Find the Function Rule table[[x,y],[-3,4],[0,-3],[1,2]]

$\begin{array}{c} x & y \\ - 3 & 4 \\ 0 & - 3 \\ 1 & 2 \end{array}$

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 $y = a x + b$ .

$4 = a (- 3) + b - 3 = a (0) + b 2 = a (1) + b$

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

Rewrite the equation as $b = - 3$ .

$b = - 3$

$4 = a (- 3) + b$

$2 = a + b$

Replace all occurrences of $b$ with $- 3$ in each equation.

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

$4 = a (- 3) - 3$

$b = - 3$

$2 = a + b$

Simplify $4 = a (- 3) - 3$ .

Simplify the left side.

Remove parentheses.

$4 = a (- 3) - 3$

$b = - 3$

$2 = a + b$

$4 = a (- 3) - 3$

$b = - 3$

$2 = a + b$

Simplify the right side.

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

$4 = - 3 a - 3$

$b = - 3$

$2 = a + b$

$4 = - 3 a - 3$

$b = - 3$

$2 = a + b$

$4 = - 3 a - 3$

$b = - 3$

$2 = a + b$

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

$2 = a - 3$

$4 = - 3 a - 3$

$b = - 3$

Simplify the left side.

Remove parentheses.

$2 = a - 3$

$4 = - 3 a - 3$

$b = - 3$

$2 = a - 3$

$4 = - 3 a - 3$

$b = - 3$

$2 = a - 3$

$4 = - 3 a - 3$

$b = - 3$

Solve for $a$ in $2 = a - 3$ .

Rewrite the equation as $a - 3 = 2$ .

$a - 3 = 2$

$4 = - 3 a - 3$

$b = - 3$

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

Add $3$ to both sides of the equation.

$a = 2 + 3$

$4 = - 3 a - 3$

$b = - 3$

Add $2$ and $3$ .

$a = 5$

$4 = - 3 a - 3$

$b = - 3$

$a = 5$

$4 = - 3 a - 3$

$b = - 3$

$a = 5$

$4 = - 3 a - 3$

$b = - 3$

Replace all occurrences of $a$ with $5$ in each equation.

Replace all occurrences of $a$ in $4 = - 3 a - 3$ with $5$ .

$4 = - 3 \cdot 5 - 3$

$a = 5$

$b = - 3$

Simplify the right side.

Simplify $- 3 \cdot 5 - 3$ .

Multiply $- 3$ by $5$ .

$4 = - 15 - 3$

$a = 5$

$b = - 3$

Subtract $3$ from $- 15$ .

$4 = - 18$

$a = 5$

$b = - 3$

$4 = - 18$

$a = 5$

$b = - 3$

$4 = - 18$

$a = 5$

$b = - 3$

$4 = - 18$

$a = 5$

$b = - 3$

Since $4 = - 18$ is not true, there is no solution.

No solution

No solution

Since $y \neq y$ for the corresponding $x$ values, the function is not linear.

The function is not linear

The function is not linear

Step 2

Check if the function rule is quadratic.

To find if the table follows a function rule, check whether the function rule could follow the form $y = a x^{2} + b x + c$ .

$y = a x^{2} + b x + c$

Build a set of $3$ equations from the table such that $y = a x^{2} + b x + c$ .

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

Solve for $c$ in $- 3 = a \cdot 0^{2} + c$ .

Rewrite the equation as $a \cdot 0^{2} + c = - 3$ .

$a \cdot 0^{2} + c = - 3$

$4 = a {(- 3)}^{2} + b (- 3) + c$

$2 = a + b + c$

Simplify $a \cdot 0^{2} + c$ .

Simplify each term.

Raising $0$ to any positive power yields $0$ .

$a \cdot 0 + c = - 3$

$4 = a {(- 3)}^{2} + b (- 3) + c$

$2 = a + b + c$

Multiply $a$ by $0$ .

$0 + c = - 3$

$4 = a {(- 3)}^{2} + b (- 3) + c$

$2 = a + b + c$

$0 + c = - 3$

$4 = a {(- 3)}^{2} + b (- 3) + c$

$2 = a + b + c$

Add $0$ and $c$ .

$c = - 3$

$4 = a {(- 3)}^{2} + b (- 3) + c$

$2 = a + b + c$

$c = - 3$

$4 = a {(- 3)}^{2} + b (- 3) + c$

$2 = a + b + c$

$c = - 3$

$4 = a {(- 3)}^{2} + b (- 3) + c$

$2 = a + b + c$

Replace all occurrences of $c$ with $- 3$ in each equation.

