Find the Function Rule table[[x,y],[-2,-0.5],[-1,-1],[1,1]]

$\begin{array}{c} x & y \\ - 2 & - 0.5 \\ - 1 & - 1 \\ 1 & 1 \end{array}$

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 $y = a x + b$ .

$- 0.5 = a (- 2) + b - 1 = a (- 1) + b 1 = a (1) + b$

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

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

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

$a + b = 1$

$- 0.5 = a (- 2) + b$

$- 1 = a (- 1) + b$

Subtract $b$ from both sides of the equation.

$a = 1 - b$

$- 0.5 = a (- 2) + b$

$- 1 = a (- 1) + b$

$a = 1 - b$

$- 0.5 = a (- 2) + b$

$- 1 = a (- 1) + b$

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

Replace all occurrences of $a$ in $- 0.5 = a (- 2) + b$ with $1 - b$ .

$- 0.5 = (1 - b) (- 2) + b$

$a = 1 - b$

$- 1 = a (- 1) + b$

Simplify the right side.

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

Simplify each term.

Apply the distributive property.

$- 0.5 = 1 \cdot - 2 - b \cdot - 2 + b$

$a = 1 - b$

$- 1 = a (- 1) + b$

Multiply $- 2$ by $1$ .

$- 0.5 = - 2 - b \cdot - 2 + b$

$a = 1 - b$

$- 1 = a (- 1) + b$

Multiply $- 2$ by $- 1$ .

$- 0.5 = - 2 + 2 b + b$

$a = 1 - b$

$- 1 = a (- 1) + b$

$- 0.5 = - 2 + 2 b + b$

$a = 1 - b$

$- 1 = a (- 1) + b$

Add $2 b$ and $b$ .

$- 0.5 = - 2 + 3 b$

$a = 1 - b$

$- 1 = a (- 1) + b$

$- 0.5 = - 2 + 3 b$

$a = 1 - b$

$- 1 = a (- 1) + b$

$- 0.5 = - 2 + 3 b$

$a = 1 - b$

$- 1 = a (- 1) + b$

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

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

$- 0.5 = - 2 + 3 b$

$a = 1 - b$

Simplify the right side.

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

Simplify each term.

Apply the distributive property.

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

$- 0.5 = - 2 + 3 b$

$a = 1 - b$

Multiply $- 1$ by $1$ .

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

$- 0.5 = - 2 + 3 b$

$a = 1 - b$

Multiply $- b \cdot - 1$ .

Multiply $- 1$ by $- 1$ .

$- 1 = - 1 + 1 b + b$

$- 0.5 = - 2 + 3 b$

$a = 1 - b$

Multiply $b$ by $1$ .

$- 1 = - 1 + b + b$

$- 0.5 = - 2 + 3 b$

$a = 1 - b$

$- 1 = - 1 + b + b$

$- 0.5 = - 2 + 3 b$

$a = 1 - b$

$- 1 = - 1 + b + b$

$- 0.5 = - 2 + 3 b$

$a = 1 - b$

Add $b$ and $b$ .

$- 1 = - 1 + 2 b$

$- 0.5 = - 2 + 3 b$

$a = 1 - b$

$- 1 = - 1 + 2 b$

$- 0.5 = - 2 + 3 b$

$a = 1 - b$

$- 1 = - 1 + 2 b$

$- 0.5 = - 2 + 3 b$

$a = 1 - b$

$- 1 = - 1 + 2 b$

$- 0.5 = - 2 + 3 b$

$a = 1 - b$

Solve for $b$ in $- 1 = - 1 + 2 b$ .

Rewrite the equation as $- 1 + 2 b = - 1$ .

$- 1 + 2 b = - 1$

$- 0.5 = - 2 + 3 b$

$a = 1 - b$

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

Add $1$ to both sides of the equation.

$2 b = - 1 + 1$

$- 0.5 = - 2 + 3 b$

$a = 1 - b$

Add $- 1$ and $1$ .

$2 b = 0$

$- 0.5 = - 2 + 3 b$

$a = 1 - b$

$2 b = 0$

$- 0.5 = - 2 + 3 b$

$a = 1 - b$

Divide each term in $2 b = 0$ by $2$ and simplify.

Divide each term in $2 b = 0$ by $2$ .

$\frac{2 b}{2} = \frac{0}{2}$

$- 0.5 = - 2 + 3 b$

$a = 1 - b$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 b}{2} = \frac{0}{2}$

$- 0.5 = - 2 + 3 b$

$a = 1 - b$

Divide $b$ by $1$ .

