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

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

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.

$4 = a {(- 1)}^{2} + b (- 1) + c, 2 = a {(0)}^{2} + b (0) + c, 4 = a {(2)}^{2} + b (2) + c$

Step 3

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

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

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

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

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

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

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

Simplify each term.

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

$a \cdot 0 + c = 2$

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

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

Multiply $a$ by $0$ .

$0 + c = 2$

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

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

$0 + c = 2$

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

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

Add $0$ and $c$ .

$c = 2$

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

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

$c = 2$

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

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

$c = 2$

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

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

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

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

$4 = a {(- 1)}^{2} + b (- 1) + 2$

$c = 2$

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

Simplify $4 = a {(- 1)}^{2} + b (- 1) + 2$ .

Simplify the left side.

Remove parentheses.

$4 = a {(- 1)}^{2} + b (- 1) + 2$

$c = 2$

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

$4 = a {(- 1)}^{2} + b (- 1) + 2$

$c = 2$

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

Simplify the right side.

Simplify each term.

Raise $- 1$ to the power of $2$ .

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

$c = 2$

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

Multiply $a$ by $1$ .

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

$c = 2$

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

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

$4 = a - 1 b + 2$

$c = 2$

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

Rewrite $- 1 b$ as $- b$ .

$4 = a - b + 2$

$c = 2$

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

$4 = a - b + 2$

$c = 2$

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

$4 = a - b + 2$

$c = 2$

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

$4 = a - b + 2$

$c = 2$

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

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

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

$4 = a - b + 2$

$c = 2$

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

Simplify the left side.

Remove parentheses.

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

$4 = a - b + 2$

$c = 2$

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

$4 = a - b + 2$

$c = 2$

Simplify the right side.

Simplify each term.

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

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

$4 = a - b + 2$

$c = 2$

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

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

$4 = a - b + 2$

$c = 2$

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

$4 = 4 a + 2 b + 2$

$4 = a - b + 2$

$c = 2$

$4 = 4 a + 2 b + 2$

$4 = a - b + 2$

$c = 2$

$4 = 4 a + 2 b + 2$

$4 = a - b + 2$

$c = 2$

$4 = 4 a + 2 b + 2$

$4 = a - b + 2$

$c = 2$

$4 = 4 a + 2 b + 2$

$4 = a - b + 2$

$c = 2$

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

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

$a - b + 2 = 4$

$4 = 4 a + 2 b + 2$

$c = 2$

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

Add $b$ to both sides of the equation.

$a + 2 = 4 + b$

$4 = 4 a + 2 b + 2$

$c = 2$

Subtract $2$ from both sides of the equation.

$a = 4 + b - 2$

$4 = 4 a + 2 b + 2$

$c = 2$

Subtract $2$ from $4$ .

$a = b + 2$

$4 = 4 a + 2 b + 2$

$c = 2$

$a = b + 2$

$4 = 4 a + 2 b + 2$

$c = 2$

$a = b + 2$

$4 = 4 a + 2 b + 2$

$c = 2$

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

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

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

$a = b + 2$

$c = 2$

Simplify the right side.

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

Simplify each term.

Apply the distributive property.

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

$a = b + 2$

$c = 2$

Multiply $4$ by $2$ .

$4 = 4 b + 8 + 2 b + 2$

$a = b + 2$

$c = 2$

$4 = 4 b + 8 + 2 b + 2$

$a = b + 2$

$c = 2$

Simplify by adding terms.

Add $4 b$ and $2 b$ .

$4 = 6 b + 8 + 2$

$a = b + 2$

$c = 2$

Add $8$ and $2$ .

$4 = 6 b + 10$

$a = b + 2$

$c = 2$

$4 = 6 b + 10$

$a = b + 2$

$c = 2$

$4 = 6 b + 10$

$a = b + 2$

$c = 2$

$4 = 6 b + 10$

$a = b + 2$

$c = 2$

$4 = 6 b + 10$

$a = b + 2$

$c = 2$

Solve for $b$ in $4 = 6 b + 10$ .

Rewrite the equation as $6 b + 10 = 4$ .

$6 b + 10 = 4$

$a = b + 2$

$c = 2$

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

Subtract $10$ from both sides of the equation.

$6 b = 4 - 10$

$a = b + 2$

$c = 2$

Subtract $10$ from $4$ .

$6 b = - 6$

$a = b + 2$

$c = 2$

$6 b = - 6$

$a = b + 2$

$c = 2$

Divide each term in $6 b = - 6$ by $6$ and simplify.

Divide each term in $6 b = - 6$ by $6$ .

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

$a = b + 2$

$c = 2$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

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

$a = b + 2$

$c = 2$

Divide $b$ by $1$ .

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

$a = b + 2$

$c = 2$

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

$a = b + 2$

$c = 2$

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

$a = b + 2$

$c = 2$

Simplify the right side.

Divide $- 6$ by $6$ .

$b = - 1$

$a = b + 2$

$c = 2$

$b = - 1$

$a = b + 2$

$c = 2$

$b = - 1$

$a = b + 2$

$c = 2$

$b = - 1$

$a = b + 2$

$c = 2$

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

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

$a = (- 1) + 2$

$b = - 1$

$c = 2$

Simplify the right side.

Add $- 1$ and $2$ .

$a = 1$

$b = - 1$

$c = 2$

$a = 1$

$b = - 1$

$c = 2$

$a = 1$

$b = - 1$

$c = 2$

List all of the solutions.

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

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

Step 4

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

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

Step 5

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