Find the Function Rule table[[x,y],[-15,9],[-10,5],[-5,1]]

$\begin{array}{c} x & y \\ - 15 & 9 \\ - 10 & 5 \\ - 5 & 1 \end{array}$

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 $y = a x + b$ .

$9 = a (- 15) + b 5 = a (- 10) + b 1 = a (- 5) + b$

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

Solve for $b$ in $9 = a (- 15) + b$ .

Rewrite the equation as $a (- 15) + b = 9$ .

$a (- 15) + b = 9$

$5 = a (- 10) + b$

$1 = a (- 5) + b$

Move $- 15$ to the left of $a$ .

$- 15 a + b = 9$

$5 = a (- 10) + b$

$1 = a (- 5) + b$

Add $15 a$ to both sides of the equation.

$b = 9 + 15 a$

$5 = a (- 10) + b$

$1 = a (- 5) + b$

$b = 9 + 15 a$

$5 = a (- 10) + b$

$1 = a (- 5) + b$

Replace all occurrences of $b$ with $9 + 15 a$ in each equation.

Replace all occurrences of $b$ in $5 = a (- 10) + b$ with $9 + 15 a$ .

$5 = a (- 10) + 9 + 15 a$

$b = 9 + 15 a$

$1 = a (- 5) + b$

Simplify $5 = a (- 10) + 9 + 15 a$ .

Simplify the left side.

Remove parentheses.

$5 = a (- 10) + 9 + 15 a$

$b = 9 + 15 a$

$1 = a (- 5) + b$

$5 = a (- 10) + 9 + 15 a$

$b = 9 + 15 a$

$1 = a (- 5) + b$

Simplify the right side.

Simplify $a (- 10) + 9 + 15 a$ .

Move $- 10$ to the left of $a$ .

$5 = - 10 a + 9 + 15 a$

$b = 9 + 15 a$

$1 = a (- 5) + b$

Add $- 10 a$ and $15 a$ .

$5 = 5 a + 9$

$b = 9 + 15 a$

$1 = a (- 5) + b$

$5 = 5 a + 9$

$b = 9 + 15 a$

$1 = a (- 5) + b$

$5 = 5 a + 9$

$b = 9 + 15 a$

$1 = a (- 5) + b$

$5 = 5 a + 9$

$b = 9 + 15 a$

$1 = a (- 5) + b$

Replace all occurrences of $b$ in $1 = a (- 5) + b$ with $9 + 15 a$ .

$1 = a (- 5) + 9 + 15 a$

$5 = 5 a + 9$

$b = 9 + 15 a$

Simplify $1 = a (- 5) + 9 + 15 a$ .

Simplify the left side.

Remove parentheses.

$1 = a (- 5) + 9 + 15 a$

$5 = 5 a + 9$

$b = 9 + 15 a$

$1 = a (- 5) + 9 + 15 a$

$5 = 5 a + 9$

$b = 9 + 15 a$

Simplify the right side.

Simplify $a (- 5) + 9 + 15 a$ .

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

$1 = - 5 a + 9 + 15 a$

$5 = 5 a + 9$

$b = 9 + 15 a$

Add $- 5 a$ and $15 a$ .

$1 = 10 a + 9$

$5 = 5 a + 9$

$b = 9 + 15 a$

$1 = 10 a + 9$

$5 = 5 a + 9$

$b = 9 + 15 a$

$1 = 10 a + 9$

$5 = 5 a + 9$

$b = 9 + 15 a$

$1 = 10 a + 9$

$5 = 5 a + 9$

$b = 9 + 15 a$

$1 = 10 a + 9$

$5 = 5 a + 9$

$b = 9 + 15 a$

Solve for $a$ in $1 = 10 a + 9$ .

Rewrite the equation as $10 a + 9 = 1$ .

$10 a + 9 = 1$

$5 = 5 a + 9$

$b = 9 + 15 a$

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

Subtract $9$ from both sides of the equation.

$10 a = 1 - 9$

$5 = 5 a + 9$

$b = 9 + 15 a$

Subtract $9$ from $1$ .