Replace all occurrences of $c$ in $4 = a {(- 3)}^{2} + b (- 3) + c$ with $- 3$ .

$4 = a {(- 3)}^{2} + b (- 3) - 3$

$c = - 3$

$2 = a + b + c$

Simplify $4 = a {(- 3)}^{2} + b (- 3) - 3$ .

Simplify the left side.

Remove parentheses.

$4 = a {(- 3)}^{2} + b (- 3) - 3$

$c = - 3$

$2 = a + b + c$

$4 = a {(- 3)}^{2} + b (- 3) - 3$

$c = - 3$

$2 = a + b + c$

Simplify the right side.

Simplify each term.

Raise $- 3$ to the power of $2$ .

$4 = a \cdot 9 + b (- 3) - 3$

$c = - 3$

$2 = a + b + c$

Move $9$ to the left of $a$ .

$4 = 9 \cdot a + b (- 3) - 3$

$c = - 3$

$2 = a + b + c$

Move $- 3$ to the left of $b$ .

$4 = 9 a - 3 b - 3$

$c = - 3$

$2 = a + b + c$

$4 = 9 a - 3 b - 3$

$c = - 3$

$2 = a + b + c$

$4 = 9 a - 3 b - 3$

$c = - 3$

$2 = a + b + c$

$4 = 9 a - 3 b - 3$

$c = - 3$

$2 = a + b + c$

Replace all occurrences of $c$ in $2 = a + b + c$ with $- 3$ .

$2 = a + b - 3$

$4 = 9 a - 3 b - 3$

$c = - 3$

Simplify the left side.

Remove parentheses.

$2 = a + b - 3$

$4 = 9 a - 3 b - 3$

$c = - 3$

$2 = a + b - 3$

$4 = 9 a - 3 b - 3$

$c = - 3$

$2 = a + b - 3$

$4 = 9 a - 3 b - 3$

$c = - 3$

Solve for $a$ in $2 = a + b - 3$ .

Rewrite the equation as $a + b - 3 = 2$ .

$a + b - 3 = 2$

$4 = 9 a - 3 b - 3$

$c = - 3$

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

Subtract $b$ from both sides of the equation.

$a - 3 = 2 - b$

$4 = 9 a - 3 b - 3$

$c = - 3$

Add $3$ to both sides of the equation.

$a = 2 - b + 3$

$4 = 9 a - 3 b - 3$

$c = - 3$

Add $2$ and $3$ .

$a = - b + 5$

$4 = 9 a - 3 b - 3$

$c = - 3$

$a = - b + 5$

$4 = 9 a - 3 b - 3$

$c = - 3$

$a = - b + 5$

$4 = 9 a - 3 b - 3$

$c = - 3$

Replace all occurrences of $a$ with $- b + 5$ in each equation.

Replace all occurrences of $a$ in $4 = 9 a - 3 b - 3$ with $- b + 5$ .

$4 = 9 (- b + 5) - 3 b - 3$

$a = - b + 5$

$c = - 3$

Simplify the right side.

Simplify $9 (- b + 5) - 3 b - 3$ .

Simplify each term.

Apply the distributive property.

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

$a = - b + 5$

$c = - 3$

Multiply $- 1$ by $9$ .

$4 = - 9 b + 9 \cdot 5 - 3 b - 3$

$a = - b + 5$

$c = - 3$

Multiply $9$ by $5$ .

$4 = - 9 b + 45 - 3 b - 3$

$a = - b + 5$

$c = - 3$

$4 = - 9 b + 45 - 3 b - 3$

$a = - b + 5$

$c = - 3$

Simplify by adding terms.

Subtract $3 b$ from $- 9 b$ .

$4 = - 12 b + 45 - 3$

$a = - b + 5$

$c = - 3$

Subtract $3$ from $45$ .

$4 = - 12 b + 42$

$a = - b + 5$

$c = - 3$

$4 = - 12 b + 42$

$a = - b + 5$

$c = - 3$

$4 = - 12 b + 42$

$a = - b + 5$

$c = - 3$

$4 = - 12 b + 42$

$a = - b + 5$

$c = - 3$

$4 = - 12 b + 42$

$a = - b + 5$

$c = - 3$

Solve for $b$ in $4 = - 12 b + 42$ .

Rewrite the equation as $- 12 b + 42 = 4$ .

$- 12 b + 42 = 4$

$a = - b + 5$

$c = - 3$

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

Subtract $42$ from both sides of the equation.

$- 12 b = 4 - 42$

$a = - b + 5$

$c = - 3$

Subtract $42$ from $4$ .

