Find the Function Rule table[[x,y],[1,1],[2,4],[3,9],[4,16]]

$\begin{array}{c} x & y \\ 1 & 1 \\ 2 & 4 \\ 3 & 9 \\ 4 & 16 \end{array}$

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 $y = a x + b$ .

$1 = a (1) + b 4 = a (2) + b 9 = a (3) + b 16 = a (4) + b$

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

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

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

$a + b = 1$

$4 = a (2) + b$

$9 = a (3) + b$

$16 = a (4) + b$

Subtract $b$ from both sides of the equation.

$a = 1 - b$

$4 = a (2) + b$

$9 = a (3) + b$

$16 = a (4) + b$

$a = 1 - b$

$4 = a (2) + b$

$9 = a (3) + b$

$16 = a (4) + b$

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

Replace all occurrences of $a$ in $4 = a (2) + b$ with $1 - b$ .

$4 = (1 - b) (2) + b$

$a = 1 - b$

$9 = a (3) + b$

$16 = a (4) + b$

Simplify the right side.

Simplify $(1 - b) (2) + b$ .

Simplify each term.

Apply the distributive property.

$4 = 1 \cdot 2 - b \cdot 2 + b$

$a = 1 - b$

$9 = a (3) + b$

$16 = a (4) + b$

Multiply $2$ by $1$ .

$4 = 2 - b \cdot 2 + b$

$a = 1 - b$

$9 = a (3) + b$

$16 = a (4) + b$

Multiply $2$ by $- 1$ .

$4 = 2 - 2 b + b$

$a = 1 - b$

$9 = a (3) + b$

$16 = a (4) + b$

$4 = 2 - 2 b + b$

$a = 1 - b$

$9 = a (3) + b$

$16 = a (4) + b$

Add $- 2 b$ and $b$ .

$4 = 2 - b$

$a = 1 - b$

$9 = a (3) + b$

$16 = a (4) + b$

$4 = 2 - b$

$a = 1 - b$

$9 = a (3) + b$

$16 = a (4) + b$

$4 = 2 - b$

$a = 1 - b$

$9 = a (3) + b$

$16 = a (4) + b$

Replace all occurrences of $a$ in $9 = a (3) + b$ with $1 - b$ .

$9 = (1 - b) (3) + b$

$4 = 2 - b$

$a = 1 - b$

$16 = a (4) + b$

Simplify the right side.

Simplify $(1 - b) (3) + b$ .

Simplify each term.

Apply the distributive property.

$9 = 1 \cdot 3 - b \cdot 3 + b$

$4 = 2 - b$

$a = 1 - b$

$16 = a (4) + b$

Multiply $3$ by $1$ .

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

$4 = 2 - b$

$a = 1 - b$

$16 = a (4) + b$

Multiply $3$ by $- 1$ .

$9 = 3 - 3 b + b$

$4 = 2 - b$

$a = 1 - b$

$16 = a (4) + b$

$9 = 3 - 3 b + b$

$4 = 2 - b$

$a = 1 - b$

$16 = a (4) + b$

Add $- 3 b$ and $b$ .

$9 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

$16 = a (4) + b$

$9 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

$16 = a (4) + b$

$9 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

$16 = a (4) + b$

Replace all occurrences of $a$ in $16 = a (4) + b$ with $1 - b$ .

$16 = (1 - b) (4) + b$

$9 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

Simplify the right side.

Simplify $(1 - b) (4) + b$ .

Simplify each term.

Apply the distributive property.

$16 = 1 \cdot 4 - b \cdot 4 + b$

$9 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

Multiply $4$ by $1$ .

$16 = 4 - b \cdot 4 + b$

$9 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

Multiply $4$ by $- 1$ .

$16 = 4 - 4 b + b$

$9 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

$16 = 4 - 4 b + b$

$9 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

Add $- 4 b$ and $b$ .

$16 = 4 - 3 b$

$9 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

$16 = 4 - 3 b$

$9 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

$16 = 4 - 3 b$

$9 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

$16 = 4 - 3 b$

$9 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

Solve for $b$ in $16 = 4 - 3 b$ .

Rewrite the equation as $4 - 3 b = 16$ .

$4 - 3 b = 16$

$9 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

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

Subtract $4$ from both sides of the equation.

$- 3 b = 16 - 4$

$9 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

Subtract $4$ from $16$ .

