Find the Function Rule table[[x,y],[1,-2],[2,-4],[3,-8]]

$\begin{array}{c} x & y \\ 1 & - 2 \\ 2 & - 4 \\ 3 & - 8 \end{array}$

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 $y = a x + b$ .

$- 2 = a (1) + b - 4 = a (2) + b - 8 = a (3) + b$

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

Solve for $a$ in $- 2 = a + b$ .

Rewrite the equation as $a + b = - 2$ .

$a + b = - 2$

$- 4 = a (2) + b$

$- 8 = a (3) + b$

Subtract $b$ from both sides of the equation.

$a = - 2 - b$

$- 4 = a (2) + b$

$- 8 = a (3) + b$

$a = - 2 - b$

$- 4 = a (2) + b$

$- 8 = a (3) + b$

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

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

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

$a = - 2 - b$

$- 8 = a (3) + b$

Simplify the right side.

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

Simplify each term.

Apply the distributive property.

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

$a = - 2 - b$

$- 8 = a (3) + b$

Multiply $- 2$ by $2$ .

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

$a = - 2 - b$

$- 8 = a (3) + b$

Multiply $2$ by $- 1$ .

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

$a = - 2 - b$

$- 8 = a (3) + b$

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

$a = - 2 - b$

$- 8 = a (3) + b$

Add $- 2 b$ and $b$ .

$- 4 = - 4 - b$

$a = - 2 - b$

$- 8 = a (3) + b$

$- 4 = - 4 - b$

$a = - 2 - b$

$- 8 = a (3) + b$

$- 4 = - 4 - b$

$a = - 2 - b$

$- 8 = a (3) + b$

Replace all occurrences of $a$ in $- 8 = a (3) + b$ with $- 2 - b$ .

$- 8 = (- 2 - b) (3) + b$

$- 4 = - 4 - b$

$a = - 2 - b$

Simplify the right side.

Simplify $(- 2 - b) (3) + b$ .

Simplify each term.

Apply the distributive property.

$- 8 = - 2 \cdot 3 - b \cdot 3 + b$

$- 4 = - 4 - b$

$a = - 2 - b$

Multiply $- 2$ by $3$ .

$- 8 = - 6 - b \cdot 3 + b$

$- 4 = - 4 - b$

$a = - 2 - b$

Multiply $3$ by $- 1$ .

$- 8 = - 6 - 3 b + b$

$- 4 = - 4 - b$

$a = - 2 - b$

$- 8 = - 6 - 3 b + b$

$- 4 = - 4 - b$

$a = - 2 - b$

Add $- 3 b$ and $b$ .

$- 8 = - 6 - 2 b$

$- 4 = - 4 - b$

$a = - 2 - b$

$- 8 = - 6 - 2 b$

$- 4 = - 4 - b$

$a = - 2 - b$

$- 8 = - 6 - 2 b$

$- 4 = - 4 - b$

$a = - 2 - b$

$- 8 = - 6 - 2 b$

$- 4 = - 4 - b$

$a = - 2 - b$

Solve for $b$ in $- 8 = - 6 - 2 b$ .

Rewrite the equation as $- 6 - 2 b = - 8$ .

$- 6 - 2 b = - 8$

$- 4 = - 4 - b$

$a = - 2 - b$

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

Add $6$ to both sides of the equation.

$- 2 b = - 8 + 6$

$- 4 = - 4 - b$

$a = - 2 - b$

Add $- 8$ and $6$ .

$- 2 b = - 2$

$- 4 = - 4 - b$

$a = - 2 - b$

$- 2 b = - 2$

$- 4 = - 4 - b$

$a = - 2 - b$

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

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

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

$- 4 = - 4 - b$

$a = - 2 - b$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

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

$- 4 = - 4 - b$

$a = - 2 - b$

Divide $b$ by $1$ .

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

$- 4 = - 4 - b$

$a = - 2 - b$

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

$- 4 = - 4 - b$

$a = - 2 - b$

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

$- 4 = - 4 - b$

$a = - 2 - b$

Simplify the right side.

Divide $- 2$ by $- 2$ .

$b = 1$

$- 4 = - 4 - b$

$a = - 2 - b$

$b = 1$

$- 4 = - 4 - b$

$a = - 2 - b$

$b = 1$

$- 4 = - 4 - b$

$a = - 2 - b$

$b = 1$

$- 4 = - 4 - b$

$a = - 2 - b$

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

Replace all occurrences of $b$ in $- 4 = - 4 - b$ with $1$ .