$- 12 b = - 38$

$a = - b + 5$

$c = - 3$

$- 12 b = - 38$

$a = - b + 5$

$c = - 3$

Divide each term in $- 12 b = - 38$ by $- 12$ and simplify.

Divide each term in $- 12 b = - 38$ by $- 12$ .

$\frac{- 12 b}{- 12} = \frac{- 38}{- 12}$

$a = - b + 5$

$c = - 3$

Simplify the left side.

Cancel the common factor of $- 12$ .

Cancel the common factor.

$\frac{- 12 b}{- 12} = \frac{- 38}{- 12}$

$a = - b + 5$

$c = - 3$

Divide $b$ by $1$ .

$b = \frac{- 38}{- 12}$

$a = - b + 5$

$c = - 3$

$b = \frac{- 38}{- 12}$

$a = - b + 5$

$c = - 3$

$b = \frac{- 38}{- 12}$

$a = - b + 5$

$c = - 3$

Simplify the right side.

Cancel the common factor of $- 38$ and $- 12$ .

Factor $- 2$ out of $- 38$ .

$b = \frac{- 2 \cdot 19}{- 12}$

$a = - b + 5$

$c = - 3$

Cancel the common factors.

Factor $- 2$ out of $- 12$ .

$b = \frac{- 2 \cdot 19}{- 2 \cdot 6}$

$a = - b + 5$

$c = - 3$

Cancel the common factor.

$b = \frac{- 2 \cdot 19}{- 2 \cdot 6}$

$a = - b + 5$

$c = - 3$

Rewrite the expression.

$b = \frac{19}{6}$

$a = - b + 5$

$c = - 3$

$b = \frac{19}{6}$

$a = - b + 5$

$c = - 3$

$b = \frac{19}{6}$

$a = - b + 5$

$c = - 3$

$b = \frac{19}{6}$

$a = - b + 5$

$c = - 3$

$b = \frac{19}{6}$

$a = - b + 5$

$c = - 3$

$b = \frac{19}{6}$

$a = - b + 5$

$c = - 3$

Replace all occurrences of $b$ with $\frac{19}{6}$ in each equation.

Replace all occurrences of $b$ in $a = - b + 5$ with $\frac{19}{6}$ .

$a = - (\frac{19}{6}) + 5$

$b = \frac{19}{6}$

$c = - 3$

Simplify the right side.

Simplify $- (\frac{19}{6}) + 5$ .

To write $5$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$a = - \frac{19}{6} + 5 \cdot \frac{6}{6}$

$b = \frac{19}{6}$

$c = - 3$

Combine $5$ and $\frac{6}{6}$ .

$a = - \frac{19}{6} + \frac{5 \cdot 6}{6}$

$b = \frac{19}{6}$

$c = - 3$

Combine the numerators over the common denominator.

$a = \frac{- 19 + 5 \cdot 6}{6}$

$b = \frac{19}{6}$

$c = - 3$

Simplify the numerator.

Multiply $5$ by $6$ .

$a = \frac{- 19 + 30}{6}$

$b = \frac{19}{6}$

$c = - 3$

Add $- 19$ and $30$ .

$a = \frac{11}{6}$

$b = \frac{19}{6}$

$c = - 3$

$a = \frac{11}{6}$

$b = \frac{19}{6}$

$c = - 3$

$a = \frac{11}{6}$

$b = \frac{19}{6}$

$c = - 3$

$a = \frac{11}{6}$

$b = \frac{19}{6}$

$c = - 3$

$a = \frac{11}{6}$

$b = \frac{19}{6}$

$c = - 3$

List all of the solutions.

$a = \frac{11}{6}, b = \frac{19}{6}, c = - 3$

$a = \frac{11}{6}, b = \frac{19}{6}, c = - 3$

Calculate the value of $y$ using each $x$ value in the table and compare this value to the given $y$ value in the table.

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = \frac{11}{6}$ , $b = \frac{19}{6}$ , $c = - 3$ , and $x = - 3$ .

Simplify each term.

Raise $- 3$ to the power of $2$ .

$y = \frac{11}{6} \cdot 9 + (\frac{19}{6}) \cdot (- 3) - 3$

Cancel the common factor of $3$ .

Factor $3$ out of $6$ .

$y = \frac{11}{3 (2)} \cdot 9 + (\frac{19}{6}) \cdot (- 3) - 3$

Factor $3$ out of $9$ .

$y = \frac{11}{3 \cdot 2} \cdot (3 \cdot 3) + (\frac{19}{6}) \cdot (- 3) - 3$

Cancel the common factor.