$- 3 b = 12$

$9 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

$- 3 b = 12$

$9 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

Divide each term in $- 3 b = 12$ by $- 3$ and simplify.

Divide each term in $- 3 b = 12$ by $- 3$ .

$\frac{- 3 b}{- 3} = \frac{12}{- 3}$

$9 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 b}{- 3} = \frac{12}{- 3}$

$9 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

Divide $b$ by $1$ .

$b = \frac{12}{- 3}$

$9 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

$b = \frac{12}{- 3}$

$9 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

$b = \frac{12}{- 3}$

$9 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

Simplify the right side.

Divide $12$ by $- 3$ .

$b = - 4$

$9 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

$b = - 4$

$9 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

$b = - 4$

$9 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

$b = - 4$

$9 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

Replace all occurrences of $b$ with $- 4$ in each equation.

Replace all occurrences of $b$ in $9 = 3 - 2 b$ with $- 4$ .

$9 = 3 - 2 \cdot - 4$

$b = - 4$

$4 = 2 - b$

$a = 1 - b$

Simplify the right side.

Simplify $3 - 2 \cdot - 4$ .

Multiply $- 2$ by $- 4$ .

$9 = 3 + 8$

$b = - 4$

$4 = 2 - b$

$a = 1 - b$

Add $3$ and $8$ .

$9 = 11$

$b = - 4$

$4 = 2 - b$

$a = 1 - b$

$9 = 11$

$b = - 4$

$4 = 2 - b$

$a = 1 - b$

$9 = 11$

$b = - 4$

$4 = 2 - b$

$a = 1 - b$

Replace all occurrences of $b$ in $4 = 2 - b$ with $- 4$ .

$4 = 2 - (- 4)$

$9 = 11$

$b = - 4$

$a = 1 - b$

Simplify the right side.

Simplify $2 - (- 4)$ .

Multiply $- 1$ by $- 4$ .

$4 = 2 + 4$

$9 = 11$

$b = - 4$

$a = 1 - b$

Add $2$ and $4$ .

$4 = 6$

$9 = 11$

$b = - 4$

$a = 1 - b$

$4 = 6$

$9 = 11$

$b = - 4$

$a = 1 - b$

$4 = 6$

$9 = 11$

$b = - 4$

$a = 1 - b$

Replace all occurrences of $b$ in $a = 1 - b$ with $- 4$ .

$a = 1 - (- 4)$

$4 = 6$

$9 = 11$

$b = - 4$

Simplify the right side.

Simplify $1 - (- 4)$ .

Multiply $- 1$ by $- 4$ .

$a = 1 + 4$

$4 = 6$

$9 = 11$

$b = - 4$

Add $1$ and $4$ .

$a = 5$

$4 = 6$

$9 = 11$

$b = - 4$

$a = 5$

$4 = 6$

$9 = 11$

$b = - 4$

$a = 5$

$4 = 6$

$9 = 11$

$b = - 4$

$a = 5$

$4 = 6$

$9 = 11$

$b = - 4$

Since $4 = 6$ is not true, there is no solution.

No solution

Since $y \neq y$ for the corresponding $x$ values, 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 $a$ in $1 = a + b + c$ .

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

$a + b + c = 1$

$4 = a \cdot 2^{2} + b (2) + c$

$9 = a \cdot 3^{2} + b (3) + c$

$16 = a \cdot 4^{2} + b (4) + c$

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

Subtract $b$ from both sides of the equation.

$a + c = 1 - b$

$4 = a \cdot 2^{2} + b (2) + c$

$9 = a \cdot 3^{2} + b (3) + c$

$16 = a \cdot 4^{2} + b (4) + c$

Subtract $c$ from both sides of the equation.

$a = 1 - b - c$

$4 = a \cdot 2^{2} + b (2) + c$

$9 = a \cdot 3^{2} + b (3) + c$

$16 = a \cdot 4^{2} + b (4) + c$

$a = 1 - b - c$

$4 = a \cdot 2^{2} + b (2) + c$

$9 = a \cdot 3^{2} + b (3) + c$

$16 = a \cdot 4^{2} + b (4) + c$

$a = 1 - b - c$

$4 = a \cdot 2^{2} + b (2) + c$

$9 = a \cdot 3^{2} + b (3) + c$

$16 = a \cdot 4^{2} + b (4) + c$

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

Replace all occurrences of $a$ in $4 = a \cdot 2^{2} + b (2) + c$ with $1 - b - c$ .