$- 4 = - 4 - (1)$

$b = 1$

$a = - 2 - b$

Simplify the right side.

Simplify $- 4 - (1)$ .

Multiply $- 1$ by $1$ .

$- 4 = - 4 - 1$

$b = 1$

$a = - 2 - b$

Subtract $1$ from $- 4$ .

$- 4 = - 5$

$b = 1$

$a = - 2 - b$

$- 4 = - 5$

$b = 1$

$a = - 2 - b$

$- 4 = - 5$

$b = 1$

$a = - 2 - b$

Replace all occurrences of $b$ in $a = - 2 - b$ with $1$ .

$a = - 2 - (1)$

$- 4 = - 5$

$b = 1$

Simplify the right side.

Simplify $- 2 - (1)$ .

Multiply $- 1$ by $1$ .

$a = - 2 - 1$

$- 4 = - 5$

$b = 1$

Subtract $1$ from $- 2$ .

$a = - 3$

$- 4 = - 5$

$b = 1$

$a = - 3$

$- 4 = - 5$

$b = 1$

$a = - 3$

$- 4 = - 5$

$b = 1$

$a = - 3$

$- 4 = - 5$

$b = 1$

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 $- 2 = a + b + c$ .

Rewrite the equation as $a + b + c = - 2$ .

$a + b + c = - 2$

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

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

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

Subtract $b$ from both sides of the equation.

$a + c = - 2 - b$

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

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

Subtract $c$ from both sides of the equation.

$a = - 2 - b - c$

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

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

$a = - 2 - b - c$

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

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

$a = - 2 - b - c$

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

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

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

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

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

$a = - 2 - b - c$

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

Simplify the right side.

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

Simplify each term.

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

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

$a = - 2 - b - c$

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

Apply the distributive property.

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

$a = - 2 - b - c$

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

Simplify.

Multiply $- 2$ by $4$ .

$- 4 = - 8 - b \cdot 4 - c \cdot 4 + b (2) + c$

$a = - 2 - b - c$

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

Multiply $4$ by $- 1$ .

$- 4 = - 8 - 4 b - c \cdot 4 + b (2) + c$

$a = - 2 - b - c$

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

Multiply $4$ by $- 1$ .

$- 4 = - 8 - 4 b - 4 c + b (2) + c$

$a = - 2 - b - c$

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

$- 4 = - 8 - 4 b - 4 c + b (2) + c$

$a = - 2 - b - c$

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

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

$- 4 = - 8 - 4 b - 4 c + 2 b + c$

$a = - 2 - b - c$

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

$- 4 = - 8 - 4 b - 4 c + 2 b + c$

$a = - 2 - b - c$

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

Simplify by adding terms.

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

$- 4 = - 8 - 2 b - 4 c + c$

$a = - 2 - b - c$

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

Add $- 4 c$ and $c$ .

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

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

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

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

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

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

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

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

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

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

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

Simplify the right side.

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

Simplify each term.

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

$- 8 = (- 2 - b - c) \cdot 9 + b (3) + c$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

Apply the distributive property.

$- 8 = - 2 \cdot 9 - b \cdot 9 - c \cdot 9 + b (3) + c$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

Simplify.

Multiply $- 2$ by $9$ .

$- 8 = - 18 - b \cdot 9 - c \cdot 9 + b (3) + c$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

Multiply $9$ by $- 1$ .

$- 8 = - 18 - 9 b - c \cdot 9 + b (3) + c$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

Multiply $9$ by $- 1$ .

$- 8 = - 18 - 9 b - 9 c + b (3) + c$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

$- 8 = - 18 - 9 b - 9 c + b (3) + c$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

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

$- 8 = - 18 - 9 b - 9 c + 3 b + c$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

$- 8 = - 18 - 9 b - 9 c + 3 b + c$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

Simplify by adding terms.

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

$- 8 = - 18 - 6 b - 9 c + c$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

Add $- 9 c$ and $c$ .

$- 8 = - 18 - 6 b - 8 c$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

$- 8 = - 18 - 6 b - 8 c$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

$- 8 = - 18 - 6 b - 8 c$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

$- 8 = - 18 - 6 b - 8 c$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

$- 8 = - 18 - 6 b - 8 c$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

Solve for $b$ in $- 8 = - 18 - 6 b - 8 c$ .

