Find the Function Rule table[[x,y],[0,5],[1,25],[2,125]]

$\begin{array}{c} x & y \\ 0 & 5 \\ 1 & 25 \\ 2 & 125 \end{array}$

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 $y = a x + b$ .

$5 = a (0) + b 25 = a (1) + b 125 = a (2) + b$

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

Rewrite the equation as $b = 5$ .

$b = 5$

$25 = a + b$

$125 = a (2) + b$

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

Replace all occurrences of $b$ in $25 = a + b$ with $5$ .

$25 = a + 5$

$b = 5$

$125 = a (2) + b$

Simplify the left side.

Remove parentheses.

$25 = a + 5$

$b = 5$

$125 = a (2) + b$

$25 = a + 5$

$b = 5$

$125 = a (2) + b$

Replace all occurrences of $b$ in $125 = a (2) + b$ with $5$ .

$125 = a (2) + 5$

$25 = a + 5$

$b = 5$

Simplify $125 = a (2) + 5$ .

Simplify the left side.

Remove parentheses.

$125 = a (2) + 5$

$25 = a + 5$

$b = 5$

$125 = a (2) + 5$

$25 = a + 5$

$b = 5$

Simplify the right side.

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

$125 = 2 a + 5$

$25 = a + 5$

$b = 5$

$125 = 2 a + 5$

$25 = a + 5$

$b = 5$

$125 = 2 a + 5$

$25 = a + 5$

$b = 5$

$125 = 2 a + 5$

$25 = a + 5$

$b = 5$

Solve for $a$ in $125 = 2 a + 5$ .

Rewrite the equation as $2 a + 5 = 125$ .

$2 a + 5 = 125$

$25 = a + 5$

$b = 5$

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

Subtract $5$ from both sides of the equation.

$2 a = 125 - 5$

$25 = a + 5$

$b = 5$

Subtract $5$ from $125$ .

$2 a = 120$

$25 = a + 5$

$b = 5$

$2 a = 120$

$25 = a + 5$

$b = 5$

Divide each term in $2 a = 120$ by $2$ and simplify.

Divide each term in $2 a = 120$ by $2$ .

$\frac{2 a}{2} = \frac{120}{2}$

$25 = a + 5$

$b = 5$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 a}{2} = \frac{120}{2}$

$25 = a + 5$

$b = 5$

Divide $a$ by $1$ .

$a = \frac{120}{2}$

$25 = a + 5$

$b = 5$

$a = \frac{120}{2}$

$25 = a + 5$

$b = 5$

$a = \frac{120}{2}$

$25 = a + 5$

$b = 5$

Simplify the right side.

Divide $120$ by $2$ .

$a = 60$

$25 = a + 5$

$b = 5$

$a = 60$

$25 = a + 5$

$b = 5$

$a = 60$

$25 = a + 5$

$b = 5$

$a = 60$

$25 = a + 5$

$b = 5$

Replace all occurrences of $a$ with $60$ in each equation.

Replace all occurrences of $a$ in $25 = a + 5$ with $60$ .

$25 = (60) + 5$

$a = 60$

$b = 5$

Simplify the right side.

Add $60$ and $5$ .

$25 = 65$

$a = 60$

$b = 5$

$25 = 65$

$a = 60$

$b = 5$

$25 = 65$

$a = 60$

$b = 5$

Since $25 = 65$ 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 $5 = a \cdot 0^{2} + c$ .

Rewrite the equation as $a \cdot 0^{2} + c = 5$ .

$a \cdot 0^{2} + c = 5$

$25 = a + b + c$

$125 = a \cdot 2^{2} + b (2) + c$

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

Simplify each term.

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

$a \cdot 0 + c = 5$

$25 = a + b + c$

$125 = a \cdot 2^{2} + b (2) + c$

Multiply $a$ by $0$ .

$0 + c = 5$

$25 = a + b + c$

$125 = a \cdot 2^{2} + b (2) + c$

$0 + c = 5$

$25 = a + b + c$

$125 = a \cdot 2^{2} + b (2) + c$

Add $0$ and $c$ .

$c = 5$

$25 = a + b + c$

$125 = a \cdot 2^{2} + b (2) + c$

$c = 5$

$25 = a + b + c$

$125 = a \cdot 2^{2} + b (2) + c$

$c = 5$

$25 = a + b + c$

$125 = a \cdot 2^{2} + b (2) + c$

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

Replace all occurrences of $c$ in $25 = a + b + c$ with $5$ .