$4 = (1 - b - c) \cdot 2^{2} + b (2) + c$

$a = 1 - b - c$

$9 = a \cdot 3^{2} + b (3) + c$

$16 = a \cdot 4^{2} + b (4) + c$

Simplify the right side.

Simplify $(1 - b - c) \cdot 2^{2} + b (2) + c$ .

Simplify each term.

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

$4 = (1 - b - c) \cdot 4 + b (2) + c$

$a = 1 - b - c$

$9 = a \cdot 3^{2} + b (3) + c$

$16 = a \cdot 4^{2} + b (4) + c$

Apply the distributive property.

$4 = 1 \cdot 4 - b \cdot 4 - c \cdot 4 + b (2) + c$

$a = 1 - b - c$

$9 = a \cdot 3^{2} + b (3) + c$

$16 = a \cdot 4^{2} + b (4) + c$

Simplify.

Multiply $4$ by $1$ .

$4 = 4 - b \cdot 4 - c \cdot 4 + b (2) + c$

$a = 1 - b - c$

$9 = a \cdot 3^{2} + b (3) + c$

$16 = a \cdot 4^{2} + b (4) + c$

Multiply $4$ by $- 1$ .

$4 = 4 - 4 b - c \cdot 4 + b (2) + c$

$a = 1 - b - c$

$9 = a \cdot 3^{2} + b (3) + c$

$16 = a \cdot 4^{2} + b (4) + c$

Multiply $4$ by $- 1$ .

$4 = 4 - 4 b - 4 c + b (2) + c$

$a = 1 - b - c$

$9 = a \cdot 3^{2} + b (3) + c$

$16 = a \cdot 4^{2} + b (4) + c$

$4 = 4 - 4 b - 4 c + b (2) + c$

$a = 1 - b - c$

$9 = a \cdot 3^{2} + b (3) + c$

$16 = a \cdot 4^{2} + b (4) + c$

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

$4 = 4 - 4 b - 4 c + 2 b + c$

$a = 1 - b - c$

$9 = a \cdot 3^{2} + b (3) + c$

$16 = a \cdot 4^{2} + b (4) + c$

$4 = 4 - 4 b - 4 c + 2 b + c$

$a = 1 - b - c$

$9 = a \cdot 3^{2} + b (3) + c$

$16 = a \cdot 4^{2} + b (4) + c$

Simplify by adding terms.

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

$4 = 4 - 2 b - 4 c + c$

$a = 1 - b - c$

$9 = a \cdot 3^{2} + b (3) + c$

$16 = a \cdot 4^{2} + b (4) + c$

Add $- 4 c$ and $c$ .

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$9 = a \cdot 3^{2} + b (3) + c$

$16 = a \cdot 4^{2} + b (4) + c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$9 = a \cdot 3^{2} + b (3) + c$

$16 = a \cdot 4^{2} + b (4) + c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$9 = a \cdot 3^{2} + b (3) + c$

$16 = a \cdot 4^{2} + b (4) + c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$9 = a \cdot 3^{2} + b (3) + c$

$16 = a \cdot 4^{2} + b (4) + c$

Replace all occurrences of $a$ in $9 = a \cdot 3^{2} + b (3) + c$ with $1 - b - c$ .

$9 = (1 - b - c) \cdot 3^{2} + b (3) + c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$16 = a \cdot 4^{2} + b (4) + c$

Simplify the right side.

Simplify $(1 - b - c) \cdot 3^{2} + b (3) + c$ .

Simplify each term.

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

$9 = (1 - b - c) \cdot 9 + b (3) + c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$16 = a \cdot 4^{2} + b (4) + c$

Apply the distributive property.

$9 = 1 \cdot 9 - b \cdot 9 - c \cdot 9 + b (3) + c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$16 = a \cdot 4^{2} + b (4) + c$

Simplify.

Multiply $9$ by $1$ .

$9 = 9 - b \cdot 9 - c \cdot 9 + b (3) + c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$16 = a \cdot 4^{2} + b (4) + c$

Multiply $9$ by $- 1$ .