Rewrite the equation as $- 18 - 6 b - 8 c = - 8$ .

$- 18 - 6 b - 8 c = - 8$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

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

Add $18$ to both sides of the equation.

$- 6 b - 8 c = - 8 + 18$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

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

$- 6 b = - 8 + 18 + 8 c$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

Add $- 8$ and $18$ .

$- 6 b = 10 + 8 c$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

$- 6 b = 10 + 8 c$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

Divide each term in $- 6 b = 10 + 8 c$ by $- 6$ and simplify.

Divide each term in $- 6 b = 10 + 8 c$ by $- 6$ .

$\frac{- 6 b}{- 6} = \frac{10}{- 6} + \frac{8 c}{- 6}$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

Simplify the left side.

Cancel the common factor of $- 6$ .

Cancel the common factor.

$\frac{- 6 b}{- 6} = \frac{10}{- 6} + \frac{8 c}{- 6}$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

Divide $b$ by $1$ .

$b = \frac{10}{- 6} + \frac{8 c}{- 6}$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

$b = \frac{10}{- 6} + \frac{8 c}{- 6}$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

$b = \frac{10}{- 6} + \frac{8 c}{- 6}$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

Simplify the right side.

Simplify each term.

Cancel the common factor of $10$ and $- 6$ .

Factor $2$ out of $10$ .

$b = \frac{2 (5)}{- 6} + \frac{8 c}{- 6}$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

Cancel the common factors.

Factor $2$ out of $- 6$ .

$b = \frac{2 \cdot 5}{2 \cdot - 3} + \frac{8 c}{- 6}$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

Cancel the common factor.

$b = \frac{2 \cdot 5}{2 \cdot - 3} + \frac{8 c}{- 6}$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

Rewrite the expression.

$b = \frac{5}{- 3} + \frac{8 c}{- 6}$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

$b = \frac{5}{- 3} + \frac{8 c}{- 6}$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

$b = \frac{5}{- 3} + \frac{8 c}{- 6}$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

Move the negative in front of the fraction.

$b = - \frac{5}{3} + \frac{8 c}{- 6}$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

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

Factor $2$ out of $8 c$ .

$b = - \frac{5}{3} + \frac{2 (4 c)}{- 6}$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

Cancel the common factors.

Factor $2$ out of $- 6$ .

$b = - \frac{5}{3} + \frac{2 (4 c)}{2 (- 3)}$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

Cancel the common factor.

$b = - \frac{5}{3} + \frac{2 (4 c)}{2 \cdot - 3}$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

Rewrite the expression.

$b = - \frac{5}{3} + \frac{4 c}{- 3}$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

$b = - \frac{5}{3} + \frac{4 c}{- 3}$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

$b = - \frac{5}{3} + \frac{4 c}{- 3}$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

Move the negative in front of the fraction.

$b = - \frac{5}{3} - \frac{4 c}{3}$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$- 4 = - 8 - 2 b - 3 c$

$a = - 2 - b - c$

Replace all occurrences of $b$ with $- \frac{5}{3} - \frac{4 c}{3}$ in each equation.

Replace all occurrences of $b$ in $- 4 = - 8 - 2 b - 3 c$ with $- \frac{5}{3} - \frac{4 c}{3}$ .

$- 4 = - 8 - 2 (- \frac{5}{3} - \frac{4 c}{3}) - 3 c$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 - b - c$

Simplify the right side.

Simplify $- 8 - 2 (- \frac{5}{3} - \frac{4 c}{3}) - 3 c$ .

Simplify each term.

Apply the distributive property.

$- 4 = - 8 - 2 (- \frac{5}{3}) - 2 (- \frac{4 c}{3}) - 3 c$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 - b - c$

Multiply $- 2 (- \frac{5}{3})$ .

Multiply $- 1$ by $- 2$ .

$- 4 = - 8 + 2 (\frac{5}{3}) - 2 (- \frac{4 c}{3}) - 3 c$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 - b - c$

Combine $2$ and $\frac{5}{3}$ .

$- 4 = - 8 + \frac{2 \cdot 5}{3} - 2 (- \frac{4 c}{3}) - 3 c$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 - b - c$

Multiply $2$ by $5$ .

