Find the Equation of the Parabola (0,4) , (1,5) , (2,4)

$(0, 4)$ , $(1, 5)$ , $(2, 4)$

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.

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

$5 = a {(1)}^{2} + b (1) + c$

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

Step 3

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

Solve for $c$ in $4 = a \cdot 0^{2} + c$ .

Rewrite the equation as $a \cdot 0^{2} + c = 4$ .

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

$5 = a + b + c$

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

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

Simplify each term.

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

$a \cdot 0 + c = 4$

$5 = a + b + c$

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

Multiply $a$ by $0$ .

$0 + c = 4$

$5 = a + b + c$

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

$0 + c = 4$

$5 = a + b + c$

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

Add $0$ and $c$ .

$c = 4$

$5 = a + b + c$

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

$c = 4$

$5 = a + b + c$

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

$c = 4$

$5 = a + b + c$

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

Replace all occurrences of $c$ with $4$ in each equation.

Replace all occurrences of $c$ in $5 = a + b + c$ with $4$ .

$5 = a + b + 4$

$c = 4$

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

Simplify the left side.

Remove parentheses.

$5 = a + b + 4$

$c = 4$

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

$5 = a + b + 4$

$c = 4$

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

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

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

$5 = a + b + 4$

$c = 4$

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

Simplify the left side.

Remove parentheses.

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

$5 = a + b + 4$

$c = 4$

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

$5 = a + b + 4$

$c = 4$

Simplify the right side.

Simplify each term.

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

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

$5 = a + b + 4$

$c = 4$

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

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

$5 = a + b + 4$

$c = 4$

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

$4 = 4 a + 2 b + 4$

$5 = a + b + 4$

$c = 4$

$4 = 4 a + 2 b + 4$

$5 = a + b + 4$

$c = 4$

$4 = 4 a + 2 b + 4$

$5 = a + b + 4$

$c = 4$

$4 = 4 a + 2 b + 4$

$5 = a + b + 4$

$c = 4$

$4 = 4 a + 2 b + 4$

$5 = a + b + 4$

$c = 4$

Solve for $a$ in $5 = a + b + 4$ .

Rewrite the equation as $a + b + 4 = 5$ .

$a + b + 4 = 5$

$4 = 4 a + 2 b + 4$

$c = 4$

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

Subtract $b$ from both sides of the equation.

$a + 4 = 5 - b$

$4 = 4 a + 2 b + 4$

$c = 4$

Subtract $4$ from both sides of the equation.

$a = 5 - b - 4$

$4 = 4 a + 2 b + 4$

$c = 4$

Subtract $4$ from $5$ .

$a = - b + 1$

$4 = 4 a + 2 b + 4$

$c = 4$

$a = - b + 1$

$4 = 4 a + 2 b + 4$

$c = 4$

$a = - b + 1$

$4 = 4 a + 2 b + 4$

$c = 4$

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

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

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

$a = - b + 1$

$c = 4$

Simplify the right side.

Simplify $4 (- b + 1) + 2 b + 4$ .

Simplify each term.

Apply the distributive property.

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

$a = - b + 1$

$c = 4$

Multiply $- 1$ by $4$ .

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

$a = - b + 1$

$c = 4$

Multiply $4$ by $1$ .

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

$a = - b + 1$

$c = 4$

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

$a = - b + 1$

$c = 4$

Simplify by adding terms.

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

$4 = - 2 b + 4 + 4$

$a = - b + 1$

$c = 4$

Add $4$ and $4$ .

$4 = - 2 b + 8$

$a = - b + 1$

$c = 4$

$4 = - 2 b + 8$

$a = - b + 1$

$c = 4$

$4 = - 2 b + 8$

$a = - b + 1$

$c = 4$

$4 = - 2 b + 8$

$a = - b + 1$

$c = 4$

$4 = - 2 b + 8$

$a = - b + 1$

$c = 4$

Solve for $b$ in $4 = - 2 b + 8$ .

Rewrite the equation as $- 2 b + 8 = 4$ .

$- 2 b + 8 = 4$

$a = - b + 1$

$c = 4$

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

Subtract $8$ from both sides of the equation.

$- 2 b = 4 - 8$

$a = - b + 1$

$c = 4$

Subtract $8$ from $4$ .

$- 2 b = - 4$

$a = - b + 1$

$c = 4$

$- 2 b = - 4$

$a = - b + 1$

$c = 4$

Divide each term in $- 2 b = - 4$ by $- 2$ and simplify.

Divide each term in $- 2 b = - 4$ by $- 2$ .

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

$a = - b + 1$

$c = 4$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

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

$a = - b + 1$

$c = 4$

Divide $b$ by $1$ .

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

$a = - b + 1$

$c = 4$

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

$a = - b + 1$

$c = 4$

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

$a = - b + 1$

$c = 4$

Simplify the right side.

Divide $- 4$ by $- 2$ .

$b = 2$

$a = - b + 1$

$c = 4$

$b = 2$

$a = - b + 1$

$c = 4$

$b = 2$

$a = - b + 1$

$c = 4$

$b = 2$

$a = - b + 1$

$c = 4$

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

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

$a = - (2) + 1$

$b = 2$

$c = 4$

Simplify the right side.

Simplify $- (2) + 1$ .

Multiply $- 1$ by $2$ .

$a = - 2 + 1$

$b = 2$

$c = 4$

Add $- 2$ and $1$ .

$a = - 1$

$b = 2$

$c = 4$

$a = - 1$

$b = 2$

$c = 4$

$a = - 1$

$b = 2$

$c = 4$

$a = - 1$

$b = 2$

$c = 4$

List all of the solutions.

$a = - 1, b = 2, c = 4$

$a = - 1, b = 2, c = 4$

Step 4

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

$y = - x^{2} + 2 x + 4$

Step 5

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