$9 = 9 - 9 b - c \cdot 9 + b (3) + c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$16 = a \cdot 4^{2} + b (4) + c$

Multiply $9$ by $- 1$ .

$9 = 9 - 9 b - 9 c + b (3) + c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$16 = a \cdot 4^{2} + b (4) + c$

$9 = 9 - 9 b - 9 c + b (3) + c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$16 = a \cdot 4^{2} + b (4) + c$

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

$9 = 9 - 9 b - 9 c + 3 b + c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$16 = a \cdot 4^{2} + b (4) + c$

$9 = 9 - 9 b - 9 c + 3 b + c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$16 = a \cdot 4^{2} + b (4) + c$

Simplify by adding terms.

Add $- 9 b$ and $3 b$ .

$9 = 9 - 6 b - 9 c + c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$16 = a \cdot 4^{2} + b (4) + c$

Add $- 9 c$ and $c$ .

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$16 = a \cdot 4^{2} + b (4) + c$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$16 = a \cdot 4^{2} + b (4) + c$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$16 = a \cdot 4^{2} + b (4) + c$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$16 = a \cdot 4^{2} + b (4) + c$

Replace all occurrences of $a$ in $16 = a \cdot 4^{2} + b (4) + c$ with $1 - b - c$ .

$16 = (1 - b - c) \cdot 4^{2} + b (4) + c$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Simplify the right side.

Simplify $(1 - b - c) \cdot 4^{2} + b (4) + c$ .

Simplify each term.

Raise $4$ to the power of $2$ .

$16 = (1 - b - c) \cdot 16 + b (4) + c$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Apply the distributive property.

$16 = 1 \cdot 16 - b \cdot 16 - c \cdot 16 + b (4) + c$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Simplify.

Multiply $16$ by $1$ .

$16 = 16 - b \cdot 16 - c \cdot 16 + b (4) + c$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Multiply $16$ by $- 1$ .

$16 = 16 - 16 b - c \cdot 16 + b (4) + c$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Multiply $16$ by $- 1$ .

$16 = 16 - 16 b - 16 c + b (4) + c$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$16 = 16 - 16 b - 16 c + b (4) + c$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

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

$16 = 16 - 16 b - 16 c + 4 b + c$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$16 = 16 - 16 b - 16 c + 4 b + c$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Simplify by adding terms.

Add $- 16 b$ and $4 b$ .

$16 = 16 - 12 b - 16 c + c$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Add $- 16 c$ and $c$ .

$16 = 16 - 12 b - 15 c$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$16 = 16 - 12 b - 15 c$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$16 = 16 - 12 b - 15 c$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$16 = 16 - 12 b - 15 c$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$16 = 16 - 12 b - 15 c$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Solve for $b$ in $16 = 16 - 12 b - 15 c$ .

Rewrite the equation as $16 - 12 b - 15 c = 16$ .

$16 - 12 b - 15 c = 16$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

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

Subtract $16$ from both sides of the equation.

$- 12 b - 15 c = 16 - 16$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

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

$- 12 b = 16 - 16 + 15 c$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Subtract $16$ from $16$ .

$- 12 b = 0 + 15 c$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Add $0$ and $15 c$ .

$- 12 b = 15 c$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$- 12 b = 15 c$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Divide each term in $- 12 b = 15 c$ by $- 12$ and simplify.

Divide each term in $- 12 b = 15 c$ by $- 12$ .

$\frac{- 12 b}{- 12} = \frac{15 c}{- 12}$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Simplify the left side.

Cancel the common factor of $- 12$ .

Cancel the common factor.

$\frac{- 12 b}{- 12} = \frac{15 c}{- 12}$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Divide $b$ by $1$ .

$b = \frac{15 c}{- 12}$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$b = \frac{15 c}{- 12}$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$b = \frac{15 c}{- 12}$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Simplify the right side.

Cancel the common factor of $15$ and $- 12$ .

Factor $3$ out of $15 c$ .

$b = \frac{3 (5 c)}{- 12}$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Cancel the common factors.

Factor $3$ out of $- 12$ .

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

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Cancel the common factor.

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

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Rewrite the expression.

$b = \frac{5 c}{- 4}$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$b = \frac{5 c}{- 4}$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$b = \frac{5 c}{- 4}$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Move the negative in front of the fraction.

