Find the Function Rule table[[x,q(x)],[4,10],[7,16],[2,6]]

$\begin{array}{c} x & q (x) \\ 4 & 10 \\ 7 & 16 \\ 2 & 6 \end{array}$

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

$10 = a (4) + b 16 = a (7) + b 6 = a (2) + b$

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

Solve for $b$ in $10 = a (4) + b$ .

Rewrite the equation as $a (4) + b = 10$ .

$a (4) + b = 10$

$16 = a (7) + b$

$6 = a (2) + b$

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

$4 a + b = 10$

$16 = a (7) + b$

$6 = a (2) + b$

Subtract $4 a$ from both sides of the equation.

$b = 10 - 4 a$

$16 = a (7) + b$

$6 = a (2) + b$

$b = 10 - 4 a$

$16 = a (7) + b$

$6 = a (2) + b$

Replace all occurrences of $b$ with $10 - 4 a$ in each equation.

Replace all occurrences of $b$ in $16 = a (7) + b$ with $10 - 4 a$ .

$16 = a (7) + 10 - 4 a$

$b = 10 - 4 a$

$6 = a (2) + b$

Simplify $16 = a (7) + 10 - 4 a$ .

Simplify the left side.

Remove parentheses.

$16 = a (7) + 10 - 4 a$

$b = 10 - 4 a$

$6 = a (2) + b$

$16 = a (7) + 10 - 4 a$

$b = 10 - 4 a$

$6 = a (2) + b$

Simplify the right side.

Simplify $a (7) + 10 - 4 a$ .

Move $7$ to the left of $a$ .

$16 = 7 a + 10 - 4 a$

$b = 10 - 4 a$

$6 = a (2) + b$

Subtract $4 a$ from $7 a$ .

$16 = 3 a + 10$

$b = 10 - 4 a$

$6 = a (2) + b$

$16 = 3 a + 10$

$b = 10 - 4 a$

$6 = a (2) + b$

$16 = 3 a + 10$

$b = 10 - 4 a$

$6 = a (2) + b$

$16 = 3 a + 10$

$b = 10 - 4 a$

$6 = a (2) + b$

Replace all occurrences of $b$ in $6 = a (2) + b$ with $10 - 4 a$ .

$6 = a (2) + 10 - 4 a$

$16 = 3 a + 10$

$b = 10 - 4 a$

Simplify $6 = a (2) + 10 - 4 a$ .

Simplify the left side.

Remove parentheses.

$6 = a (2) + 10 - 4 a$

$16 = 3 a + 10$

$b = 10 - 4 a$

$6 = a (2) + 10 - 4 a$

$16 = 3 a + 10$

$b = 10 - 4 a$

Simplify the right side.

Simplify $a (2) + 10 - 4 a$ .

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

$6 = 2 a + 10 - 4 a$

$16 = 3 a + 10$

$b = 10 - 4 a$

Subtract $4 a$ from $2 a$ .

$6 = - 2 a + 10$

$16 = 3 a + 10$

$b = 10 - 4 a$

$6 = - 2 a + 10$

$16 = 3 a + 10$

$b = 10 - 4 a$

$6 = - 2 a + 10$

$16 = 3 a + 10$

$b = 10 - 4 a$

$6 = - 2 a + 10$

$16 = 3 a + 10$

$b = 10 - 4 a$

$6 = - 2 a + 10$

$16 = 3 a + 10$

$b = 10 - 4 a$

Solve for $a$ in $6 = - 2 a + 10$ .

Rewrite the equation as $- 2 a + 10 = 6$ .

$- 2 a + 10 = 6$

$16 = 3 a + 10$

$b = 10 - 4 a$

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

Subtract $10$ from both sides of the equation.

$- 2 a = 6 - 10$

$16 = 3 a + 10$

$b = 10 - 4 a$

Subtract $10$ from $6$ .

$- 2 a = - 4$

$16 = 3 a + 10$

$b = 10 - 4 a$

$- 2 a = - 4$

$16 = 3 a + 10$

$b = 10 - 4 a$

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

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

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

$16 = 3 a + 10$

$b = 10 - 4 a$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

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

$16 = 3 a + 10$

$b = 10 - 4 a$

Divide $a$ by $1$ .

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

$16 = 3 a + 10$

$b = 10 - 4 a$

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

$16 = 3 a + 10$

$b = 10 - 4 a$

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

$16 = 3 a + 10$

$b = 10 - 4 a$

Simplify the right side.

Divide $- 4$ by $- 2$ .

$a = 2$

$16 = 3 a + 10$

$b = 10 - 4 a$

$a = 2$

$16 = 3 a + 10$

$b = 10 - 4 a$

$a = 2$

$16 = 3 a + 10$

$b = 10 - 4 a$

$a = 2$

$16 = 3 a + 10$

$b = 10 - 4 a$

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

Replace all occurrences of $a$ in $16 = 3 a + 10$ with $2$ .

$16 = 3 (2) + 10$

$a = 2$

$b = 10 - 4 a$

Simplify the right side.

Simplify $3 (2) + 10$ .

Multiply $3$ by $2$ .

$16 = 6 + 10$

$a = 2$

$b = 10 - 4 a$

Add $6$ and $10$ .

$16 = 16$

$a = 2$

$b = 10 - 4 a$

$16 = 16$

$a = 2$

$b = 10 - 4 a$

$16 = 16$

$a = 2$

$b = 10 - 4 a$

Replace all occurrences of $a$ in $b = 10 - 4 a$ with $2$ .

$b = 10 - 4 \cdot 2$

$16 = 16$

$a = 2$

Simplify the right side.

Simplify $10 - 4 \cdot 2$ .

Multiply $- 4$ by $2$ .

$b = 10 - 8$

$16 = 16$

$a = 2$

Subtract $8$ from $10$ .

$b = 2$

$16 = 16$

$a = 2$

$b = 2$

$16 = 16$

$a = 2$

$b = 2$

$16 = 16$

$a = 2$

$b = 2$

$16 = 16$

$a = 2$

Remove any equations from the system that are always true.

$b = 2$

$a = 2$

List all of the solutions.

$b = 2, a = 2$

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 $a = 2$ , $b = 2$ , and $x = 4$ .

Multiply $2$ by $4$ .

$y = 8 + 2$

Add $8$ and $2$ .

$y = 10$

If the table has a linear function rule, $y = q (x)$ for the corresponding $x$ value, $x = 4$ . This check passes since $y = 10$ and $q (x) = 10$ .

$10 = 10$

Calculate the value of $y$ when $a = 2$ , $b = 2$ , and $x = 7$ .

Multiply $2$ by $7$ .

$y = 14 + 2$

Add $14$ and $2$ .

$y = 16$

If the table has a linear function rule, $y = q (x)$ for the corresponding $x$ value, $x = 7$ . This check passes since $y = 16$ and $q (x) = 16$ .

$16 = 16$

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

Multiply $2$ by $2$ .

$y = 4 + 2$

Add $4$ and $2$ .

$y = 6$

If the table has a linear function rule, $y = q (x)$ for the corresponding $x$ value, $x = 2$ . This check passes since $y = 6$ and $q (x) = 6$ .

$6 = 6$

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 $y = 2 x + 2$ .

$y = 2 x + 2$

Find the Function Rule table[[x,q(x)],[4,10],[7,16],[2,6]]

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