Find the Function Rule table[[x,y],[1,0],[2,3],[3,8]]

$\begin{array}{c} x & y \\ 1 & 0 \\ 2 & 3 \\ 3 & 8 \end{array}$

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 $y = a x + b$ .

$0 = a (1) + b 3 = a (2) + b 8 = a (3) + b$

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

Solve for $a$ in $0 = a + b$ .

Rewrite the equation as $a + b = 0$ .

$a + b = 0$

$3 = a (2) + b$

$8 = a (3) + b$

Subtract $b$ from both sides of the equation.

$a = - b$

$3 = a (2) + b$

$8 = a (3) + b$

$a = - b$

$3 = a (2) + b$

$8 = a (3) + b$

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

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

$3 = (- b) (2) + b$

$a = - b$

$8 = a (3) + b$

Simplify the right side.

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

Multiply $2$ by $- 1$ .

$3 = - 2 b + b$

$a = - b$

$8 = a (3) + b$

Add $- 2 b$ and $b$ .

$3 = - b$

$a = - b$

$8 = a (3) + b$

$3 = - b$

$a = - b$

$8 = a (3) + b$

$3 = - b$

$a = - b$

$8 = a (3) + b$

Replace all occurrences of $a$ in $8 = a (3) + b$ with $- b$ .

$8 = (- b) (3) + b$

$3 = - b$

$a = - b$

Simplify the right side.

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

Multiply $3$ by $- 1$ .

$8 = - 3 b + b$

$3 = - b$

$a = - b$

Add $- 3 b$ and $b$ .

$8 = - 2 b$

$3 = - b$

$a = - b$

$8 = - 2 b$

$3 = - b$

$a = - b$

$8 = - 2 b$

$3 = - b$

$a = - b$

$8 = - 2 b$

$3 = - b$

$a = - b$

Solve for $b$ in $8 = - 2 b$ .

Rewrite the equation as $- 2 b = 8$ .

$- 2 b = 8$

$3 = - b$

$a = - b$

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

Divide each term in $- 2 b = 8$ by $- 2$ .

$\frac{- 2 b}{- 2} = \frac{8}{- 2}$

$3 = - b$

$a = - b$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 b}{- 2} = \frac{8}{- 2}$

$3 = - b$

$a = - b$

Divide $b$ by $1$ .

$b = \frac{8}{- 2}$

$3 = - b$

$a = - b$

$b = \frac{8}{- 2}$

$3 = - b$

$a = - b$

$b = \frac{8}{- 2}$

$3 = - b$

$a = - b$

Simplify the right side.

Divide $8$ by $- 2$ .

$b = - 4$

$3 = - b$

$a = - b$

$b = - 4$

$3 = - b$

$a = - b$

$b = - 4$

$3 = - b$

$a = - b$

$b = - 4$

$3 = - b$

$a = - b$

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

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

$3 = - (- 4)$

$b = - 4$

$a = - b$

Simplify the right side.

Multiply $- 1$ by $- 4$ .

$3 = 4$

$b = - 4$

$a = - b$

$3 = 4$

$b = - 4$

$a = - b$

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

$a = - (- 4)$

$3 = 4$

$b = - 4$

Simplify the right side.

Multiply $- 1$ by $- 4$ .

$a = 4$

$3 = 4$

$b = - 4$

$a = 4$

$3 = 4$

$b = - 4$

$a = 4$

$3 = 4$

$b = - 4$

Since $3 = 4$ 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 $0 = a + b + c$ .

Rewrite the equation as $a + b + c = 0$ .

$a + b + c = 0$

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

$8 = a \cdot 3^{2} + b (3) + c$

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

Subtract $b$ from both sides of the equation.

$a + c = - b$

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

$8 = a \cdot 3^{2} + b (3) + c$

Subtract $c$ from both sides of the equation.

$a = - b - c$

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

$8 = a \cdot 3^{2} + b (3) + c$

$a = - b - c$

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

$8 = a \cdot 3^{2} + b (3) + c$

$a = - b - c$

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

$8 = a \cdot 3^{2} + b (3) + c$

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

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

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

$a = - b - c$

$8 = a \cdot 3^{2} + b (3) + c$

Simplify the right side.

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

Simplify each term.