$b = - \frac{5 c}{4}$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$b = - \frac{5 c}{4}$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$b = - \frac{5 c}{4}$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$b = - \frac{5 c}{4}$

$9 = 9 - 6 b - 8 c$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

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

Replace all occurrences of $b$ in $9 = 9 - 6 b - 8 c$ with $- \frac{5 c}{4}$ .

$9 = 9 - 6 (- \frac{5 c}{4}) - 8 c$

$b = - \frac{5 c}{4}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Simplify the right side.

Simplify $9 - 6 (- \frac{5 c}{4}) - 8 c$ .

Simplify each term.

Cancel the common factor of $2$ .

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

$9 = 9 - 6 \frac{- 5 c}{4} - 8 c$

$b = - \frac{5 c}{4}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Factor $2$ out of $- 6$ .

$9 = 9 + 2 (- 3) (\frac{- 5 c}{4}) - 8 c$

$b = - \frac{5 c}{4}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Factor $2$ out of $4$ .

$9 = 9 + 2 \cdot (- 3 \frac{- 5 c}{2 \cdot 2}) - 8 c$

$b = - \frac{5 c}{4}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Cancel the common factor.

$9 = 9 + 2 \cdot (- 3 \frac{- 5 c}{2 \cdot 2}) - 8 c$

$b = - \frac{5 c}{4}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Rewrite the expression.

$9 = 9 - 3 \frac{- 5 c}{2} - 8 c$

$b = - \frac{5 c}{4}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$9 = 9 - 3 \frac{- 5 c}{2} - 8 c$

$b = - \frac{5 c}{4}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Combine $- 3$ and $\frac{- 5 c}{2}$ .

$9 = 9 + \frac{- 3 (- 5 c)}{2} - 8 c$

$b = - \frac{5 c}{4}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Multiply $- 5$ by $- 3$ .

$9 = 9 + \frac{15 c}{2} - 8 c$

$b = - \frac{5 c}{4}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$9 = 9 + \frac{15 c}{2} - 8 c$

$b = - \frac{5 c}{4}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

To write $- 8 c$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$9 = 9 + \frac{15 c}{2} - 8 c \cdot \frac{2}{2}$

$b = - \frac{5 c}{4}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Simplify terms.

Combine $- 8 c$ and $\frac{2}{2}$ .

$9 = 9 + \frac{15 c}{2} + \frac{- 8 c \cdot 2}{2}$

$b = - \frac{5 c}{4}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Combine the numerators over the common denominator.

$9 = 9 + \frac{15 c - 8 c \cdot 2}{2}$

$b = - \frac{5 c}{4}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$9 = 9 + \frac{15 c - 8 c \cdot 2}{2}$

$b = - \frac{5 c}{4}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Simplify each term.

Simplify the numerator.

Factor $c$ out of $15 c - 8 c \cdot 2$ .

Factor $c$ out of $15 c$ .

$9 = 9 + \frac{c \cdot 15 - 8 c \cdot 2}{2}$

$b = - \frac{5 c}{4}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Factor $c$ out of $- 8 c \cdot 2$ .

$9 = 9 + \frac{c \cdot 15 + c (- 8 \cdot 2)}{2}$

$b = - \frac{5 c}{4}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Factor $c$ out of $c \cdot 15 + c (- 8 \cdot 2)$ .

$9 = 9 + \frac{c (15 - 8 \cdot 2)}{2}$

$b = - \frac{5 c}{4}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$9 = 9 + \frac{c (15 - 8 \cdot 2)}{2}$

$b = - \frac{5 c}{4}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Multiply $- 8$ by $2$ .

$9 = 9 + \frac{c (15 - 16)}{2}$

$b = - \frac{5 c}{4}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Subtract $16$ from $15$ .

$9 = 9 + \frac{c \cdot - 1}{2}$

$b = - \frac{5 c}{4}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$9 = 9 + \frac{c \cdot - 1}{2}$

$b = - \frac{5 c}{4}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Move $- 1$ to the left of $c$ .

$9 = 9 + \frac{- 1 \cdot c}{2}$

$b = - \frac{5 c}{4}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Move the negative in front of the fraction.

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$4 = 4 - 2 b - 3 c$

$a = 1 - b - c$

Replace all occurrences of $b$ in $4 = 4 - 2 b - 3 c$ with $- \frac{5 c}{4}$ .

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

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1 - b - c$

Simplify the right side.