$25 = a + b + 5$

$c = 5$

$125 = a \cdot 2^{2} + b (2) + c$

Simplify the left side.

Remove parentheses.

$25 = a + b + 5$

$c = 5$

$125 = a \cdot 2^{2} + b (2) + c$

$25 = a + b + 5$

$c = 5$

$125 = a \cdot 2^{2} + b (2) + c$

Replace all occurrences of $c$ in $125 = a \cdot 2^{2} + b (2) + c$ with $5$ .

$125 = a \cdot 2^{2} + b (2) + 5$

$25 = a + b + 5$

$c = 5$

Simplify $125 = a \cdot 2^{2} + b (2) + 5$ .

Simplify the left side.

Remove parentheses.

$125 = a \cdot 2^{2} + b (2) + 5$

$25 = a + b + 5$

$c = 5$

$125 = a \cdot 2^{2} + b (2) + 5$

$25 = a + b + 5$

$c = 5$

Simplify the right side.

Simplify each term.

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

$125 = a \cdot 4 + b (2) + 5$

$25 = a + b + 5$

$c = 5$

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

$125 = 4 \cdot a + b (2) + 5$

$25 = a + b + 5$

$c = 5$

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

$125 = 4 a + 2 b + 5$

$25 = a + b + 5$

$c = 5$

$125 = 4 a + 2 b + 5$

$25 = a + b + 5$

$c = 5$

$125 = 4 a + 2 b + 5$

$25 = a + b + 5$

$c = 5$

$125 = 4 a + 2 b + 5$

$25 = a + b + 5$

$c = 5$

$125 = 4 a + 2 b + 5$

$25 = a + b + 5$

$c = 5$

Solve for $a$ in $25 = a + b + 5$ .

Rewrite the equation as $a + b + 5 = 25$ .

$a + b + 5 = 25$

$125 = 4 a + 2 b + 5$

$c = 5$

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

Subtract $b$ from both sides of the equation.

$a + 5 = 25 - b$

$125 = 4 a + 2 b + 5$

$c = 5$

Subtract $5$ from both sides of the equation.

$a = 25 - b - 5$

$125 = 4 a + 2 b + 5$

$c = 5$

Subtract $5$ from $25$ .

$a = - b + 20$

$125 = 4 a + 2 b + 5$

$c = 5$

$a = - b + 20$

$125 = 4 a + 2 b + 5$

$c = 5$

$a = - b + 20$

$125 = 4 a + 2 b + 5$

$c = 5$

Replace all occurrences of $a$ with $- b + 20$ in each equation.

Replace all occurrences of $a$ in $125 = 4 a + 2 b + 5$ with $- b + 20$ .

$125 = 4 (- b + 20) + 2 b + 5$

$a = - b + 20$

$c = 5$

Simplify the right side.

Simplify $4 (- b + 20) + 2 b + 5$ .

Simplify each term.

Apply the distributive property.

$125 = 4 (- b) + 4 \cdot 20 + 2 b + 5$

$a = - b + 20$

$c = 5$

Multiply $- 1$ by $4$ .

$125 = - 4 b + 4 \cdot 20 + 2 b + 5$

$a = - b + 20$

$c = 5$

Multiply $4$ by $20$ .

$125 = - 4 b + 80 + 2 b + 5$

$a = - b + 20$

$c = 5$

$125 = - 4 b + 80 + 2 b + 5$

$a = - b + 20$

$c = 5$

Simplify by adding terms.

Add $- 4 b$ and $2 b$ .

$125 = - 2 b + 80 + 5$

$a = - b + 20$

$c = 5$

Add $80$ and $5$ .

$125 = - 2 b + 85$

$a = - b + 20$

$c = 5$

$125 = - 2 b + 85$

$a = - b + 20$

$c = 5$

$125 = - 2 b + 85$

$a = - b + 20$

$c = 5$

$125 = - 2 b + 85$

$a = - b + 20$

$c = 5$

$125 = - 2 b + 85$

$a = - b + 20$

$c = 5$

Solve for $b$ in $125 = - 2 b + 85$ .

Rewrite the equation as $- 2 b + 85 = 125$ .