$- 4 = - 8 + \frac{10}{3} - 2 (- \frac{4 c}{3}) - 3 c$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 - b - c$

$- 4 = - 8 + \frac{10}{3} - 2 (- \frac{4 c}{3}) - 3 c$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 - b - c$

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

Multiply $- 1$ by $- 2$ .

$- 4 = - 8 + \frac{10}{3} + 2 (\frac{4 c}{3}) - 3 c$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 - b - c$

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

$- 4 = - 8 + \frac{10}{3} + \frac{2 (4 c)}{3} - 3 c$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 - b - c$

Multiply $4$ by $2$ .

$- 4 = - 8 + \frac{10}{3} + \frac{8 c}{3} - 3 c$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 - b - c$

$- 4 = - 8 + \frac{10}{3} + \frac{8 c}{3} - 3 c$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 - b - c$

$- 4 = - 8 + \frac{10}{3} + \frac{8 c}{3} - 3 c$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 - b - c$

To write $- 8$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$- 4 = - 8 \cdot \frac{3}{3} + \frac{10}{3} + \frac{8 c}{3} - 3 c$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 - b - c$

Combine $- 8$ and $\frac{3}{3}$ .

$- 4 = \frac{- 8 \cdot 3}{3} + \frac{10}{3} + \frac{8 c}{3} - 3 c$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 - b - c$

Combine the numerators over the common denominator.

$- 4 = \frac{- 8 \cdot 3 + 10}{3} + \frac{8 c}{3} - 3 c$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 - b - c$

Simplify the numerator.

Multiply $- 8$ by $3$ .

$- 4 = \frac{- 24 + 10}{3} + \frac{8 c}{3} - 3 c$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 - b - c$

Add $- 24$ and $10$ .

$- 4 = \frac{- 14}{3} + \frac{8 c}{3} - 3 c$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 - b - c$

$- 4 = \frac{- 14}{3} + \frac{8 c}{3} - 3 c$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 - b - c$

Move the negative in front of the fraction.

$- 4 = - \frac{14}{3} + \frac{8 c}{3} - 3 c$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 - b - c$

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

$- 4 = - \frac{14}{3} + \frac{8 c}{3} - 3 c \cdot \frac{3}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 - b - c$

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

$- 4 = - \frac{14}{3} + \frac{8 c}{3} + \frac{- 3 c \cdot 3}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 - b - c$

Combine the numerators over the common denominator.

$- 4 = - \frac{14}{3} + \frac{8 c - 3 c \cdot 3}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 - b - c$

Combine the numerators over the common denominator.

$- 4 = \frac{- 14 + 8 c - 3 c \cdot 3}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 - b - c$

Multiply $3$ by $- 3$ .

$- 4 = \frac{- 14 + 8 c - 9 c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 - b - c$

Subtract $9 c$ from $8 c$ .

$- 4 = \frac{- 14 - c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 - b - c$

Rewrite $- 14$ as $- 1 (14)$ .

$- 4 = \frac{- 1 \cdot 14 - c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 - b - c$

Factor $- 1$ out of $- c$ .

$- 4 = \frac{- 1 \cdot 14 - (c)}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 - b - c$

Factor $- 1$ out of $- 1 (14) - (c)$ .

$- 4 = \frac{- 1 (14 + c)}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 - b - c$

Move the negative in front of the fraction.

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 - b - c$

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 - b - c$

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 - b - c$

Replace all occurrences of $b$ in $a = - 2 - b - c$ with $- \frac{5}{3} - \frac{4 c}{3}$ .

$a = - 2 - (- \frac{5}{3} - \frac{4 c}{3}) - c$

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

Simplify the right side.

Simplify $- 2 - (- \frac{5}{3} - \frac{4 c}{3}) - c$ .

Simplify each term.

Apply the distributive property.

$a = - 2 + \frac{5}{3} + \frac{4 c}{3} - c$

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

Multiply $- - \frac{5}{3}$ .

Multiply $- 1$ by $- 1$ .

$a = - 2 + 1 (\frac{5}{3}) + \frac{4 c}{3} - c$

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

Multiply $\frac{5}{3}$ by $1$ .

$a = - 2 + \frac{5}{3} + \frac{4 c}{3} - c$

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 + \frac{5}{3} + \frac{4 c}{3} - c$

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

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

Multiply $- 1$ by $- 1$ .