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

$3 = (- b - c) \cdot 4 + b (2) + c$

$a = - b - c$

$8 = a \cdot 3^{2} + b (3) + c$

Apply the distributive property.

$3 = - b \cdot 4 - c \cdot 4 + b (2) + c$

$a = - b - c$

$8 = a \cdot 3^{2} + b (3) + c$

Multiply $4$ by $- 1$ .

$3 = - 4 b - c \cdot 4 + b (2) + c$

$a = - b - c$

$8 = a \cdot 3^{2} + b (3) + c$

Multiply $4$ by $- 1$ .

$3 = - 4 b - 4 c + b (2) + c$

$a = - b - c$

$8 = a \cdot 3^{2} + b (3) + c$

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

$3 = - 4 b - 4 c + 2 b + c$

$a = - b - c$

$8 = a \cdot 3^{2} + b (3) + c$

$3 = - 4 b - 4 c + 2 b + c$

$a = - b - c$

$8 = a \cdot 3^{2} + b (3) + c$

Simplify by adding terms.

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

$3 = - 2 b - 4 c + c$

$a = - b - c$

$8 = a \cdot 3^{2} + b (3) + c$

Add $- 4 c$ and $c$ .

$3 = - 2 b - 3 c$

$a = - b - c$

$8 = a \cdot 3^{2} + b (3) + c$

$3 = - 2 b - 3 c$

$a = - b - c$

$8 = a \cdot 3^{2} + b (3) + c$

$3 = - 2 b - 3 c$

$a = - b - c$

$8 = a \cdot 3^{2} + b (3) + c$

$3 = - 2 b - 3 c$

$a = - b - c$

$8 = a \cdot 3^{2} + b (3) + c$

Replace all occurrences of $a$ in $8 = a \cdot 3^{2} + b (3) + c$ with $- b - c$ .

$8 = (- b - c) \cdot 3^{2} + b (3) + c$

$3 = - 2 b - 3 c$

$a = - b - c$

Simplify the right side.

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

Simplify each term.

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

$8 = (- b - c) \cdot 9 + b (3) + c$

$3 = - 2 b - 3 c$

$a = - b - c$

Apply the distributive property.

$8 = - b \cdot 9 - c \cdot 9 + b (3) + c$

$3 = - 2 b - 3 c$

$a = - b - c$

Multiply $9$ by $- 1$ .

$8 = - 9 b - c \cdot 9 + b (3) + c$

$3 = - 2 b - 3 c$

$a = - b - c$

Multiply $9$ by $- 1$ .

$8 = - 9 b - 9 c + b (3) + c$

$3 = - 2 b - 3 c$

$a = - b - c$

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

$8 = - 9 b - 9 c + 3 b + c$

$3 = - 2 b - 3 c$

$a = - b - c$

$8 = - 9 b - 9 c + 3 b + c$

$3 = - 2 b - 3 c$

$a = - b - c$

Simplify by adding terms.

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

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

$3 = - 2 b - 3 c$

$a = - b - c$

Add $- 9 c$ and $c$ .

$8 = - 6 b - 8 c$

$3 = - 2 b - 3 c$

$a = - b - c$

$8 = - 6 b - 8 c$

$3 = - 2 b - 3 c$

$a = - b - c$

$8 = - 6 b - 8 c$

$3 = - 2 b - 3 c$

$a = - b - c$

$8 = - 6 b - 8 c$

$3 = - 2 b - 3 c$

$a = - b - c$

$8 = - 6 b - 8 c$

$3 = - 2 b - 3 c$

$a = - b - c$

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

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

$- 6 b - 8 c = 8$

$3 = - 2 b - 3 c$

$a = - b - c$

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

$- 6 b = 8 + 8 c$

$3 = - 2 b - 3 c$

$a = - b - c$

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

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

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

$3 = - 2 b - 3 c$

$a = - b - c$

Simplify the left side.

Cancel the common factor of $- 6$ .

Cancel the common factor.

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

$3 = - 2 b - 3 c$

$a = - b - c$

Divide $b$ by $1$ .

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

$3 = - 2 b - 3 c$

$a = - b - c$

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

$3 = - 2 b - 3 c$

$a = - b - c$

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

$3 = - 2 b - 3 c$

$a = - b - c$

Simplify the right side.

Simplify each term.

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

Factor $2$ out of $8$ .