$10 a = - 8$

$5 = 5 a + 9$

$b = 9 + 15 a$

$10 a = - 8$

$5 = 5 a + 9$

$b = 9 + 15 a$

Divide each term in $10 a = - 8$ by $10$ and simplify.

Divide each term in $10 a = - 8$ by $10$ .

$\frac{10 a}{10} = \frac{- 8}{10}$

$5 = 5 a + 9$

$b = 9 + 15 a$

Simplify the left side.

Cancel the common factor of $10$ .

Cancel the common factor.

$\frac{10 a}{10} = \frac{- 8}{10}$

$5 = 5 a + 9$

$b = 9 + 15 a$

Divide $a$ by $1$ .

$a = \frac{- 8}{10}$

$5 = 5 a + 9$

$b = 9 + 15 a$

$a = \frac{- 8}{10}$

$5 = 5 a + 9$

$b = 9 + 15 a$

$a = \frac{- 8}{10}$

$5 = 5 a + 9$

$b = 9 + 15 a$

Simplify the right side.

Cancel the common factor of $- 8$ and $10$ .

Factor $2$ out of $- 8$ .

$a = \frac{2 (- 4)}{10}$

$5 = 5 a + 9$

$b = 9 + 15 a$

Cancel the common factors.

Factor $2$ out of $10$ .

$a = \frac{2 \cdot - 4}{2 \cdot 5}$

$5 = 5 a + 9$

$b = 9 + 15 a$

Cancel the common factor.

$a = \frac{2 \cdot - 4}{2 \cdot 5}$

$5 = 5 a + 9$

$b = 9 + 15 a$

Rewrite the expression.

$a = \frac{- 4}{5}$

$5 = 5 a + 9$

$b = 9 + 15 a$

$a = \frac{- 4}{5}$

$5 = 5 a + 9$

$b = 9 + 15 a$

$a = \frac{- 4}{5}$

$5 = 5 a + 9$

$b = 9 + 15 a$

Move the negative in front of the fraction.

$a = - \frac{4}{5}$

$5 = 5 a + 9$

$b = 9 + 15 a$

$a = - \frac{4}{5}$

$5 = 5 a + 9$

$b = 9 + 15 a$

$a = - \frac{4}{5}$

$5 = 5 a + 9$

$b = 9 + 15 a$

$a = - \frac{4}{5}$

$5 = 5 a + 9$

$b = 9 + 15 a$

Replace all occurrences of $a$ with $- \frac{4}{5}$ in each equation.

Replace all occurrences of $a$ in $5 = 5 a + 9$ with $- \frac{4}{5}$ .

$5 = 5 (- \frac{4}{5}) + 9$

$a = - \frac{4}{5}$

$b = 9 + 15 a$

Simplify the right side.

Simplify $5 (- \frac{4}{5}) + 9$ .

Cancel the common factor of $5$ .

Move the leading negative in $- \frac{4}{5}$ into the numerator.

$5 = 5 (\frac{- 4}{5}) + 9$

$a = - \frac{4}{5}$

$b = 9 + 15 a$

Cancel the common factor.

$5 = 5 (\frac{- 4}{5}) + 9$

$a = - \frac{4}{5}$

$b = 9 + 15 a$

Rewrite the expression.

$5 = - 4 + 9$

$a = - \frac{4}{5}$

$b = 9 + 15 a$

$5 = - 4 + 9$

$a = - \frac{4}{5}$

$b = 9 + 15 a$

Add $- 4$ and $9$ .

$5 = 5$

$a = - \frac{4}{5}$

$b = 9 + 15 a$

$5 = 5$

$a = - \frac{4}{5}$

$b = 9 + 15 a$

$5 = 5$

$a = - \frac{4}{5}$

$b = 9 + 15 a$

Replace all occurrences of $a$ in $b = 9 + 15 a$ with $- \frac{4}{5}$ .

$b = 9 + 15 (- \frac{4}{5})$

$5 = 5$

$a = - \frac{4}{5}$

Simplify the right side.

Simplify $9 + 15 (- \frac{4}{5})$ .

Simplify each term.

Cancel the common factor of $5$ .

Move the leading negative in $- \frac{4}{5}$ into the numerator.

