Find the Function Rule table[[x,y],[-1,6],[0,5],[1,6]]

$\begin{array}{c} x & y \\ - 1 & 6 \\ 0 & 5 \\ 1 & 6 \end{array}$

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 $y = a x + b$ .

$6 = a (- 1) + b 5 = a (0) + b 6 = a (1) + b$

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

Rewrite the equation as $b = 5$ .

$b = 5$

$6 = a (- 1) + b$

$6 = a + b$

Replace all occurrences of $b$ with $5$ in each equation.

Replace all occurrences of $b$ in $6 = a (- 1) + b$ with $5$ .

$6 = a (- 1) + 5$

$b = 5$

$6 = a + b$

Simplify $6 = a (- 1) + 5$ .

Simplify the left side.

Remove parentheses.

$6 = a (- 1) + 5$

$b = 5$

$6 = a + b$

$6 = a (- 1) + 5$

$b = 5$

$6 = a + b$

Simplify the right side.

Simplify each term.

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

$6 = - 1 a + 5$

$b = 5$

$6 = a + b$

Rewrite $- 1 a$ as $- a$ .

$6 = - a + 5$

$b = 5$

$6 = a + b$

$6 = - a + 5$

$b = 5$

$6 = a + b$

$6 = - a + 5$

$b = 5$

$6 = a + b$

$6 = - a + 5$

$b = 5$

$6 = a + b$

Replace all occurrences of $b$ in $6 = a + b$ with $5$ .

$6 = a + 5$

$6 = - a + 5$

$b = 5$

Simplify the left side.

Remove parentheses.

$6 = a + 5$

$6 = - a + 5$

$b = 5$

$6 = a + 5$

$6 = - a + 5$

$b = 5$

$6 = a + 5$

$6 = - a + 5$

$b = 5$

Solve for $a$ in $6 = a + 5$ .

Rewrite the equation as $a + 5 = 6$ .

$a + 5 = 6$

$6 = - a + 5$

$b = 5$

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

Subtract $5$ from both sides of the equation.

$a = 6 - 5$

$6 = - a + 5$

$b = 5$

Subtract $5$ from $6$ .

$a = 1$

$6 = - a + 5$

$b = 5$

$a = 1$

$6 = - a + 5$

$b = 5$

$a = 1$

$6 = - a + 5$

$b = 5$

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

Replace all occurrences of $a$ in $6 = - a + 5$ with $1$ .

$6 = - (1) + 5$

$a = 1$

$b = 5$

Simplify the right side.

Simplify $- (1) + 5$ .

Multiply $- 1$ by $1$ .

$6 = - 1 + 5$

$a = 1$

$b = 5$

Add $- 1$ and $5$ .

$6 = 4$

$a = 1$

$b = 5$

$6 = 4$

$a = 1$

$b = 5$

$6 = 4$

$a = 1$

$b = 5$

$6 = 4$

$a = 1$

$b = 5$

Since $6 = 4$ is not true, there is no solution.

No solution

No solution

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

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 $5 = a \cdot 0^{2} + c$ .

Rewrite the equation as $a \cdot 0^{2} + c = 5$ .

$a \cdot 0^{2} + c = 5$

$6 = a {(- 1)}^{2} + b (- 1) + c$

$6 = a + b + c$

Simplify $a \cdot 0^{2} + c$ .

Simplify each term.

Raising $0$ to any positive power yields $0$ .

$a \cdot 0 + c = 5$

$6 = a {(- 1)}^{2} + b (- 1) + c$

$6 = a + b + c$

Multiply $a$ by $0$ .

$0 + c = 5$

$6 = a {(- 1)}^{2} + b (- 1) + c$

$6 = a + b + c$

$0 + c = 5$

$6 = a {(- 1)}^{2} + b (- 1) + c$

$6 = a + b + c$

Add $0$ and $c$ .

$c = 5$

$6 = a {(- 1)}^{2} + b (- 1) + c$

$6 = a + b + c$

$c = 5$

$6 = a {(- 1)}^{2} + b (- 1) + c$

$6 = a + b + c$

$c = 5$

$6 = a {(- 1)}^{2} + b (- 1) + c$

$6 = a + b + c$

Replace all occurrences of $c$ with $5$ in each equation.

Replace all occurrences of $c$ in $6 = a {(- 1)}^{2} + b (- 1) + c$ with $5$ .

$6 = a {(- 1)}^{2} + b (- 1) + 5$

$c = 5$

$6 = a + b + c$

Simplify $6 = a {(- 1)}^{2} + b (- 1) + 5$ .

Simplify the left side.

Remove parentheses.

$6 = a {(- 1)}^{2} + b (- 1) + 5$

$c = 5$

$6 = a + b + c$

$6 = a {(- 1)}^{2} + b (- 1) + 5$

$c = 5$

$6 = a + b + c$

Simplify the right side.

Simplify each term.

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

$6 = a \cdot 1 + b (- 1) + 5$

$c = 5$

$6 = a + b + c$

Multiply $a$ by $1$ .

$6 = a + b (- 1) + 5$

$c = 5$

$6 = a + b + c$

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

$6 = a - 1 b + 5$

$c = 5$

$6 = a + b + c$

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

$6 = a - b + 5$

$c = 5$

$6 = a + b + c$

$6 = a - b + 5$

$c = 5$

$6 = a + b + c$

$6 = a - b + 5$

$c = 5$

$6 = a + b + c$

$6 = a - b + 5$

$c = 5$

$6 = a + b + c$

Replace all occurrences of $c$ in $6 = a + b + c$ with $5$ .

