Find the Function Rule table[[x,y],[2,1],[4,0],[6,-1]]

$\begin{array}{c} x & y \\ 2 & 1 \\ 4 & 0 \\ 6 & - 1 \end{array}$

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 $y = a x + b$ .

$1 = a (2) + b 0 = a (4) + b - 1 = a (6) + b$

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

Solve for $b$ in $1 = a (2) + b$ .

Rewrite the equation as $a (2) + b = 1$ .

$a (2) + b = 1$

$0 = a (4) + b$

$- 1 = a (6) + b$

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

$2 a + b = 1$

$0 = a (4) + b$

$- 1 = a (6) + b$

Subtract $2 a$ from both sides of the equation.

$b = 1 - 2 a$

$0 = a (4) + b$

$- 1 = a (6) + b$

$b = 1 - 2 a$

$0 = a (4) + b$

$- 1 = a (6) + b$

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

Replace all occurrences of $b$ in $0 = a (4) + b$ with $1 - 2 a$ .

$0 = a (4) + 1 - 2 a$

$b = 1 - 2 a$

$- 1 = a (6) + b$

Simplify $0 = a (4) + 1 - 2 a$ .

Simplify the left side.

Remove parentheses.

$0 = a (4) + 1 - 2 a$

$b = 1 - 2 a$

$- 1 = a (6) + b$

$0 = a (4) + 1 - 2 a$

$b = 1 - 2 a$

$- 1 = a (6) + b$

Simplify the right side.

Simplify $a (4) + 1 - 2 a$ .

Move $4$ to the left of $a$ .

$0 = 4 a + 1 - 2 a$

$b = 1 - 2 a$

$- 1 = a (6) + b$

Subtract $2 a$ from $4 a$ .

$0 = 2 a + 1$

$b = 1 - 2 a$

$- 1 = a (6) + b$

$0 = 2 a + 1$

$b = 1 - 2 a$

$- 1 = a (6) + b$

$0 = 2 a + 1$

$b = 1 - 2 a$

$- 1 = a (6) + b$

$0 = 2 a + 1$

$b = 1 - 2 a$

$- 1 = a (6) + b$

Replace all occurrences of $b$ in $- 1 = a (6) + b$ with $1 - 2 a$ .

$- 1 = a (6) + 1 - 2 a$

$0 = 2 a + 1$

$b = 1 - 2 a$

Simplify $- 1 = a (6) + 1 - 2 a$ .

Simplify the left side.

Remove parentheses.

$- 1 = a (6) + 1 - 2 a$

$0 = 2 a + 1$

$b = 1 - 2 a$

$- 1 = a (6) + 1 - 2 a$

$0 = 2 a + 1$

$b = 1 - 2 a$

Simplify the right side.

Simplify $a (6) + 1 - 2 a$ .

Move $6$ to the left of $a$ .

$- 1 = 6 a + 1 - 2 a$

$0 = 2 a + 1$

$b = 1 - 2 a$

Subtract $2 a$ from $6 a$ .

$- 1 = 4 a + 1$

$0 = 2 a + 1$

$b = 1 - 2 a$

$- 1 = 4 a + 1$

$0 = 2 a + 1$

$b = 1 - 2 a$

$- 1 = 4 a + 1$

$0 = 2 a + 1$

$b = 1 - 2 a$

$- 1 = 4 a + 1$

$0 = 2 a + 1$

$b = 1 - 2 a$

$- 1 = 4 a + 1$

$0 = 2 a + 1$

$b = 1 - 2 a$

Solve for $a$ in $- 1 = 4 a + 1$ .

Rewrite the equation as $4 a + 1 = - 1$ .

$4 a + 1 = - 1$

$0 = 2 a + 1$

$b = 1 - 2 a$

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

Subtract $1$ from both sides of the equation.

$4 a = - 1 - 1$

$0 = 2 a + 1$

$b = 1 - 2 a$

Subtract $1$ from $- 1$ .

$4 a = - 2$

$0 = 2 a + 1$

$b = 1 - 2 a$

$4 a = - 2$

$0 = 2 a + 1$

$b = 1 - 2 a$

Divide each term in $4 a = - 2$ by $4$ and simplify.

Divide each term in $4 a = - 2$ by $4$ .

$\frac{4 a}{4} = \frac{- 2}{4}$

$0 = 2 a + 1$

$b = 1 - 2 a$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 a}{4} = \frac{- 2}{4}$

$0 = 2 a + 1$

$b = 1 - 2 a$

Divide $a$ by $1$ .

$a = \frac{- 2}{4}$

$0 = 2 a + 1$

$b = 1 - 2 a$

$a = \frac{- 2}{4}$

$0 = 2 a + 1$

$b = 1 - 2 a$

$a = \frac{- 2}{4}$

$0 = 2 a + 1$

$b = 1 - 2 a$

Simplify the right side.

Cancel the common factor of $- 2$ and $4$ .

