$a = 1$

$c = 0$

$b = 0$

$a = 1$

$c = 0$

$b = 0$

$a = 1$

$c = 0$

$b = 0$

$a = 1$

$c = 0$

$b = 0$

$a = 1$

$c = 0$

List all of the solutions.

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

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

Simplify each term.

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

$y = {(1)}^{2} + (0) \cdot (1) + 0$

One to any power is one.

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

Multiply $0$ by $1$ .

$y = 1 + 0 + 0$

Simplify by adding numbers.

Add $1$ and $0$ .

$y = 1 + 0$

Add $1$ and $0$ .

$y = 1$

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

$1 = 1$

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

Simplify each term.

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

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

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

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

Multiply $0$ by $2$ .

$y = 4 + 0 + 0$

Simplify by adding numbers.

Add $4$ and $0$ .

$y = 4 + 0$

Add $4$ and $0$ .

$y = 4$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 2$ . This check passes since $y = 4$ and $y = 4$ .

$4 = 4$

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

Simplify each term.

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

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

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

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

Multiply $0$ by $3$ .

$y = 9 + 0 + 0$

Simplify by adding numbers.

Add $9$ and $0$ .

$y = 9 + 0$

Add $9$ and $0$ .

$y = 9$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 3$ . 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

Step 3

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

$y = x^{2}$

Find the Function Rule table[[x,y],[1,1],[2,4],[3,9]]

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