$6 = a + b + 5$

$6 = a - b + 5$

$c = 5$

Simplify the left side.

Remove parentheses.

$6 = a + b + 5$

$6 = a - b + 5$

$c = 5$

$6 = a + b + 5$

$6 = a - b + 5$

$c = 5$

$6 = a + b + 5$

$6 = a - b + 5$

$c = 5$

Solve for $a$ in $6 = a + b + 5$ .

Rewrite the equation as $a + b + 5 = 6$ .

$a + b + 5 = 6$

$6 = a - b + 5$

$c = 5$

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

Subtract $b$ from both sides of the equation.

$a + 5 = 6 - b$

$6 = a - b + 5$

$c = 5$

Subtract $5$ from both sides of the equation.

$a = 6 - b - 5$

$6 = a - b + 5$

$c = 5$

Subtract $5$ from $6$ .

$a = - b + 1$

$6 = a - b + 5$

$c = 5$

$a = - b + 1$

$6 = a - b + 5$

$c = 5$

$a = - b + 1$

$6 = a - b + 5$

$c = 5$

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

Replace all occurrences of $a$ in $6 = a - b + 5$ with $- b + 1$ .

$6 = (- b + 1) - b + 5$

$a = - b + 1$

$c = 5$

Simplify the right side.

Simplify $(- b + 1) - b + 5$ .

Subtract $b$ from $- b$ .

$6 = - 2 b + 1 + 5$

$a = - b + 1$

$c = 5$

Add $1$ and $5$ .

$6 = - 2 b + 6$

$a = - b + 1$

$c = 5$

$6 = - 2 b + 6$

$a = - b + 1$

$c = 5$

$6 = - 2 b + 6$

$a = - b + 1$

$c = 5$

$6 = - 2 b + 6$

$a = - b + 1$

$c = 5$

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

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

$- 2 b + 6 = 6$

$a = - b + 1$

$c = 5$

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

Subtract $6$ from both sides of the equation.

$- 2 b = 6 - 6$

$a = - b + 1$

$c = 5$

Subtract $6$ from $6$ .

$- 2 b = 0$

$a = - b + 1$

$c = 5$

$- 2 b = 0$

$a = - b + 1$

$c = 5$

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

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

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

$a = - b + 1$

$c = 5$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

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

$a = - b + 1$

$c = 5$

Divide $b$ by $1$ .

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

$a = - b + 1$

$c = 5$

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

$a = - b + 1$

$c = 5$

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

$a = - b + 1$

$c = 5$

Simplify the right side.

Divide $0$ by $- 2$ .

$b = 0$

$a = - b + 1$

$c = 5$

$b = 0$

$a = - b + 1$

$c = 5$

$b = 0$

$a = - b + 1$

$c = 5$

$b = 0$

$a = - b + 1$

$c = 5$

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

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

$a = - (0) + 1$

$b = 0$

$c = 5$

Simplify the right side.

Simplify $- (0) + 1$ .

Multiply $- 1$ by $0$ .

$a = 0 + 1$

$b = 0$

$c = 5$

Add $0$ and $1$ .

$a = 1$

$b = 0$

$c = 5$

$a = 1$

$b = 0$

$c = 5$

$a = 1$

$b = 0$

$c = 5$

$a = 1$

$b = 0$

$c = 5$

List all of the solutions.

$a = 1, b = 0, c = 5$

$a = 1, b = 0, c = 5$

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 $a = 1$ , $b = 0$ , $c = 5$ , and $x = - 1$ .

Simplify each term.

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

$y = {(- 1)}^{2} + (0) \cdot (- 1) + 5$

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

$y = 1 + (0) \cdot (- 1) + 5$

Multiply $0$ by $- 1$ .

$y = 1 + 0 + 5$

$y = 1 + 0 + 5$

Simplify by adding numbers.

Add $1$ and $0$ .

$y = 1 + 5$

Add $1$ and $5$ .

$y = 6$

$y = 6$

$y = 6$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = - 1$ . This check passes since $y = 6$ and $y = 6$ .

$6 = 6$

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = 1$ , $b = 0$ , $c = 5$ , and $x = 0$ .

Simplify each term.

Multiply ${(0)}^{2}$ by $1$ .

$y = {(0)}^{2} + (0) \cdot (0) + 5$

Raising $0$ to any positive power yields $0$ .

$y = 0 + (0) \cdot (0) + 5$

Multiply $0$ by $0$ .

$y = 0 + 0 + 5$

$y = 0 + 0 + 5$

Simplify by adding numbers.

Add $0$ and $0$ .

$y = 0 + 5$

Add $0$ and $5$ .

$y = 5$

$y = 5$

$y = 5$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 0$ . This check passes since $y = 5$ and $y = 5$ .

$5 = 5$

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = 1$ , $b = 0$ , $c = 5$ , and $x = 1$ .

Simplify each term.

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

$y = {(1)}^{2} + (0) \cdot (1) + 5$

One to any power is one.

$y = 1 + (0) \cdot (1) + 5$

Multiply $0$ by $1$ .

$y = 1 + 0 + 5$

$y = 1 + 0 + 5$

Simplify by adding numbers.

Add $1$ and $0$ .

$y = 1 + 5$

Add $1$ and $5$ .

$y = 6$

$y = 6$

$y = 6$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 1$ . This check passes since $y = 6$ and $y = 6$ .

$6 = 6$

Since $y = y$ for the corresponding $x$ values, the function is quadratic.

The function is quadratic

The function is quadratic

The function is quadratic

Step 3

Since all $y = y$ , the function is quadratic and follows the form $y = x^{2} + 5$ .

$y = x^{2} + 5$

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