Simplify $4 - 2 (- \frac{5 c}{4}) - 3 c$ .

Simplify each term.

Cancel the common factor of $2$ .

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

$4 = 4 - 2 \frac{- 5 c}{4} - 3 c$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1 - b - c$

Factor $2$ out of $- 2$ .

$4 = 4 + 2 (- 1) (\frac{- 5 c}{4}) - 3 c$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1 - b - c$

Factor $2$ out of $4$ .

$4 = 4 + 2 \cdot (- 1 \frac{- 5 c}{2 \cdot 2}) - 3 c$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1 - b - c$

Cancel the common factor.

$4 = 4 + 2 \cdot (- 1 \frac{- 5 c}{2 \cdot 2}) - 3 c$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1 - b - c$

Rewrite the expression.

$4 = 4 - 1 \frac{- 5 c}{2} - 3 c$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1 - b - c$

$4 = 4 - 1 \frac{- 5 c}{2} - 3 c$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1 - b - c$

Move the negative in front of the fraction.

$4 = 4 - 1 (- \frac{(5) c}{2}) - 3 c$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1 - b - c$

Multiply $- 1 (- \frac{5 c}{2})$ .

Multiply $- 1$ by $- 1$ .

$4 = 4 + 1 (\frac{5 c}{2}) - 3 c$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1 - b - c$

Multiply $\frac{5 c}{2}$ by $1$ .

$4 = 4 + \frac{5 c}{2} - 3 c$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1 - b - c$

$4 = 4 + \frac{5 c}{2} - 3 c$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1 - b - c$

$4 = 4 + \frac{5 c}{2} - 3 c$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1 - b - c$

To write $- 3 c$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$4 = 4 + \frac{5 c}{2} - 3 c \cdot \frac{2}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1 - b - c$

Simplify terms.

Combine $- 3 c$ and $\frac{2}{2}$ .

$4 = 4 + \frac{5 c}{2} + \frac{- 3 c \cdot 2}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1 - b - c$

Combine the numerators over the common denominator.

$4 = 4 + \frac{5 c - 3 c \cdot 2}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1 - b - c$

$4 = 4 + \frac{5 c - 3 c \cdot 2}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1 - b - c$

Simplify each term.

Simplify the numerator.

Factor $c$ out of $5 c - 3 c \cdot 2$ .

Factor $c$ out of $5 c$ .

$4 = 4 + \frac{c \cdot 5 - 3 c \cdot 2}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1 - b - c$

Factor $c$ out of $- 3 c \cdot 2$ .

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

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1 - b - c$

Factor $c$ out of $c \cdot 5 + c (- 3 \cdot 2)$ .

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

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1 - b - c$

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

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1 - b - c$

Multiply $- 3$ by $2$ .

$4 = 4 + \frac{c (5 - 6)}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1 - b - c$

Subtract $6$ from $5$ .

$4 = 4 + \frac{c \cdot - 1}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1 - b - c$

$4 = 4 + \frac{c \cdot - 1}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1 - b - c$

Move $- 1$ to the left of $c$ .

$4 = 4 + \frac{- 1 \cdot c}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1 - b - c$

Move the negative in front of the fraction.

$4 = 4 - \frac{c}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1 - b - c$

$4 = 4 - \frac{c}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1 - b - c$

$4 = 4 - \frac{c}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1 - b - c$

$4 = 4 - \frac{c}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1 - b - c$

Replace all occurrences of $b$ in $a = 1 - b - c$ with $- \frac{5 c}{4}$ .

$a = 1 - (- \frac{5 c}{4}) - c$

$4 = 4 - \frac{c}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

Simplify the right side.

Simplify $1 - (- \frac{5 c}{4}) - c$ .

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

Multiply $- 1$ by $- 1$ .

$a = 1 + 1 (\frac{5 c}{4}) - c$

$4 = 4 - \frac{c}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

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

$a = 1 + \frac{5 c}{4} - c$

$4 = 4 - \frac{c}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1 + \frac{5 c}{4} - c$

$4 = 4 - \frac{c}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

To write $- c$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$a = 1 + \frac{5 c}{4} - c \cdot \frac{4}{4}$

$4 = 4 - \frac{c}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

Simplify terms.

Combine $- c$ and $\frac{4}{4}$ .

