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

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

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 $y = a x + b$ .

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

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

Solve for $a$ in $0 = a + b$ .

Rewrite the equation as $a + b = 0$ .

$a + b = 0$

$1 = a (2) + b$

$2 = a (3) + b$

Subtract $b$ from both sides of the equation.

$a = - b$

$1 = a (2) + b$

$2 = a (3) + b$

$a = - b$

$1 = a (2) + b$

$2 = a (3) + b$

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

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

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

$a = - b$

$2 = a (3) + b$

Simplify the right side.

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

Multiply $2$ by $- 1$ .

$1 = - 2 b + b$

$a = - b$

$2 = a (3) + b$

Add $- 2 b$ and $b$ .

$1 = - b$

$a = - b$

$2 = a (3) + b$

$1 = - b$

$a = - b$

$2 = a (3) + b$

$1 = - b$

$a = - b$

$2 = a (3) + b$

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

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

$1 = - b$

$a = - b$

Simplify the right side.

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

Multiply $3$ by $- 1$ .

$2 = - 3 b + b$

$1 = - b$

$a = - b$

Add $- 3 b$ and $b$ .

$2 = - 2 b$

$1 = - b$

$a = - b$

$2 = - 2 b$

$1 = - b$

$a = - b$

$2 = - 2 b$

$1 = - b$

$a = - b$

$2 = - 2 b$

$1 = - b$

$a = - b$

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

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

$- 2 b = 2$

$1 = - b$

$a = - b$

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

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

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

$1 = - b$

$a = - b$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

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

$1 = - b$

$a = - b$

Divide $b$ by $1$ .

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

$1 = - b$

$a = - b$

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

$1 = - b$

$a = - b$

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

$1 = - b$

$a = - b$

Simplify the right side.

Divide $2$ by $- 2$ .

$b = - 1$

$1 = - b$

$a = - b$

$b = - 1$

$1 = - b$

$a = - b$

$b = - 1$

$1 = - b$

$a = - b$

$b = - 1$

$1 = - b$

$a = - b$

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

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

$1 = - (- 1)$

$b = - 1$

$a = - b$

Simplify the right side.

Multiply $- 1$ by $- 1$ .

$1 = 1$

$b = - 1$

$a = - b$

$1 = 1$

$b = - 1$

$a = - b$

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

$a = - (- 1)$

$1 = 1$

$b = - 1$

Simplify the right side.

Multiply $- 1$ by $- 1$ .

$a = 1$

$1 = 1$

$b = - 1$

$a = 1$

$1 = 1$

$b = - 1$

$a = 1$

$1 = 1$

$b = - 1$

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 $x = 1$ .

Multiply $1$ by $1$ .

$y = 1 - 1$

Subtract $1$ from $1$ .

$y = 0$

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

$0 = 0$

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

Multiply $2$ by $1$ .

$y = 2 - 1$

Subtract $1$ from $2$ .

$y = 1$

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

Multiply $3$ by $1$ .

$y = 3 - 1$

Subtract $1$ from $3$ .

$y = 2$

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

$2 = 2$

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],[1,0],[2,1],[3,2]]

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