Find the Function Rule table[[x,q(x)],[2,1],[3,2],[4,3],[5,4]]

$\begin{array}{c} x & q (x) \\ 2 & 1 \\ 3 & 2 \\ 4 & 3 \\ 5 & 4 \end{array}$

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 $q (x) = a x + b$ .

$1 = a (2) + b 2 = a (3) + b 3 = a (4) + b 4 = a (5) + b$

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

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

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

$a (2) + b = 1$

$2 = a (3) + b$

$3 = a (4) + b$

$4 = a (5) + b$

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

$2 a + b = 1$

$2 = a (3) + b$

$3 = a (4) + b$

$4 = a (5) + b$

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

$b = 1 - 2 a$

$2 = a (3) + b$

$3 = a (4) + b$

$4 = a (5) + b$

$b = 1 - 2 a$

$2 = a (3) + b$

$3 = a (4) + b$

$4 = a (5) + b$

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

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

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

$b = 1 - 2 a$

$3 = a (4) + b$

$4 = a (5) + b$

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

Simplify the left side.

Remove parentheses.

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

$b = 1 - 2 a$

$3 = a (4) + b$

$4 = a (5) + b$

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

$b = 1 - 2 a$

$3 = a (4) + b$

$4 = a (5) + b$

Simplify the right side.

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

Move $3$ to the left of $a$ .

$2 = 3 a + 1 - 2 a$

$b = 1 - 2 a$

$3 = a (4) + b$

$4 = a (5) + b$

Subtract $2 a$ from $3 a$ .

$2 = a + 1$

$b = 1 - 2 a$

$3 = a (4) + b$

$4 = a (5) + b$

$2 = a + 1$

$b = 1 - 2 a$

$3 = a (4) + b$

$4 = a (5) + b$

$2 = a + 1$

$b = 1 - 2 a$

$3 = a (4) + b$

$4 = a (5) + b$

$2 = a + 1$

$b = 1 - 2 a$

$3 = a (4) + b$

$4 = a (5) + b$

Replace all occurrences of $b$ in $3 = a (4) + b$ with $1 - 2 a$ .

$3 = a (4) + 1 - 2 a$

$2 = a + 1$

$b = 1 - 2 a$

$4 = a (5) + b$

Simplify $3 = a (4) + 1 - 2 a$ .

Simplify the left side.

Remove parentheses.

$3 = a (4) + 1 - 2 a$

$2 = a + 1$

$b = 1 - 2 a$

$4 = a (5) + b$

$3 = a (4) + 1 - 2 a$

$2 = a + 1$

$b = 1 - 2 a$

$4 = a (5) + b$

Simplify the right side.

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

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

$3 = 4 a + 1 - 2 a$

$2 = a + 1$

$b = 1 - 2 a$

$4 = a (5) + b$

Subtract $2 a$ from $4 a$ .

$3 = 2 a + 1$

$2 = a + 1$

$b = 1 - 2 a$

$4 = a (5) + b$

$3 = 2 a + 1$

$2 = a + 1$

$b = 1 - 2 a$

$4 = a (5) + b$

$3 = 2 a + 1$

$2 = a + 1$

$b = 1 - 2 a$

$4 = a (5) + b$

$3 = 2 a + 1$

$2 = a + 1$

$b = 1 - 2 a$

$4 = a (5) + b$

Replace all occurrences of $b$ in $4 = a (5) + b$ with $1 - 2 a$ .

$4 = a (5) + 1 - 2 a$

$3 = 2 a + 1$

$2 = a + 1$

$b = 1 - 2 a$

Simplify $4 = a (5) + 1 - 2 a$ .

Simplify the left side.

Remove parentheses.

$4 = a (5) + 1 - 2 a$

$3 = 2 a + 1$

$2 = a + 1$

$b = 1 - 2 a$

$4 = a (5) + 1 - 2 a$

$3 = 2 a + 1$

$2 = a + 1$

$b = 1 - 2 a$

Simplify the right side.

Simplify $a (5) + 1 - 2 a$ .

Move $5$ to the left of $a$ .

$4 = 5 a + 1 - 2 a$

$3 = 2 a + 1$

$2 = a + 1$

$b = 1 - 2 a$

Subtract $2 a$ from $5 a$ .

$4 = 3 a + 1$

$3 = 2 a + 1$

$2 = a + 1$

$b = 1 - 2 a$

$4 = 3 a + 1$

$3 = 2 a + 1$

$2 = a + 1$

$b = 1 - 2 a$

$4 = 3 a + 1$

$3 = 2 a + 1$

$2 = a + 1$

$b = 1 - 2 a$

$4 = 3 a + 1$

$3 = 2 a + 1$

$2 = a + 1$

$b = 1 - 2 a$

$4 = 3 a + 1$

$3 = 2 a + 1$

$2 = a + 1$

$b = 1 - 2 a$

Solve for $a$ in $4 = 3 a + 1$ .

Rewrite the equation as $3 a + 1 = 4$ .

