Find the Function Rule table[[x,y],[0,10],[1,12],[2,14]]

$\begin{array}{c} x & y \\ 0 & 10 \\ 1 & 12 \\ 2 & 14 \end{array}$

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 $y = a x + b$ .

$10 = a (0) + b 12 = a (1) + b 14 = a (2) + b$

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

Rewrite the equation as $b = 10$ .

$b = 10$

$12 = a + b$

$14 = a (2) + b$

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

Replace all occurrences of $b$ in $12 = a + b$ with $10$ .

$12 = a + 10$

$b = 10$

$14 = a (2) + b$

Simplify the left side.

Remove parentheses.

$12 = a + 10$

$b = 10$

$14 = a (2) + b$

$12 = a + 10$

$b = 10$

$14 = a (2) + b$

Replace all occurrences of $b$ in $14 = a (2) + b$ with $10$ .

$14 = a (2) + 10$

$12 = a + 10$

$b = 10$

Simplify $14 = a (2) + 10$ .

Simplify the left side.

Remove parentheses.

$14 = a (2) + 10$

$12 = a + 10$

$b = 10$

$14 = a (2) + 10$

$12 = a + 10$

$b = 10$

Simplify the right side.

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

$14 = 2 a + 10$

$12 = a + 10$

$b = 10$

$14 = 2 a + 10$

$12 = a + 10$

$b = 10$

$14 = 2 a + 10$

$12 = a + 10$

$b = 10$

$14 = 2 a + 10$

$12 = a + 10$

$b = 10$

Solve for $a$ in $14 = 2 a + 10$ .

Rewrite the equation as $2 a + 10 = 14$ .

$2 a + 10 = 14$

$12 = a + 10$

$b = 10$

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

Subtract $10$ from both sides of the equation.

$2 a = 14 - 10$

$12 = a + 10$

$b = 10$

Subtract $10$ from $14$ .

$2 a = 4$

$12 = a + 10$

$b = 10$

$2 a = 4$

$12 = a + 10$

$b = 10$

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

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

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

$12 = a + 10$

$b = 10$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

$12 = a + 10$

$b = 10$

Divide $a$ by $1$ .

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

$12 = a + 10$

$b = 10$

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

$12 = a + 10$

$b = 10$

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

$12 = a + 10$

$b = 10$

Simplify the right side.

Divide $4$ by $2$ .

$a = 2$

$12 = a + 10$

$b = 10$

$a = 2$

$12 = a + 10$

$b = 10$

$a = 2$

$12 = a + 10$

$b = 10$

$a = 2$

$12 = a + 10$

$b = 10$

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

Replace all occurrences of $a$ in $12 = a + 10$ with $2$ .

$12 = (2) + 10$

$a = 2$

$b = 10$

Simplify the right side.

Add $2$ and $10$ .

$12 = 12$

$a = 2$

$b = 10$

$12 = 12$

$a = 2$

$b = 10$

$12 = 12$

$a = 2$

$b = 10$

Remove any equations from the system that are always true.

$a = 2$

$b = 10$

List all of the solutions.

$a = 2, b = 10$

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 $a = 2$ , $b = 10$ , and $x = 0$ .

Multiply $2$ by $0$ .

$y = 0 + 10$

Add $0$ and $10$ .

$y = 10$

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 0$ . This check passes since $y = 10$ and $y = 10$ .

$10 = 10$

Calculate the value of $y$ when $a = 2$ , $b = 10$ , and $x = 1$ .

Multiply $2$ by $1$ .

$y = 2 + 10$

Add $2$ and $10$ .

$y = 12$

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 1$ . This check passes since $y = 12$ and $y = 12$ .

$12 = 12$

Calculate the value of $y$ when $a = 2$ , $b = 10$ , and $x = 2$ .

Multiply $2$ by $2$ .

$y = 4 + 10$

Add $4$ and $10$ .

$y = 14$

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 2$ . This check passes since $y = 14$ and $y = 14$ .

$14 = 14$

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 $y = 2 x + 10$ .

$y = 2 x + 10$

Find the Function Rule table[[x,y],[0,10],[1,12],[2,14]]

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