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

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

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 $y = a x + b$ .

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

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

Rewrite the equation as $b = - 1$ .

$b = - 1$

$- 2 = a + b$

$- 3 = a (2) + b$

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

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

$- 2 = a - 1$

$b = - 1$

$- 3 = a (2) + b$

Simplify the left side.

Remove parentheses.

$- 2 = a - 1$

$b = - 1$

$- 3 = a (2) + b$

$- 2 = a - 1$

$b = - 1$

$- 3 = a (2) + b$

Replace all occurrences of $b$ in $- 3 = a (2) + b$ with $- 1$ .

$- 3 = a (2) - 1$

$- 2 = a - 1$

$b = - 1$

Simplify $- 3 = a (2) - 1$ .

Simplify the left side.

Remove parentheses.

$- 3 = a (2) - 1$

$- 2 = a - 1$

$b = - 1$

$- 3 = a (2) - 1$

$- 2 = a - 1$

$b = - 1$

Simplify the right side.

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

$- 3 = 2 a - 1$

$- 2 = a - 1$

$b = - 1$

$- 3 = 2 a - 1$

$- 2 = a - 1$

$b = - 1$

$- 3 = 2 a - 1$

$- 2 = a - 1$

$b = - 1$

$- 3 = 2 a - 1$

$- 2 = a - 1$

$b = - 1$

Solve for $a$ in $- 3 = 2 a - 1$ .

Rewrite the equation as $2 a - 1 = - 3$ .

$2 a - 1 = - 3$

$- 2 = a - 1$

$b = - 1$

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

Add $1$ to both sides of the equation.

$2 a = - 3 + 1$

$- 2 = a - 1$

$b = - 1$

Add $- 3$ and $1$ .

$2 a = - 2$

$- 2 = a - 1$

$b = - 1$

$2 a = - 2$

$- 2 = a - 1$

$b = - 1$

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

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

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

$- 2 = a - 1$

$b = - 1$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

$- 2 = a - 1$

$b = - 1$

Divide $a$ by $1$ .

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

$- 2 = a - 1$

$b = - 1$

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

$- 2 = a - 1$

$b = - 1$

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

$- 2 = a - 1$

$b = - 1$

Simplify the right side.

Divide $- 2$ by $2$ .

$a = - 1$

$- 2 = a - 1$

$b = - 1$

$a = - 1$

$- 2 = a - 1$

$b = - 1$

$a = - 1$

$- 2 = a - 1$

$b = - 1$

$a = - 1$

$- 2 = a - 1$

$b = - 1$

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

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

$- 2 = (- 1) - 1$

$a = - 1$

$b = - 1$

Simplify the right side.

Subtract $1$ from $- 1$ .

$- 2 = - 2$

$a = - 1$

$b = - 1$

$- 2 = - 2$

$a = - 1$

$b = - 1$

$- 2 = - 2$

$a = - 1$

$b = - 1$

Remove any equations from the system that are always true.

$a = - 1$

$b = - 1$

List all of the solutions.

$a = - 1, b = - 1$

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

Multiply $- 1$ by $0$ .

$y = 0 - 1$

Subtract $1$ from $0$ .

$y = - 1$

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

$- 1 = - 1$

Calculate the value of $y$ when $a = - 1$ , $b = - 1$ , and $x = 1$ .

Multiply $- 1$ by $1$ .

$y = - 1 - 1$

Subtract $1$ from $- 1$ .

$y = - 2$

If the table has a linear 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 $y$ when $a = - 1$ , $b = - 1$ , and $x = 2$ .

Multiply $- 1$ by $2$ .

$y = - 2 - 1$

Subtract $1$ from $- 2$ .

$y = - 3$

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

$- 3 = - 3$

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 $y = - x - 1$ .

$y = - x - 1$

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

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