$b = 9 + 15 (\frac{- 4}{5})$

$5 = 5$

$a = - \frac{4}{5}$

Factor $5$ out of $15$ .

$b = 9 + 5 (3) (\frac{- 4}{5})$

$5 = 5$

$a = - \frac{4}{5}$

Cancel the common factor.

$b = 9 + 5 \cdot (3 (\frac{- 4}{5}))$

$5 = 5$

$a = - \frac{4}{5}$

Rewrite the expression.

$b = 9 + 3 \cdot - 4$

$5 = 5$

$a = - \frac{4}{5}$

$b = 9 + 3 \cdot - 4$

$5 = 5$

$a = - \frac{4}{5}$

Multiply $3$ by $- 4$ .

$b = 9 - 12$

$5 = 5$

$a = - \frac{4}{5}$

$b = 9 - 12$

$5 = 5$

$a = - \frac{4}{5}$

Subtract $12$ from $9$ .

$b = - 3$

$5 = 5$

$a = - \frac{4}{5}$

$b = - 3$

$5 = 5$

$a = - \frac{4}{5}$

$b = - 3$

$5 = 5$

$a = - \frac{4}{5}$

$b = - 3$

$5 = 5$

$a = - \frac{4}{5}$

Remove any equations from the system that are always true.

$b = - 3$

$a = - \frac{4}{5}$

List all of the solutions.

$b = - 3, a = - \frac{4}{5}$

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 $a = - \frac{4}{5}$ , $b = - 3$ , and $x = - 15$ .

Simplify each term.

Cancel the common factor of $5$ .

Move the leading negative in $- \frac{4}{5}$ into the numerator.

$y = \frac{- 4}{5} \cdot - 15 - 3$

Factor $5$ out of $- 15$ .

$y = \frac{- 4}{5} \cdot (5 (- 3)) - 3$

Cancel the common factor.

$y = \frac{- 4}{5} \cdot (5 \cdot - 3) - 3$

Rewrite the expression.

$y = - 4 \cdot - 3 - 3$

Multiply $- 4$ by $- 3$ .

$y = 12 - 3$

Subtract $3$ from $12$ .

$y = 9$

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = - 15$ . This check passes since $y = 9$ and $y = 9$ .

$9 = 9$

Calculate the value of $y$ when $a = - \frac{4}{5}$ , $b = - 3$ , and $x = - 10$ .

Simplify each term.

Cancel the common factor of $5$ .

Move the leading negative in $- \frac{4}{5}$ into the numerator.

$y = \frac{- 4}{5} \cdot - 10 - 3$

Factor $5$ out of $- 10$ .

$y = \frac{- 4}{5} \cdot (5 (- 2)) - 3$

Cancel the common factor.

$y = \frac{- 4}{5} \cdot (5 \cdot - 2) - 3$

Rewrite the expression.

$y = - 4 \cdot - 2 - 3$

Multiply $- 4$ by $- 2$ .

$y = 8 - 3$

Subtract $3$ from $8$ .

$y = 5$

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = - 10$ . This check passes since $y = 5$ and $y = 5$ .

$5 = 5$

Calculate the value of $y$ when $a = - \frac{4}{5}$ , $b = - 3$ , and $x = - 5$ .

Simplify each term.

Cancel the common factor of $5$ .

Move the leading negative in $- \frac{4}{5}$ into the numerator.

$y = \frac{- 4}{5} \cdot - 5 - 3$

Factor $5$ out of $- 5$ .

$y = \frac{- 4}{5} \cdot (5 (- 1)) - 3$

Cancel the common factor.

$y = \frac{- 4}{5} \cdot (5 \cdot - 1) - 3$

Rewrite the expression.

$y = - 4 \cdot - 1 - 3$

Multiply $- 4$ by $- 1$ .

$y = 4 - 3$

Subtract $3$ from $4$ .

$y = 1$

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = - 5$ . This check passes since $y = 1$ and $y = 1$ .

$1 = 1$

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 $y = - \frac{4 x}{5} - 3$ .

$y = - \frac{4 x}{5} - 3$

Find the Function Rule table[[x,y],[-15,9],[-10,5],[-5,1]]

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