$a = - 2 + \frac{5}{3} + 1 (\frac{4 c}{3}) - c$

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

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

$a = - 2 + \frac{5}{3} + \frac{4 c}{3} - c$

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 + \frac{5}{3} + \frac{4 c}{3} - c$

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 2 + \frac{5}{3} + \frac{4 c}{3} - c$

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

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

$a = - 2 \cdot \frac{3}{3} + \frac{5}{3} + \frac{4 c}{3} - c$

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

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

$a = \frac{- 2 \cdot 3}{3} + \frac{5}{3} + \frac{4 c}{3} - c$

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

Combine the numerators over the common denominator.

$a = \frac{- 2 \cdot 3 + 5}{3} + \frac{4 c}{3} - c$

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

Simplify the numerator.

Multiply $- 2$ by $3$ .

$a = \frac{- 6 + 5}{3} + \frac{4 c}{3} - c$

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

Add $- 6$ and $5$ .

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

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

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

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

Move the negative in front of the fraction.

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

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

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

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

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

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

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

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

Combine the numerators over the common denominator.

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

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

Combine the numerators over the common denominator.

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

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

Multiply $3$ by $- 1$ .

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

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

Subtract $3 c$ from $4 c$ .

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

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

Rewrite $- 1$ as $- 1 (1)$ .

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

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

Factor $- 1$ out of $c$ .

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

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

Factor $- 1$ out of $- 1 (1) - 1 (- c)$ .

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

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

Move the negative in front of the fraction.

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

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

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

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

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

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

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

$- 4 = - \frac{14 + c}{3}$

$b = - \frac{5}{3} - \frac{4 c}{3}$

Solve for $c$ in $- 4 = - \frac{14 + c}{3}$ .

Rewrite the equation as $- \frac{14 + c}{3} = - 4$ .

$- \frac{14 + c}{3} = - 4$

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

$b = - \frac{5}{3} - \frac{4 c}{3}$

Multiply both sides of the equation by $- 3$ .

$- 3 (- \frac{14 + c}{3}) = - 3 \cdot - 4$

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

$b = - \frac{5}{3} - \frac{4 c}{3}$

Simplify both sides of the equation.

Simplify the left side.

Simplify $- 3 (- \frac{14 + c}{3})$ .

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{14 + c}{3}$ into the numerator.

$- 3 \frac{- (14 + c)}{3} = - 3 \cdot - 4$

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

$b = - \frac{5}{3} - \frac{4 c}{3}$

Factor $3$ out of $- 3$ .

$3 (- 1) (\frac{- (14 + c)}{3}) = - 3 \cdot - 4$

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

$b = - \frac{5}{3} - \frac{4 c}{3}$

Cancel the common factor.

$3 \cdot (- 1 \frac{- (14 + c)}{3}) = - 3 \cdot - 4$

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

$b = - \frac{5}{3} - \frac{4 c}{3}$

Rewrite the expression.

$14 + c = - 3 \cdot - 4$

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

$b = - \frac{5}{3} - \frac{4 c}{3}$

$14 + c = - 3 \cdot - 4$

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

$b = - \frac{5}{3} - \frac{4 c}{3}$

Multiply.

Multiply $- 1$ by $- 1$ .

$1 (14 + c) = - 3 \cdot - 4$

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

$b = - \frac{5}{3} - \frac{4 c}{3}$

Multiply $14 + c$ by $1$ .

$14 + c = - 3 \cdot - 4$

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

$b = - \frac{5}{3} - \frac{4 c}{3}$

$14 + c = - 3 \cdot - 4$

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

$b = - \frac{5}{3} - \frac{4 c}{3}$

$14 + c = - 3 \cdot - 4$

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

$b = - \frac{5}{3} - \frac{4 c}{3}$

$14 + c = - 3 \cdot - 4$

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

$b = - \frac{5}{3} - \frac{4 c}{3}$

Simplify the right side.

Multiply $- 3$ by $- 4$ .

$14 + c = 12$

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

$b = - \frac{5}{3} - \frac{4 c}{3}$

$14 + c = 12$

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

$b = - \frac{5}{3} - \frac{4 c}{3}$

$14 + c = 12$

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

$b = - \frac{5}{3} - \frac{4 c}{3}$

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

Subtract $14$ from both sides of the equation.

$c = 12 - 14$

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

$b = - \frac{5}{3} - \frac{4 c}{3}$

Subtract $14$ from $12$ .