$- 2 b + 85 = 125$

$a = - b + 20$

$c = 5$

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

Subtract $85$ from both sides of the equation.

$- 2 b = 125 - 85$

$a = - b + 20$

$c = 5$

Subtract $85$ from $125$ .

$- 2 b = 40$

$a = - b + 20$

$c = 5$

$- 2 b = 40$

$a = - b + 20$

$c = 5$

Divide each term in $- 2 b = 40$ by $- 2$ and simplify.

Divide each term in $- 2 b = 40$ by $- 2$ .

$\frac{- 2 b}{- 2} = \frac{40}{- 2}$

$a = - b + 20$

$c = 5$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 b}{- 2} = \frac{40}{- 2}$

$a = - b + 20$

$c = 5$

Divide $b$ by $1$ .

$b = \frac{40}{- 2}$

$a = - b + 20$

$c = 5$

$b = \frac{40}{- 2}$

$a = - b + 20$

$c = 5$

$b = \frac{40}{- 2}$

$a = - b + 20$

$c = 5$

Simplify the right side.

Divide $40$ by $- 2$ .

$b = - 20$

$a = - b + 20$

$c = 5$

$b = - 20$

$a = - b + 20$

$c = 5$

$b = - 20$

$a = - b + 20$

$c = 5$

$b = - 20$

$a = - b + 20$

$c = 5$

Replace all occurrences of $b$ with $- 20$ in each equation.

Replace all occurrences of $b$ in $a = - b + 20$ with $- 20$ .

$a = - (- 20) + 20$

$b = - 20$

$c = 5$

Simplify the right side.

Simplify $- (- 20) + 20$ .

Multiply $- 1$ by $- 20$ .

$a = 20 + 20$

$b = - 20$

$c = 5$

Add $20$ and $20$ .

$a = 40$

$b = - 20$

$c = 5$

$a = 40$

$b = - 20$

$c = 5$

$a = 40$

$b = - 20$

$c = 5$

$a = 40$

$b = - 20$

$c = 5$

List all of the solutions.

$a = 40, b = - 20, c = 5$

$a = 40, b = - 20, c = 5$

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 $a = 40$ , $b = - 20$ , $c = 5$ , and $x = 0$ .

Simplify each term.

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

$y = 40 \cdot 0 + (- 20) \cdot (0) + 5$

Multiply $40$ by $0$ .

$y = 0 + (- 20) \cdot (0) + 5$

Multiply $- 20$ by $0$ .

$y = 0 + 0 + 5$

$y = 0 + 0 + 5$

Simplify by adding numbers.

Add $0$ and $0$ .

$y = 0 + 5$

Add $0$ and $5$ .

$y = 5$

$y = 5$

$y = 5$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 0$ . This check passes since $y = 5$ and $y = 5$ .

$5 = 5$

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = 40$ , $b = - 20$ , $c = 5$ , and $x = 1$ .

Simplify each term.

One to any power is one.

$y = 40 \cdot 1 + (- 20) \cdot (1) + 5$

Multiply $40$ by $1$ .

$y = 40 + (- 20) \cdot (1) + 5$

Multiply $- 20$ by $1$ .

$y = 40 - 20 + 5$

$y = 40 - 20 + 5$

Simplify by adding and subtracting.

Subtract $20$ from $40$ .

$y = 20 + 5$

Add $20$ and $5$ .

$y = 25$

$y = 25$

$y = 25$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 1$ . This check passes since $y = 25$ and $y = 25$ .

$25 = 25$

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = 40$ , $b = - 20$ , $c = 5$ , and $x = 2$ .

Simplify each term.

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

$y = 40 \cdot 4 + (- 20) \cdot (2) + 5$

Multiply $40$ by $4$ .

$y = 160 + (- 20) \cdot (2) + 5$

Multiply $- 20$ by $2$ .

$y = 160 - 40 + 5$

$y = 160 - 40 + 5$

Simplify by adding and subtracting.

Subtract $40$ from $160$ .

$y = 120 + 5$

Add $120$ and $5$ .

$y = 125$

$y = 125$

$y = 125$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 2$ . This check passes since $y = 125$ and $y = 125$ .

$125 = 125$

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 $y = 40 x^{2} - 20 x + 5$ .

$y = 40 x^{2} - 20 x + 5$

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