$y = \frac{11}{3 \cdot 2} \cdot (3 \cdot 3) + (\frac{19}{6}) \cdot (- 3) - 3$

Rewrite the expression.

$y = \frac{11}{2} \cdot 3 + (\frac{19}{6}) \cdot (- 3) - 3$

$y = \frac{11}{2} \cdot 3 + (\frac{19}{6}) \cdot (- 3) - 3$

Combine $\frac{11}{2}$ and $3$ .

$y = \frac{11 \cdot 3}{2} + (\frac{19}{6}) \cdot (- 3) - 3$

Multiply $11$ by $3$ .

$y = \frac{33}{2} + (\frac{19}{6}) \cdot (- 3) - 3$

Cancel the common factor of $3$ .

Factor $3$ out of $6$ .

$y = \frac{33}{2} + \frac{19}{3 (2)} \cdot - 3 - 3$

Factor $3$ out of $- 3$ .

$y = \frac{33}{2} + \frac{19}{3 \cdot 2} \cdot (3 \cdot - 1) - 3$

Cancel the common factor.

$y = \frac{33}{2} + \frac{19}{3 \cdot 2} \cdot (3 \cdot - 1) - 3$

Rewrite the expression.

$y = \frac{33}{2} + \frac{19}{2} \cdot - 1 - 3$

$y = \frac{33}{2} + \frac{19}{2} \cdot - 1 - 3$

Combine $\frac{19}{2}$ and $- 1$ .

$y = \frac{33}{2} + \frac{19 \cdot - 1}{2} - 3$

Multiply $19$ by $- 1$ .

$y = \frac{33}{2} + \frac{- 19}{2} - 3$

Move the negative in front of the fraction.

$y = \frac{33}{2} - \frac{19}{2} - 3$

$y = \frac{33}{2} - \frac{19}{2} - 3$

Combine fractions.

Combine the numerators over the common denominator.

$y = - 3 + \frac{33 - 19}{2}$

Simplify the expression.

Subtract $19$ from $33$ .

$y = - 3 + \frac{14}{2}$

Divide $14$ by $2$ .

$y = - 3 + 7$

Add $- 3$ and $7$ .

$y = 4$

$y = 4$

$y = 4$

$y = 4$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = - 3$ . This check passes since $y = 4$ and $y = 4$ .

$4 = 4$

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = \frac{11}{6}$ , $b = \frac{19}{6}$ , $c = - 3$ , and $x = 0$ .

Simplify each term.

Raising $0$ to any positive power yields $0$ .

$y = \frac{11}{6} \cdot 0 + (\frac{19}{6}) \cdot (0) - 3$

Multiply $\frac{11}{6}$ by $0$ .

$y = 0 + (\frac{19}{6}) \cdot (0) - 3$

Multiply $\frac{19}{6}$ by $0$ .

$y = 0 + 0 - 3$

$y = 0 + 0 - 3$

Simplify by adding and subtracting.

Add $0$ and $0$ .

$y = 0 - 3$

Subtract $3$ from $0$ .

$y = - 3$

$y = - 3$

$y = - 3$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 0$ . This check passes since $y = - 3$ and $y = - 3$ .

$- 3 = - 3$

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = \frac{11}{6}$ , $b = \frac{19}{6}$ , $c = - 3$ , and $x = 1$ .

Simplify each term.

One to any power is one.

$y = \frac{11}{6} \cdot 1 + (\frac{19}{6}) \cdot (1) - 3$

Multiply $\frac{11}{6}$ by $1$ .

$y = \frac{11}{6} + (\frac{19}{6}) \cdot (1) - 3$

Multiply $\frac{19}{6}$ by $1$ .

$y = \frac{11}{6} + \frac{19}{6} - 3$

$y = \frac{11}{6} + \frac{19}{6} - 3$

Combine fractions.

Combine the numerators over the common denominator.

$y = - 3 + \frac{11 + 19}{6}$

Simplify the expression.

Add $11$ and $19$ .

$y = - 3 + \frac{30}{6}$

Divide $30$ by $6$ .

$y = - 3 + 5$

Add $- 3$ and $5$ .

$y = 2$

$y = 2$

$y = 2$

$y = 2$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 1$ . This check passes since $y = 2$ and $y = 2$ .

$2 = 2$

Since $y = y$ for the corresponding $x$ values, the function is quadratic.

The function is quadratic

The function is quadratic

The function is quadratic

Step 3

Since all $y = y$ , the function is quadratic and follows the form $y = \frac{11 x^{2}}{6} + \frac{19 x}{6} - 3$ .

$y = \frac{11 x^{2}}{6} + \frac{19 x}{6} - 3$

Related Questions