$b = \frac{0}{2}$

$- 0.5 = - 2 + 3 b$

$a = 1 - b$

$b = \frac{0}{2}$

$- 0.5 = - 2 + 3 b$

$a = 1 - b$

$b = \frac{0}{2}$

$- 0.5 = - 2 + 3 b$

$a = 1 - b$

Simplify the right side.

Divide $0$ by $2$ .

$b = 0$

$- 0.5 = - 2 + 3 b$

$a = 1 - b$

$b = 0$

$- 0.5 = - 2 + 3 b$

$a = 1 - b$

$b = 0$

$- 0.5 = - 2 + 3 b$

$a = 1 - b$

$b = 0$

$- 0.5 = - 2 + 3 b$

$a = 1 - b$

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

Replace all occurrences of $b$ in $- 0.5 = - 2 + 3 b$ with $0$ .

$- 0.5 = - 2 + 3 (0)$

$b = 0$

$a = 1 - b$

Simplify the right side.

Simplify $- 2 + 3 (0)$ .

Multiply $3$ by $0$ .

$- 0.5 = - 2 + 0$

$b = 0$

$a = 1 - b$

Add $- 2$ and $0$ .

$- 0.5 = - 2$

$b = 0$

$a = 1 - b$

$- 0.5 = - 2$

$b = 0$

$a = 1 - b$

$- 0.5 = - 2$

$b = 0$

$a = 1 - b$

Replace all occurrences of $b$ in $a = 1 - b$ with $0$ .

$a = 1 - (0)$

$- 0.5 = - 2$

$b = 0$

Simplify the right side.

Subtract $0$ from $1$ .

$a = 1$

$- 0.5 = - 2$

$b = 0$

$a = 1$

$- 0.5 = - 2$

$b = 0$

$a = 1$

$- 0.5 = - 2$

$b = 0$

Since $- 0.5 = - 2$ 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 $a$ in $1 = a + b + c$ .

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

$a + b + c = 1$

$- 0.5 = a {(- 2)}^{2} + b (- 2) + c$

$- 1 = a {(- 1)}^{2} + b (- 1) + c$

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

Subtract $b$ from both sides of the equation.

$a + c = 1 - b$

$- 0.5 = a {(- 2)}^{2} + b (- 2) + c$

$- 1 = a {(- 1)}^{2} + b (- 1) + c$

Subtract $c$ from both sides of the equation.

$a = 1 - b - c$

$- 0.5 = a {(- 2)}^{2} + b (- 2) + c$

$- 1 = a {(- 1)}^{2} + b (- 1) + c$

$a = 1 - b - c$

$- 0.5 = a {(- 2)}^{2} + b (- 2) + c$

$- 1 = a {(- 1)}^{2} + b (- 1) + c$

$a = 1 - b - c$

$- 0.5 = a {(- 2)}^{2} + b (- 2) + c$

$- 1 = a {(- 1)}^{2} + b (- 1) + c$

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

Replace all occurrences of $a$ in $- 0.5 = a {(- 2)}^{2} + b (- 2) + c$ with $1 - b - c$ .

$- 0.5 = (1 - b - c) {(- 2)}^{2} + b (- 2) + c$

$a = 1 - b - c$

$- 1 = a {(- 1)}^{2} + b (- 1) + c$

Simplify the right side.

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

Simplify each term.

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

$- 0.5 = (1 - b - c) \cdot 4 + b (- 2) + c$

$a = 1 - b - c$

$- 1 = a {(- 1)}^{2} + b (- 1) + c$

Apply the distributive property.

$- 0.5 = 1 \cdot 4 - b \cdot 4 - c \cdot 4 + b (- 2) + c$

$a = 1 - b - c$

$- 1 = a {(- 1)}^{2} + b (- 1) + c$

Simplify.

Multiply $4$ by $1$ .

$- 0.5 = 4 - b \cdot 4 - c \cdot 4 + b (- 2) + c$

$a = 1 - b - c$

$- 1 = a {(- 1)}^{2} + b (- 1) + c$

Multiply $4$ by $- 1$ .

$- 0.5 = 4 - 4 b - c \cdot 4 + b (- 2) + c$

$a = 1 - b - c$

$- 1 = a {(- 1)}^{2} + b (- 1) + c$

Multiply $4$ by $- 1$ .