$a = 1 + \frac{5 c}{4} + \frac{- c \cdot 4}{4}$

$4 = 4 - \frac{c}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

Combine the numerators over the common denominator.

$a = 1 + \frac{5 c - c \cdot 4}{4}$

$4 = 4 - \frac{c}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1 + \frac{5 c - c \cdot 4}{4}$

$4 = 4 - \frac{c}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

Simplify each term.

Simplify the numerator.

Factor $c$ out of $5 c - c \cdot 4$ .

Factor $c$ out of $5 c$ .

$a = 1 + \frac{c \cdot 5 - c \cdot 4}{4}$

$4 = 4 - \frac{c}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

Factor $c$ out of $- c \cdot 4$ .

$a = 1 + \frac{c \cdot 5 + c (- 1 \cdot 4)}{4}$

$4 = 4 - \frac{c}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

Factor $c$ out of $c \cdot 5 + c (- 1 \cdot 4)$ .

$a = 1 + \frac{c (5 - 1 \cdot 4)}{4}$

$4 = 4 - \frac{c}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1 + \frac{c (5 - 1 \cdot 4)}{4}$

$4 = 4 - \frac{c}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

Multiply $- 1$ by $4$ .

$a = 1 + \frac{c (5 - 4)}{4}$

$4 = 4 - \frac{c}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

Subtract $4$ from $5$ .

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

$4 = 4 - \frac{c}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

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

$4 = 4 - \frac{c}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

Multiply $c$ by $1$ .

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

$4 = 4 - \frac{c}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

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

$4 = 4 - \frac{c}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

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

$4 = 4 - \frac{c}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

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

$4 = 4 - \frac{c}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

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

$4 = 4 - \frac{c}{2}$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

Solve for $c$ in $4 = 4 - \frac{c}{2}$ .

Rewrite the equation as $4 - \frac{c}{2} = 4$ .

$4 - \frac{c}{2} = 4$

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

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

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

Subtract $4$ from both sides of the equation.

$- \frac{c}{2} = 4 - 4$

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

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

Subtract $4$ from $4$ .

$- \frac{c}{2} = 0$

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

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$- \frac{c}{2} = 0$

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

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

Set the numerator equal to zero.

$c = 0$

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

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$c = 0$

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

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

Replace all occurrences of $c$ with $0$ in each equation.

Replace all occurrences of $c$ in $a = 1 + \frac{c}{4}$ with $0$ .

$a = 1 + \frac{0}{4}$

$c = 0$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

Simplify the right side.

Simplify $1 + \frac{0}{4}$ .

Divide $0$ by $4$ .

$a = 1 + 0$

$c = 0$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

Add $1$ and $0$ .

$a = 1$

$c = 0$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1$

$c = 0$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

$a = 1$

$c = 0$

$9 = 9 - \frac{c}{2}$

$b = - \frac{5 c}{4}$

Replace all occurrences of $c$ in $9 = 9 - \frac{c}{2}$ with $0$ .

$9 = 9 - \frac{0}{2}$

$a = 1$

$c = 0$

$b = - \frac{5 c}{4}$

Simplify the right side.

Simplify $9 - \frac{0}{2}$ .

Simplify each term.

Divide $0$ by $2$ .

$9 = 9 - 0$

$a = 1$

$c = 0$

$b = - \frac{5 c}{4}$

Multiply $- 1$ by $0$ .

$9 = 9 + 0$

$a = 1$

$c = 0$

$b = - \frac{5 c}{4}$

$9 = 9 + 0$

$a = 1$

$c = 0$

$b = - \frac{5 c}{4}$

Add $9$ and $0$ .

$9 = 9$

$a = 1$

$c = 0$

$b = - \frac{5 c}{4}$

$9 = 9$

$a = 1$

$c = 0$

$b = - \frac{5 c}{4}$

$9 = 9$

$a = 1$

$c = 0$

$b = - \frac{5 c}{4}$

Replace all occurrences of $c$ in $b = - \frac{5 c}{4}$ with $0$ .

$b = - \frac{5 (0)}{4}$

$9 = 9$

$a = 1$

$c = 0$

Simplify the right side.

Simplify $- \frac{5 (0)}{4}$ .

Cancel the common factor of $0$ and $4$ .

Factor $4$ out of $5 (0)$ .

