Find the Equation of the Parabola (0,0) , (2,2) , (4,0)

$(0, 0)$ , $(2, 2)$ , $(4, 0)$

Step 1

Use the standard form of a quadratic equation $y = a x^{2} + b x + c$ as the starting point for finding the equation through the three points.

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

Step 2

Create a system of equations by substituting the $x$ and $y$ values of each point into the standard formula of a quadratic equation to create the three equation system.

$0 = a {(0)}^{2} + b (0) + c$

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

$0 = a {(4)}^{2} + b (4) + c$

Step 3

Solve the system of equations to find the values of $a$ , $b$ , and $c$ .

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

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

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

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

$0 = a \cdot 4^{2} + b (4) + c$

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

Simplify each term.

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

$a \cdot 0 + c = 0$

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

$0 = a \cdot 4^{2} + b (4) + c$

Multiply $a$ by $0$ .

$0 + c = 0$

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

$0 = a \cdot 4^{2} + b (4) + c$

$0 + c = 0$

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

$0 = a \cdot 4^{2} + b (4) + c$

Add $0$ and $c$ .

$c = 0$

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

$0 = a \cdot 4^{2} + b (4) + c$

$c = 0$

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

$0 = a \cdot 4^{2} + b (4) + c$

$c = 0$

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

$0 = a \cdot 4^{2} + b (4) + c$

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

Replace all occurrences of $c$ in $2 = a \cdot 2^{2} + b (2) + c$ with $0$ .

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

$c = 0$

$0 = a \cdot 4^{2} + b (4) + c$

Simplify $2 = a \cdot 2^{2} + b (2) + 0$ .

Simplify the left side.

Remove parentheses.

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

$c = 0$

$0 = a \cdot 4^{2} + b (4) + c$

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

$c = 0$

$0 = a \cdot 4^{2} + b (4) + c$

Simplify the right side.

Simplify $a \cdot 2^{2} + b (2) + 0$ .

Add $a \cdot 2^{2} + b (2)$ and $0$ .

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

$c = 0$

$0 = a \cdot 4^{2} + b (4) + c$

Simplify each term.

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

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

$c = 0$

$0 = a \cdot 4^{2} + b (4) + c$

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

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

$c = 0$

$0 = a \cdot 4^{2} + b (4) + c$

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

$2 = 4 a + 2 b$

$c = 0$

$0 = a \cdot 4^{2} + b (4) + c$

$2 = 4 a + 2 b$

$c = 0$

$0 = a \cdot 4^{2} + b (4) + c$

$2 = 4 a + 2 b$

$c = 0$

$0 = a \cdot 4^{2} + b (4) + c$

$2 = 4 a + 2 b$

$c = 0$

$0 = a \cdot 4^{2} + b (4) + c$

$2 = 4 a + 2 b$

$c = 0$

$0 = a \cdot 4^{2} + b (4) + c$

Replace all occurrences of $c$ in $0 = a \cdot 4^{2} + b (4) + c$ with $0$ .

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

$2 = 4 a + 2 b$

$c = 0$

Simplify $0 = a \cdot 4^{2} + b (4) + 0$ .

Simplify the left side.

Remove parentheses.

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

$2 = 4 a + 2 b$

$c = 0$

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

$2 = 4 a + 2 b$

$c = 0$

Simplify the right side.

Simplify $a \cdot 4^{2} + b (4) + 0$ .

Add $a \cdot 4^{2} + b (4)$ and $0$ .

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

$2 = 4 a + 2 b$

$c = 0$

Simplify each term.

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

$0 = a \cdot 16 + b (4)$

$2 = 4 a + 2 b$

$c = 0$

Move $16$ to the left of $a$ .

$0 = 16 \cdot a + b (4)$

$2 = 4 a + 2 b$

$c = 0$

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

$0 = 16 a + 4 b$

$2 = 4 a + 2 b$

$c = 0$

$0 = 16 a + 4 b$

$2 = 4 a + 2 b$

$c = 0$

$0 = 16 a + 4 b$

$2 = 4 a + 2 b$

$c = 0$

$0 = 16 a + 4 b$

$2 = 4 a + 2 b$

$c = 0$

$0 = 16 a + 4 b$

$2 = 4 a + 2 b$

$c = 0$

$0 = 16 a + 4 b$

$2 = 4 a + 2 b$

$c = 0$

Solve for $a$ in $0 = 16 a + 4 b$ .