$- 0.5 = 4 - 4 b - 4 c + b (- 2) + c$

$a = 1 - b - c$

$- 1 = a {(- 1)}^{2} + b (- 1) + c$

$- 0.5 = 4 - 4 b - 4 c + b (- 2) + c$

$a = 1 - b - c$

$- 1 = a {(- 1)}^{2} + b (- 1) + c$

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

$- 0.5 = 4 - 4 b - 4 c - 2 b + c$

$a = 1 - b - c$

$- 1 = a {(- 1)}^{2} + b (- 1) + c$

$- 0.5 = 4 - 4 b - 4 c - 2 b + c$

$a = 1 - b - c$

$- 1 = a {(- 1)}^{2} + b (- 1) + c$

Simplify by adding terms.

Subtract $2 b$ from $- 4 b$ .

$- 0.5 = 4 - 6 b - 4 c + c$

$a = 1 - b - c$

$- 1 = a {(- 1)}^{2} + b (- 1) + c$

Add $- 4 c$ and $c$ .

$- 0.5 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$- 1 = a {(- 1)}^{2} + b (- 1) + c$

$- 0.5 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$- 1 = a {(- 1)}^{2} + b (- 1) + c$

$- 0.5 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$- 1 = a {(- 1)}^{2} + b (- 1) + c$

$- 0.5 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$- 1 = a {(- 1)}^{2} + b (- 1) + c$

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

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

$- 0.5 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Simplify the right side.

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

Simplify each term.

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

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

$- 0.5 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Multiply $1 - b - c$ by $1$ .

$- 1 = 1 - b - c + b (- 1) + c$

$- 0.5 = 4 - 6 b - 3 c$

$a = 1 - b - c$

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

$- 1 = 1 - b - c - 1 b + c$

$- 0.5 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Rewrite $- 1 b$ as $- b$ .

$- 1 = 1 - b - c - b + c$

$- 0.5 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$- 1 = 1 - b - c - b + c$

$- 0.5 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Simplify by adding terms.

Combine the opposite terms in $1 - b - c - b + c$ .

Add $- c$ and $c$ .

$- 1 = 1 - b - b + 0$

$- 0.5 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Add $1 - b - b$ and $0$ .

$- 1 = 1 - b - b$

$- 0.5 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$- 1 = 1 - b - b$

$- 0.5 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Subtract $b$ from $- b$ .

$- 1 = 1 - 2 b$

$- 0.5 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$- 1 = 1 - 2 b$

$- 0.5 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$- 1 = 1 - 2 b$

$- 0.5 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$- 1 = 1 - 2 b$

$- 0.5 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$- 1 = 1 - 2 b$

$- 0.5 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Solve for $b$ in $- 1 = 1 - 2 b$ .

Rewrite the equation as $1 - 2 b = - 1$ .

$1 - 2 b = - 1$

$- 0.5 = 4 - 6 b - 3 c$

$a = 1 - b - c$

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

Subtract $1$ from both sides of the equation.

$- 2 b = - 1 - 1$

$- 0.5 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Subtract $1$ from $- 1$ .

$- 2 b = - 2$

$- 0.5 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$- 2 b = - 2$

$- 0.5 = 4 - 6 b - 3 c$

$a = 1 - b - c$

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

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

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

$- 0.5 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

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

$- 0.5 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Divide $b$ by $1$ .

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

$- 0.5 = 4 - 6 b - 3 c$

$a = 1 - b - c$

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

$- 0.5 = 4 - 6 b - 3 c$

$a = 1 - b - c$

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

$- 0.5 = 4 - 6 b - 3 c$

$a = 1 - b - c$

Simplify the right side.

Divide $- 2$ by $- 2$ .

$b = 1$

$- 0.5 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$b = 1$

$- 0.5 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$b = 1$

$- 0.5 = 4 - 6 b - 3 c$

$a = 1 - b - c$

$b = 1$

$- 0.5 = 4 - 6 b - 3 c$

$a = 1 - b - c$

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

Replace all occurrences of $b$ in $- 0.5 = 4 - 6 b - 3 c$ with $1$ .

$- 0.5 = 4 - 6 \cdot 1 - 3 c$

$b = 1$

$a = 1 - b - c$

Simplify the right side.

Simplify $4 - 6 \cdot 1 - 3 c$ .

Multiply $- 6$ by $1$ .

$- 0.5 = 4 - 6 - 3 c$

$b = 1$

$a = 1 - b - c$

Subtract $6$ from $4$ .

$- 0.5 = - 2 - 3 c$

$b = 1$

$a = 1 - b - c$

$- 0.5 = - 2 - 3 c$

$b = 1$

$a = 1 - b - c$

$- 0.5 = - 2 - 3 c$

$b = 1$

$a = 1 - b - c$

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

$a = 1 - (1) - c$

$- 0.5 = - 2 - 3 c$

$b = 1$

Simplify the right side.

Simplify $1 - (1) - c$ .

