Find the Function Rule table[[x,y],[1,1],[5,6]]

$\begin{array}{c} x & y \\ 1 & 1 \\ 5 & 6 \end{array}$

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 $y = a x + b$ .

$1 = a (1) + b 6 = a (5) + b$

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

Solve for $a$ in $1 = a + b$ .

Rewrite the equation as $a + b = 1$ .

$a + b = 1$

$6 = a (5) + b$

Subtract $b$ from both sides of the equation.

$a = 1 - b$

$6 = a (5) + b$

$a = 1 - b$

$6 = a (5) + b$

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

Replace all occurrences of $a$ in $6 = a (5) + b$ with $1 - b$ .

$6 = (1 - b) (5) + b$

$a = 1 - b$

Simplify the right side.

Simplify $(1 - b) (5) + b$ .

Simplify each term.

Apply the distributive property.

$6 = 1 \cdot 5 - b \cdot 5 + b$

$a = 1 - b$

Multiply $5$ by $1$ .

$6 = 5 - b \cdot 5 + b$

$a = 1 - b$

Multiply $5$ by $- 1$ .

$6 = 5 - 5 b + b$

$a = 1 - b$

$6 = 5 - 5 b + b$

$a = 1 - b$

Add $- 5 b$ and $b$ .

$6 = 5 - 4 b$

$a = 1 - b$

$6 = 5 - 4 b$

$a = 1 - b$

$6 = 5 - 4 b$

$a = 1 - b$

$6 = 5 - 4 b$

$a = 1 - b$

Solve for $b$ in $6 = 5 - 4 b$ .

Rewrite the equation as $5 - 4 b = 6$ .

$5 - 4 b = 6$

$a = 1 - b$

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

Subtract $5$ from both sides of the equation.

$- 4 b = 6 - 5$

$a = 1 - b$

Subtract $5$ from $6$ .

$- 4 b = 1$

$a = 1 - b$

$- 4 b = 1$

$a = 1 - b$

Divide each term in $- 4 b = 1$ by $- 4$ and simplify.

Divide each term in $- 4 b = 1$ by $- 4$ .

$\frac{- 4 b}{- 4} = \frac{1}{- 4}$

$a = 1 - b$

Simplify the left side.

Cancel the common factor of $- 4$ .

Cancel the common factor.

$\frac{- 4 b}{- 4} = \frac{1}{- 4}$

$a = 1 - b$

Divide $b$ by $1$ .

$b = \frac{1}{- 4}$

$a = 1 - b$

$b = \frac{1}{- 4}$

$a = 1 - b$

$b = \frac{1}{- 4}$

$a = 1 - b$

Simplify the right side.

Move the negative in front of the fraction.

$b = - \frac{1}{4}$

$a = 1 - b$

$b = - \frac{1}{4}$

$a = 1 - b$

$b = - \frac{1}{4}$

$a = 1 - b$

$b = - \frac{1}{4}$

$a = 1 - b$

Replace all occurrences of $b$ with $- \frac{1}{4}$ in each equation.

Replace all occurrences of $b$ in $a = 1 - b$ with $- \frac{1}{4}$ .

$a = 1 - (- \frac{1}{4})$

$b = - \frac{1}{4}$

Simplify the right side.

Simplify $1 - (- \frac{1}{4})$ .

Multiply $- (- \frac{1}{4})$ .

Multiply $- 1$ by $- 1$ .

$a = 1 + 1 (\frac{1}{4})$

$b = - \frac{1}{4}$

Multiply $\frac{1}{4}$ by $1$ .

$a = 1 + \frac{1}{4}$

$b = - \frac{1}{4}$

$a = 1 + \frac{1}{4}$

$b = - \frac{1}{4}$

Write $1$ as a fraction with a common denominator.

$a = \frac{4}{4} + \frac{1}{4}$

$b = - \frac{1}{4}$

Combine the numerators over the common denominator.

$a = \frac{4 + 1}{4}$

$b = - \frac{1}{4}$

Add $4$ and $1$ .

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

$b = - \frac{1}{4}$

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

$b = - \frac{1}{4}$

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

$b = - \frac{1}{4}$

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

$b = - \frac{1}{4}$

List all of the solutions.

$a = \frac{5}{4}, b = - \frac{1}{4}$

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 $a = \frac{5}{4}$ , $b = - \frac{1}{4}$ , and $x = 1$ .

Multiply $\frac{5}{4}$ by $1$ .

$y = \frac{5}{4} - \frac{1}{4}$

Combine fractions.

Combine the numerators over the common denominator.

$y = \frac{5 - 1}{4}$

Simplify the expression.

Subtract $1$ from $5$ .

$y = \frac{4}{4}$

Divide $4$ by $4$ .

$y = 1$

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

$1 = 1$

Calculate the value of $y$ when $a = \frac{5}{4}$ , $b = - \frac{1}{4}$ , and $x = 5$ .

Multiply $(\frac{5}{4}) (5)$ .

Combine $\frac{5}{4}$ and $5$ .

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

Multiply $5$ by $5$ .

$y = \frac{25}{4} - \frac{1}{4}$

Combine fractions.

Combine the numerators over the common denominator.

$y = \frac{25 - 1}{4}$

Simplify the expression.

Subtract $1$ from $25$ .

$y = \frac{24}{4}$

Divide $24$ by $4$ .

$y = 6$

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 5$ . This check passes since $y = 6$ and $y = 6$ .

$6 = 6$

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 $y = \frac{5 x}{4} - \frac{1}{4}$ .

$y = \frac{5 x}{4} - \frac{1}{4}$

Find the Function Rule table[[x,y],[1,1],[5,6]]

Related Questions

Support LCMGCF.com by disabling your adblocker.