Rewrite the equation as $16 a + 4 b = 0$ .

$16 a + 4 b = 0$

$2 = 4 a + 2 b$

$c = 0$

Subtract $4 b$ from both sides of the equation.

$16 a = - 4 b$

$2 = 4 a + 2 b$

$c = 0$

Divide each term in $16 a = - 4 b$ by $16$ and simplify.

Divide each term in $16 a = - 4 b$ by $16$ .

$\frac{16 a}{16} = \frac{- 4 b}{16}$

$2 = 4 a + 2 b$

$c = 0$

Simplify the left side.

Cancel the common factor of $16$ .

Cancel the common factor.

$\frac{16 a}{16} = \frac{- 4 b}{16}$

$2 = 4 a + 2 b$

$c = 0$

Divide $a$ by $1$ .

$a = \frac{- 4 b}{16}$

$2 = 4 a + 2 b$

$c = 0$

$a = \frac{- 4 b}{16}$

$2 = 4 a + 2 b$

$c = 0$

$a = \frac{- 4 b}{16}$

$2 = 4 a + 2 b$

$c = 0$

Simplify the right side.

Cancel the common factor of $- 4$ and $16$ .

Factor $4$ out of $- 4 b$ .

$a = \frac{4 (- b)}{16}$

$2 = 4 a + 2 b$

$c = 0$

Cancel the common factors.

Factor $4$ out of $16$ .

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

$2 = 4 a + 2 b$

$c = 0$

Cancel the common factor.

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

$2 = 4 a + 2 b$

$c = 0$

Rewrite the expression.

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

$2 = 4 a + 2 b$

$c = 0$

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

$2 = 4 a + 2 b$

$c = 0$

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

$2 = 4 a + 2 b$

$c = 0$

Move the negative in front of the fraction.

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

$2 = 4 a + 2 b$

$c = 0$

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

$2 = 4 a + 2 b$

$c = 0$

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

$2 = 4 a + 2 b$

$c = 0$

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

$2 = 4 a + 2 b$

$c = 0$

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

Replace all occurrences of $a$ in $2 = 4 a + 2 b$ with $- \frac{b}{4}$ .

$2 = 4 (- \frac{b}{4}) + 2 b$

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

$c = 0$

Simplify the right side.

Simplify $4 (- \frac{b}{4}) + 2 b$ .

Cancel the common factor of $4$ .

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

$2 = 4 (\frac{- b}{4}) + 2 b$

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

$c = 0$

Cancel the common factor.

$2 = 4 (\frac{- b}{4}) + 2 b$

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

$c = 0$

Rewrite the expression.

$2 = - b + 2 b$

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

$c = 0$

$2 = - b + 2 b$

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

$c = 0$

Add $- b$ and $2 b$ .

$2 = b$

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

$c = 0$

$2 = b$

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

$c = 0$

$2 = b$

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

$c = 0$

$2 = b$

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

$c = 0$

Rewrite the equation as $b = 2$ .

$b = 2$

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

$c = 0$

Replace all occurrences of $b$ with $2$ in each equation.

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

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

$b = 2$

$c = 0$

Simplify the right side.

Cancel the common factor of $2$ and $4$ .

Factor $2$ out of $2$ .

$a = - \frac{2 (1)}{4}$

$b = 2$

$c = 0$

Cancel the common factors.

Factor $2$ out of $4$ .

$a = - \frac{2 \cdot 1}{2 \cdot 2}$

$b = 2$

$c = 0$

Cancel the common factor.

$a = - \frac{2 \cdot 1}{2 \cdot 2}$

$b = 2$

$c = 0$

Rewrite the expression.

$a = - \frac{1}{2}$

$b = 2$

$c = 0$

$a = - \frac{1}{2}$

$b = 2$

$c = 0$

$a = - \frac{1}{2}$

$b = 2$

$c = 0$

$a = - \frac{1}{2}$

$b = 2$

$c = 0$

$a = - \frac{1}{2}$

$b = 2$

$c = 0$

List all of the solutions.

$a = - \frac{1}{2}, b = 2, c = 0$

$a = - \frac{1}{2}, b = 2, c = 0$

Step 4

Substitute the actual values of $a$ , $b$ , and $c$ into the formula for a quadratic equation to find the resulting equation.

$y = - \frac{x^{2}}{2} + 2 x$

Step 5

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