$b = \frac{2 (4)}{- 6} + \frac{8 c}{- 6}$

$3 = - 2 b - 3 c$

$a = - b - c$

Cancel the common factors.

Factor $2$ out of $- 6$ .

$b = \frac{2 \cdot 4}{2 \cdot - 3} + \frac{8 c}{- 6}$

$3 = - 2 b - 3 c$

$a = - b - c$

Cancel the common factor.

$b = \frac{2 \cdot 4}{2 \cdot - 3} + \frac{8 c}{- 6}$

$3 = - 2 b - 3 c$

$a = - b - c$

Rewrite the expression.

$b = \frac{4}{- 3} + \frac{8 c}{- 6}$

$3 = - 2 b - 3 c$

$a = - b - c$

$b = \frac{4}{- 3} + \frac{8 c}{- 6}$

$3 = - 2 b - 3 c$

$a = - b - c$

$b = \frac{4}{- 3} + \frac{8 c}{- 6}$

$3 = - 2 b - 3 c$

$a = - b - c$

Move the negative in front of the fraction.

$b = - \frac{4}{3} + \frac{8 c}{- 6}$

$3 = - 2 b - 3 c$

$a = - b - c$

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

Factor $2$ out of $8 c$ .

$b = - \frac{4}{3} + \frac{2 (4 c)}{- 6}$

$3 = - 2 b - 3 c$

$a = - b - c$

Cancel the common factors.

Factor $2$ out of $- 6$ .

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

$3 = - 2 b - 3 c$

$a = - b - c$

Cancel the common factor.

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

$3 = - 2 b - 3 c$

$a = - b - c$

Rewrite the expression.

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

$3 = - 2 b - 3 c$

$a = - b - c$

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

$3 = - 2 b - 3 c$

$a = - b - c$

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

$3 = - 2 b - 3 c$

$a = - b - c$

Move the negative in front of the fraction.

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

$3 = - 2 b - 3 c$

$a = - b - c$

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

$3 = - 2 b - 3 c$

$a = - b - c$

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

$3 = - 2 b - 3 c$

$a = - b - c$

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

$3 = - 2 b - 3 c$

$a = - b - c$

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

$3 = - 2 b - 3 c$

$a = - b - c$

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

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

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

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

$a = - b - c$

Simplify the right side.

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

Simplify each term.

Apply the distributive property.

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

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

$a = - b - c$

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

Multiply $- 1$ by $- 2$ .

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

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

$a = - b - c$

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

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

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

$a = - b - c$

Multiply $2$ by $4$ .

$3 = \frac{8}{3} - 2 (- \frac{4 c}{3}) - 3 c$

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

$a = - b - c$

$3 = \frac{8}{3} - 2 (- \frac{4 c}{3}) - 3 c$

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

$a = - b - c$

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

Multiply $- 1$ by $- 2$ .

$3 = \frac{8}{3} + 2 (\frac{4 c}{3}) - 3 c$

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

$a = - b - c$

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

$3 = \frac{8}{3} + \frac{2 (4 c)}{3} - 3 c$

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

$a = - b - c$

Multiply $4$ by $2$ .

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

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

$a = - b - c$

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

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

$a = - b - c$

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

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

$a = - b - c$

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

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

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

$a = - b - c$

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

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

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

$a = - b - c$

Combine the numerators over the common denominator.

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

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

$a = - b - c$

Combine the numerators over the common denominator.

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

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

$a = - b - c$

Multiply $3$ by $- 3$ .

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

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

$a = - b - c$

Subtract $9 c$ from $8 c$ .

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

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

$a = - b - c$

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

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

$a = - b - c$

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

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

$a = - b - c$

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

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

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

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

Simplify the right side.

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

Simplify each term.

Apply the distributive property.

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

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

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

Multiply $- - \frac{4}{3}$ .

Multiply $- 1$ by $- 1$ .

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

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

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

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

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

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

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

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

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

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

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

Multiply $- 1$ by $- 1$ .

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

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

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

Combine the numerators over the common denominator.

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

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

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

Multiply $3$ by $- 1$ .

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

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

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

Subtract $3 c$ from $4 c$ .

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

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

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

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

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

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

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

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

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

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

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

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

Solve for $c$ in $3 = \frac{8 - c}{3}$ .