$c = - 2$

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

$b = - \frac{5}{3} - \frac{4 c}{3}$

$c = - 2$

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

$b = - \frac{5}{3} - \frac{4 c}{3}$

$c = - 2$

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

$b = - \frac{5}{3} - \frac{4 c}{3}$

Replace all occurrences of $c$ with $- 2$ in each equation.

Replace all occurrences of $c$ in $a = - \frac{1 - c}{3}$ with $- 2$ .

$a = - \frac{1 - (- 2)}{3}$

$c = - 2$

$b = - \frac{5}{3} - \frac{4 c}{3}$

Simplify the right side.

Simplify $- \frac{1 - (- 2)}{3}$ .

Simplify the numerator.

Multiply $- 1$ by $- 2$ .

$a = - \frac{1 + 2}{3}$

$c = - 2$

$b = - \frac{5}{3} - \frac{4 c}{3}$

Add $1$ and $2$ .

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

$c = - 2$

$b = - \frac{5}{3} - \frac{4 c}{3}$

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

$c = - 2$

$b = - \frac{5}{3} - \frac{4 c}{3}$

Reduce the expression by cancelling the common factors.

Cancel the common factor of $3$ .

Cancel the common factor.

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

$c = - 2$

$b = - \frac{5}{3} - \frac{4 c}{3}$

Rewrite the expression.

$a = - 1 \cdot 1$

$c = - 2$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 1 \cdot 1$

$c = - 2$

$b = - \frac{5}{3} - \frac{4 c}{3}$

Multiply $- 1$ by $1$ .

$a = - 1$

$c = - 2$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 1$

$c = - 2$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 1$

$c = - 2$

$b = - \frac{5}{3} - \frac{4 c}{3}$

$a = - 1$

$c = - 2$

$b = - \frac{5}{3} - \frac{4 c}{3}$

Replace all occurrences of $c$ in $b = - \frac{5}{3} - \frac{4 c}{3}$ with $- 2$ .

$b = - \frac{5}{3} - \frac{4 (- 2)}{3}$

$a = - 1$

$c = - 2$

Simplify the right side.

Simplify $- \frac{5}{3} - \frac{4 (- 2)}{3}$ .

Combine the numerators over the common denominator.

$b = \frac{- 5 - 4 \cdot - 2}{3}$

$a = - 1$

$c = - 2$

Simplify the expression.

Multiply $- 4$ by $- 2$ .

$b = \frac{- 5 + 8}{3}$

$a = - 1$

$c = - 2$

Add $- 5$ and $8$ .

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

$a = - 1$

$c = - 2$

Divide $3$ by $3$ .

$b = 1$

$a = - 1$

$c = - 2$

$b = 1$

$a = - 1$

$c = - 2$

$b = 1$

$a = - 1$

$c = - 2$

$b = 1$

$a = - 1$

$c = - 2$

$b = 1$

$a = - 1$

$c = - 2$

List all of the solutions.

$b = 1, a = - 1, c = - 2$

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 $a = - 1$ , $b = 1$ , $c = - 2$ , and $x = 1$ .

Simplify each term.

One to any power is one.

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

Multiply $- 1$ by $1$ .

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

Multiply $1$ by $1$ .

$y = - 1 + 1 - 2$

Simplify by adding and subtracting.

Add $- 1$ and $1$ .

$y = 0 - 2$

Subtract $2$ from $0$ .

$y = - 2$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 1$ . This check passes since $y = - 2$ and $y = - 2$ .

$- 2 = - 2$

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

Simplify each term.

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

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

Multiply $- 1$ by $4$ .

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

Multiply $2$ by $1$ .

$y = - 4 + 2 - 2$

Simplify by adding and subtracting.

Add $- 4$ and $2$ .

$y = - 2 - 2$

Subtract $2$ from $- 2$ .

$y = - 4$

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 $a = - 1$ , $b = 1$ , $c = - 2$ , and $x = 3$ .

Simplify each term.

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

$y = - 1 \cdot 9 + (1) \cdot (3) - 2$

Multiply $- 1$ by $9$ .

$y = - 9 + (1) \cdot (3) - 2$

Multiply $3$ by $1$ .

$y = - 9 + 3 - 2$

Simplify by adding and subtracting.

Add $- 9$ and $3$ .

$y = - 6 - 2$

Subtract $2$ from $- 6$ .

$y = - 8$

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

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

Find the Function Rule table[[x,y],[1,-2],[2,-4],[3,-8]]

Related Questions

Support LCMGCF.com by disabling your adblocker.