$b = - \frac{4 (5 \cdot (0))}{4}$

$9 = 9$

$a = 1$

$c = 0$

Cancel the common factors.

Factor $4$ out of $4$ .

$b = - \frac{4 (5 \cdot (0))}{4 (1)}$

$9 = 9$

$a = 1$

$c = 0$

Cancel the common factor.

$b = - \frac{4 (5 \cdot (0))}{4 \cdot 1}$

$9 = 9$

$a = 1$

$c = 0$

Rewrite the expression.

$b = - \frac{5 \cdot (0)}{1}$

$9 = 9$

$a = 1$

$c = 0$

Divide $5 \cdot (0)$ by $1$ .

$b = - (5 \cdot (0))$

$9 = 9$

$a = 1$

$c = 0$

$b = - (5 \cdot (0))$

$9 = 9$

$a = 1$

$c = 0$

$b = - (5 \cdot (0))$

$9 = 9$

$a = 1$

$c = 0$

Multiply $- (5 \cdot (0))$ .

Multiply $5$ by $0$ .

$b = - 0$

$9 = 9$

$a = 1$

$c = 0$

Multiply $- 1$ by $0$ .

$b = 0$

$9 = 9$

$a = 1$

$c = 0$

$b = 0$

$9 = 9$

$a = 1$

$c = 0$

$b = 0$

$9 = 9$

$a = 1$

$c = 0$

$b = 0$

$9 = 9$

$a = 1$

$c = 0$

$b = 0$

$9 = 9$

$a = 1$

$c = 0$

Remove any equations from the system that are always true.

$b = 0$

$a = 1$

$c = 0$

List all of the solutions.

$b = 0, a = 1, c = 0$

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 $a = 1$ , $b = 0$ , $c = 0$ , and $x = 1$ .

Simplify each term.

Multiply ${(1)}^{2}$ by $1$ .

$y = {(1)}^{2} + (0) \cdot (1) + 0$

One to any power is one.

$y = 1 + (0) \cdot (1) + 0$

Multiply $0$ by $1$ .

$y = 1 + 0 + 0$

Simplify by adding numbers.

Add $1$ and $0$ .

$y = 1 + 0$

Add $1$ and $0$ .

$y = 1$

If the table has a quadratic 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 $y$ such that $y = a x^{2} + b$ when $a = 1$ , $b = 0$ , $c = 0$ , and $x = 2$ .

Simplify each term.

Multiply ${(2)}^{2}$ by $1$ .

$y = {(2)}^{2} + (0) \cdot (2) + 0$

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

$y = 4 + (0) \cdot (2) + 0$

Multiply $0$ by $2$ .

$y = 4 + 0 + 0$

Simplify by adding numbers.

Add $4$ and $0$ .

$y = 4 + 0$

Add $4$ and $0$ .

$y = 4$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 2$ . This check passes since $y = 4$ and $y = 4$ .

$4 = 4$

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = 1$ , $b = 0$ , $c = 0$ , and $x = 3$ .

Simplify each term.

Multiply ${(3)}^{2}$ by $1$ .

$y = {(3)}^{2} + (0) \cdot (3) + 0$

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

$y = 9 + (0) \cdot (3) + 0$

Multiply $0$ by $3$ .

$y = 9 + 0 + 0$

Simplify by adding numbers.

Add $9$ and $0$ .

$y = 9 + 0$

Add $9$ and $0$ .

$y = 9$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 3$ . This check passes since $y = 9$ and $y = 9$ .

$9 = 9$

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = 1$ , $b = 0$ , $c = 0$ , and $x = 4$ .

Simplify each term.

Multiply ${(4)}^{2}$ by $1$ .

$y = {(4)}^{2} + (0) \cdot (4) + 0$

Raise $4$ to the power of $2$ .

$y = 16 + (0) \cdot (4) + 0$

Multiply $0$ by $4$ .

$y = 16 + 0 + 0$

Simplify by adding numbers.

Add $16$ and $0$ .

$y = 16 + 0$

Add $16$ and $0$ .

$y = 16$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 4$ . This check passes since $y = 16$ and $y = 16$ .

$16 = 16$

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

The function is quadratic

Step 3

Since all $y = y$ , the function is quadratic and follows the form $y = x^{2}$ .

$y = x^{2}$

Find the Function Rule table[[x,y],[1,1],[2,4],[3,9],[4,16]]

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