Rewrite the equation as $\frac{8 - c}{3} = 3$ .

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

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

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

Multiply both sides by $3$ .

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

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

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

Simplify.

Simplify the left side.

Simplify $\frac{8 - c}{3} \cdot 3$ .

Cancel the common factor of $3$ .

Cancel the common factor.

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

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

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

Rewrite the expression.

$8 - c = 3 \cdot 3$

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

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

$8 - c = 3 \cdot 3$

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

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

Reorder $8$ and $- c$ .

$- c + 8 = 3 \cdot 3$

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

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

$- c + 8 = 3 \cdot 3$

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

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

$- c + 8 = 3 \cdot 3$

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

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

Simplify the right side.

Multiply $3$ by $3$ .

$- c + 8 = 9$

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

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

$- c + 8 = 9$

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

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

$- c + 8 = 9$

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

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

Solve for $c$ .

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

Subtract $8$ from both sides of the equation.

$- c = 9 - 8$

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

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

Subtract $8$ from $9$ .

$- c = 1$

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

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

$- c = 1$

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

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

Divide each term in $- c = 1$ by $- 1$ and simplify.

Divide each term in $- c = 1$ by $- 1$ .

$\frac{- c}{- 1} = \frac{1}{- 1}$

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{c}{1} = \frac{1}{- 1}$

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

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

Divide $c$ by $1$ .

$c = \frac{1}{- 1}$

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

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

$c = \frac{1}{- 1}$

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

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

Simplify the right side.

Divide $1$ by $- 1$ .

$c = - 1$

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

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

$c = - 1$

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

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

$c = - 1$

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

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

$c = - 1$

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

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

$c = - 1$

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

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

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

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

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

$c = - 1$

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

Simplify $a = \frac{4 - 1}{3}$ .

Simplify the left side.

Remove parentheses.

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

$c = - 1$

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

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

$c = - 1$

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

Simplify the right side.

Simplify $\frac{4 - 1}{3}$ .

Subtract $1$ from $4$ .

$a = \frac{3}{3}$

$c = - 1$

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

Divide $3$ by $3$ .

$a = 1$

$c = - 1$

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

$a = 1$

$c = - 1$

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

$a = 1$

$c = - 1$

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

$a = 1$

$c = - 1$

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

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

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

$a = 1$

$c = - 1$

Simplify the right side.

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

Combine the numerators over the common denominator.

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

$a = 1$

$c = - 1$

Simplify the expression.

Multiply $- 4$ by $- 1$ .

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

$a = 1$

$c = - 1$

Add $- 4$ and $4$ .

$b = \frac{0}{3}$

$a = 1$

$c = - 1$

Divide $0$ by $3$ .

$b = 0$

$a = 1$

$c = - 1$

$b = 0$

$a = 1$

$c = - 1$

$b = 0$

$a = 1$

$c = - 1$

$b = 0$

$a = 1$

$c = - 1$

$b = 0$

$a = 1$

$c = - 1$

List all of the solutions.

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

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

Simplify each term.

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

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

One to any power is one.

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

Multiply $0$ by $1$ .

$y = 1 + 0 - 1$

Simplify by adding and subtracting.

Add $1$ and $0$ .

$y = 1 - 1$

Subtract $1$ from $1$ .

$y = 0$

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

$0 = 0$

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

Simplify each term.

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

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

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

$y = 4 + (0) \cdot (2) - 1$

Multiply $0$ by $2$ .

$y = 4 + 0 - 1$

Simplify by adding and subtracting.

Add $4$ and $0$ .

$y = 4 - 1$

Subtract $1$ from $4$ .

$y = 3$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 2$ . This check passes since $y = 3$ and $y = 3$ .

$3 = 3$

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

Simplify each term.

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

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

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

$y = 9 + (0) \cdot (3) - 1$

Multiply $0$ by $3$ .

$y = 9 + 0 - 1$

Simplify by adding and subtracting.

Add $9$ and $0$ .

$y = 9 - 1$

Subtract $1$ from $9$ .

$y = 8$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 3$ . This check passes since $y = 8$ and $y = 8$ .

$8 = 8$

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 $y = x^{2} - 1$ .

$y = x^{2} - 1$

Find the Function Rule table[[x,y],[1,0],[2,3],[3,8]]

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