Factor $2$ out of $- 2$ .

$a = \frac{2 (- 1)}{4}$

$0 = 2 a + 1$

$b = 1 - 2 a$

Cancel the common factors.

Factor $2$ out of $4$ .

$a = \frac{2 \cdot - 1}{2 \cdot 2}$

$0 = 2 a + 1$

$b = 1 - 2 a$

Cancel the common factor.

$a = \frac{2 \cdot - 1}{2 \cdot 2}$

$0 = 2 a + 1$

$b = 1 - 2 a$

Rewrite the expression.

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

$0 = 2 a + 1$

$b = 1 - 2 a$

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

$0 = 2 a + 1$

$b = 1 - 2 a$

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

$0 = 2 a + 1$

$b = 1 - 2 a$

Move the negative in front of the fraction.

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

$0 = 2 a + 1$

$b = 1 - 2 a$

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

$0 = 2 a + 1$

$b = 1 - 2 a$

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

$0 = 2 a + 1$

$b = 1 - 2 a$

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

$0 = 2 a + 1$

$b = 1 - 2 a$

Replace all occurrences of $a$ with $- \frac{1}{2}$ in each equation.

Replace all occurrences of $a$ in $0 = 2 a + 1$ with $- \frac{1}{2}$ .

$0 = 2 (- \frac{1}{2}) + 1$

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

$b = 1 - 2 a$

Simplify the right side.

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

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$0 = 2 (\frac{- 1}{2}) + 1$

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

$b = 1 - 2 a$

Cancel the common factor.

$0 = 2 (\frac{- 1}{2}) + 1$

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

$b = 1 - 2 a$

Rewrite the expression.

$0 = - 1 + 1$

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

$b = 1 - 2 a$

$0 = - 1 + 1$

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

$b = 1 - 2 a$

Add $- 1$ and $1$ .

$0 = 0$

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

$b = 1 - 2 a$

$0 = 0$

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

$b = 1 - 2 a$

$0 = 0$

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

$b = 1 - 2 a$

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

$b = 1 - 2 (- \frac{1}{2})$

$0 = 0$

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

Simplify the right side.

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

Simplify each term.

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$b = 1 - 2 (\frac{- 1}{2})$

$0 = 0$

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

Factor $2$ out of $- 2$ .

$b = 1 + 2 (- 1) (\frac{- 1}{2})$

$0 = 0$

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

Cancel the common factor.

$b = 1 + 2 \cdot (- 1 (\frac{- 1}{2}))$

$0 = 0$

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

Rewrite the expression.

$b = 1 - 1 \cdot - 1$

$0 = 0$

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

$b = 1 - 1 \cdot - 1$

$0 = 0$

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

Multiply $- 1$ by $- 1$ .

$b = 1 + 1$

$0 = 0$

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

$b = 1 + 1$

$0 = 0$

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

Add $1$ and $1$ .

$b = 2$

$0 = 0$

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

$b = 2$

$0 = 0$

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

$b = 2$

$0 = 0$

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

$b = 2$

$0 = 0$

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

Remove any equations from the system that are always true.

$b = 2$

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

List all of the solutions.

$b = 2, a = - \frac{1}{2}$

Calculate the value of $y$ using each $x$ value in the relation and compare this value to the given $y$ value in the relation.

Calculate the value of $y$ when $a = - \frac{1}{2}$ , $b = 2$ , and $x = 2$ .

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$y = \frac{- 1}{2} \cdot 2 + 2$

Cancel the common factor.

$y = \frac{- 1}{2} \cdot 2 + 2$

Rewrite the expression.

$y = - 1 + 2$

Add $- 1$ and $2$ .

$y = 1$

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

$1 = 1$

Calculate the value of $y$ when $a = - \frac{1}{2}$ , $b = 2$ , and $x = 4$ .

Simplify each term.

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$y = \frac{- 1}{2} \cdot 4 + 2$

Factor $2$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

$y = - 1 \cdot 2 + 2$

Multiply $- 1$ by $2$ .

$y = - 2 + 2$

Add $- 2$ and $2$ .

$y = 0$

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 4$ . This check passes since $y = 0$ and $y = 0$ .

$0 = 0$

Calculate the value of $y$ when $a = - \frac{1}{2}$ , $b = 2$ , and $x = 6$ .

Simplify each term.

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$y = \frac{- 1}{2} \cdot 6 + 2$

Factor $2$ out of $6$ .

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

Cancel the common factor.

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

Rewrite the expression.

$y = - 1 \cdot 3 + 2$

Multiply $- 1$ by $3$ .

$y = - 3 + 2$

Add $- 3$ and $2$ .

$y = - 1$

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

$- 1 = - 1$

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

The function is linear

Step 2

Since all $y = y$ , the function is linear and follows the form $y = - \frac{x}{2} + 2$ .

$y = - \frac{x}{2} + 2$

Find the Function Rule table[[x,y],[2,1],[4,0],[6,-1]]

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