Multiply $- 1$ by $1$ .

$a = 1 - 1 - c$

$- 0.5 = - 2 - 3 c$

$b = 1$

Subtract $1$ from $1$ .

$a = 0 - c$

$- 0.5 = - 2 - 3 c$

$b = 1$

Subtract $c$ from $0$ .

$a = - c$

$- 0.5 = - 2 - 3 c$

$b = 1$

$a = - c$

$- 0.5 = - 2 - 3 c$

$b = 1$

$a = - c$

$- 0.5 = - 2 - 3 c$

$b = 1$

$a = - c$

$- 0.5 = - 2 - 3 c$

$b = 1$

Solve for $c$ in $- 0.5 = - 2 - 3 c$ .

Rewrite the equation as $- 2 - 3 c = - 0.5$ .

$- 2 - 3 c = - 0.5$

$a = - c$

$b = 1$

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

Add $2$ to both sides of the equation.

$- 3 c = - 0.5 + 2$

$a = - c$

$b = 1$

Add $- 0.5$ and $2$ .

$- 3 c = 1.5$

$a = - c$

$b = 1$

$- 3 c = 1.5$

$a = - c$

$b = 1$

Divide each term in $- 3 c = 1.5$ by $- 3$ and simplify.

Divide each term in $- 3 c = 1.5$ by $- 3$ .

$\frac{- 3 c}{- 3} = \frac{1.5}{- 3}$

$a = - c$

$b = 1$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 c}{- 3} = \frac{1.5}{- 3}$

$a = - c$

$b = 1$

Divide $c$ by $1$ .

$c = \frac{1.5}{- 3}$

$a = - c$

$b = 1$

$c = \frac{1.5}{- 3}$

$a = - c$

$b = 1$

$c = \frac{1.5}{- 3}$

$a = - c$

$b = 1$

Simplify the right side.

Divide $1.5$ by $- 3$ .

$c = - 0.5$

$a = - c$

$b = 1$

$c = - 0.5$

$a = - c$

$b = 1$

$c = - 0.5$

$a = - c$

$b = 1$

$c = - 0.5$

$a = - c$

$b = 1$

Replace all occurrences of $c$ with $- 0.5$ in each equation.

Replace all occurrences of $c$ in $a = - c$ with $- 0.5$ .

$a = - (- 0.5)$

$c = - 0.5$

$b = 1$

Simplify the right side.

Multiply $- 1$ by $- 0.5$ .

$a = 0.5$

$c = - 0.5$

$b = 1$

$a = 0.5$

$c = - 0.5$

$b = 1$

$a = 0.5$

$c = - 0.5$

$b = 1$

List all of the solutions.

$a = 0.5, c = - 0.5, b = 1$

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 $a = 0.5$ , $b = 1$ , $c = - 0.5$ , and $x = - 2$ .

Simplify each term.

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

$y = 0.5 \cdot 4 + (1) \cdot (- 2) - 0.5$

Multiply $0.5$ by $4$ .

$y = 2 + (1) \cdot (- 2) - 0.5$

Multiply $- 2$ by $1$ .

$y = 2 - 2 - 0.5$

Simplify by subtracting numbers.

Subtract $2$ from $2$ .

$y = 0 - 0.5$

Subtract $0.5$ from $0$ .

$y = - 0.5$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = - 2$ . This check passes since $y = - 0.5$ and $y = - 0.5$ .

$- 0.5 = - 0.5$

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = 0.5$ , $b = 1$ , $c = - 0.5$ , and $x = - 1$ .

Simplify each term.

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

$y = 0.5 \cdot 1 + (1) \cdot (- 1) - 0.5$

Multiply $0.5$ by $1$ .

$y = 0.5 + (1) \cdot (- 1) - 0.5$

Multiply $- 1$ by $1$ .

$y = 0.5 - 1 - 0.5$

Simplify by subtracting numbers.

Subtract $1$ from $0.5$ .

$y = - 0.5 - 0.5$

Subtract $0.5$ from $- 0.5$ .

$y = - 1$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = - 1$ . This check does not pass, since $y = - 1$ and $y = - 1$ . The function rule can\'t be quadratic.

$- 1 \neq - 1$

Since $y \neq y$ for the corresponding $x$ values, the function is not quadratic.

The function is not quadratic

Step 3

There are no values of $a$ , $b$ , or $c$ in the equations $y = a x + b$ or $y = a x^{2} + b x + c$ that work for every pair of $x$ and $y$ .

The table does not have a function rule that is linear or quadratic.

Find the Function Rule table[[x,y],[-2,-0.5],[-1,-1],[1,1]]

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