$3 a + 1 = 4$

$3 = 2 a + 1$

$2 = a + 1$

$b = 1 - 2 a$

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

Subtract $1$ from both sides of the equation.

$3 a = 4 - 1$

$3 = 2 a + 1$

$2 = a + 1$

$b = 1 - 2 a$

Subtract $1$ from $4$ .

$3 a = 3$

$3 = 2 a + 1$

$2 = a + 1$

$b = 1 - 2 a$

$3 a = 3$

$3 = 2 a + 1$

$2 = a + 1$

$b = 1 - 2 a$

Divide each term in $3 a = 3$ by $3$ and simplify.

Divide each term in $3 a = 3$ by $3$ .

$\frac{3 a}{3} = \frac{3}{3}$

$3 = 2 a + 1$

$2 = a + 1$

$b = 1 - 2 a$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 a}{3} = \frac{3}{3}$

$3 = 2 a + 1$

$2 = a + 1$

$b = 1 - 2 a$

Divide $a$ by $1$ .

$a = \frac{3}{3}$

$3 = 2 a + 1$

$2 = a + 1$

$b = 1 - 2 a$

$a = \frac{3}{3}$

$3 = 2 a + 1$

$2 = a + 1$

$b = 1 - 2 a$

$a = \frac{3}{3}$

$3 = 2 a + 1$

$2 = a + 1$

$b = 1 - 2 a$

Simplify the right side.

Divide $3$ by $3$ .

$a = 1$

$3 = 2 a + 1$

$2 = a + 1$

$b = 1 - 2 a$

$a = 1$

$3 = 2 a + 1$

$2 = a + 1$

$b = 1 - 2 a$

$a = 1$

$3 = 2 a + 1$

$2 = a + 1$

$b = 1 - 2 a$

$a = 1$

$3 = 2 a + 1$

$2 = a + 1$

$b = 1 - 2 a$

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

Replace all occurrences of $a$ in $3 = 2 a + 1$ with $1$ .

$3 = 2 (1) + 1$

$a = 1$

$2 = a + 1$

$b = 1 - 2 a$

Simplify the right side.

Simplify $2 (1) + 1$ .

Multiply $2$ by $1$ .

$3 = 2 + 1$

$a = 1$

$2 = a + 1$

$b = 1 - 2 a$

Add $2$ and $1$ .

$3 = 3$

$a = 1$

$2 = a + 1$

$b = 1 - 2 a$

$3 = 3$

$a = 1$

$2 = a + 1$

$b = 1 - 2 a$

$3 = 3$

$a = 1$

$2 = a + 1$

$b = 1 - 2 a$

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

$2 = (1) + 1$

$3 = 3$

$a = 1$

$b = 1 - 2 a$

Simplify the right side.

Add $1$ and $1$ .

$2 = 2$

$3 = 3$

$a = 1$

$b = 1 - 2 a$

$2 = 2$

$3 = 3$

$a = 1$

$b = 1 - 2 a$

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

$b = 1 - 2 \cdot 1$

$2 = 2$

$3 = 3$

$a = 1$

Simplify the right side.

Simplify $1 - 2 \cdot 1$ .

Multiply $- 2$ by $1$ .

$b = 1 - 2$

$2 = 2$

$3 = 3$

$a = 1$

Subtract $2$ from $1$ .

$b = - 1$

$2 = 2$

$3 = 3$

$a = 1$

$b = - 1$

$2 = 2$

$3 = 3$

$a = 1$

$b = - 1$

$2 = 2$

$3 = 3$

$a = 1$

$b = - 1$

$2 = 2$

$3 = 3$

$a = 1$

Remove any equations from the system that are always true.

$b = - 1$

$a = 1$

List all of the solutions.

$b = - 1, a = 1$

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

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, $y = q (x)$ for the corresponding $x$ value, $x = 2$ . This check passes since $y = 1$ and $q (x) = 1$ .

$1 = 1$

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, $y = q (x)$ for the corresponding $x$ value, $x = 3$ . This check passes since $y = 2$ and $q (x) = 2$ .

$2 = 2$

Calculate the value of $y$ when $a = 1$ , $b = - 1$ , and $x = 4$ .

Multiply $4$ by $1$ .

$y = 4 - 1$

Subtract $1$ from $4$ .

$y = 3$

If the table has a linear function rule, $y = q (x)$ for the corresponding $x$ value, $x = 4$ . This check passes since $y = 3$ and $q (x) = 3$ .

$3 = 3$

Calculate the value of $y$ when $a = 1$ , $b = - 1$ , and $x = 5$ .

Multiply $5$ by $1$ .

$y = 5 - 1$

Subtract $1$ from $5$ .

$y = 4$

If the table has a linear function rule, $y = q (x)$ for the corresponding $x$ value, $x = 5$ . This check passes since $y = 4$ and $q (x) = 4$ .

$4 = 4$

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

The function is linear

Step 2

Since all $y = q (x)$ , the function is linear and follows the form $y = x - 1$ .

$y = x - 1$

Find the Function Rule table[[x,q(x)],[2,1],[3,2],[4,3],[5,4]]

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