Find the Equation of the Parabola (0,5) , (2,11) , (4,1)

$(0, 5)$ , $(2, 11)$ , $(4, 1)$

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.

$5 = a {(0)}^{2} + b (0) + c$

$11 = a {(2)}^{2} + b (2) + c$

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

Step 3

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

Solve for $c$ in $5 = a \cdot 0^{2} + c$ .

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

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

$11 = a \cdot 2^{2} + b (2) + c$

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

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

Simplify each term.

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

$a \cdot 0 + c = 5$

$11 = a \cdot 2^{2} + b (2) + c$

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

Multiply $a$ by $0$ .

$0 + c = 5$

$11 = a \cdot 2^{2} + b (2) + c$

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

$0 + c = 5$

$11 = a \cdot 2^{2} + b (2) + c$

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

Add $0$ and $c$ .

$c = 5$

$11 = a \cdot 2^{2} + b (2) + c$

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

$c = 5$

$11 = a \cdot 2^{2} + b (2) + c$

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

$c = 5$

$11 = a \cdot 2^{2} + b (2) + c$

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

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

Replace all occurrences of $c$ in $11 = a \cdot 2^{2} + b (2) + c$ with $5$ .

$11 = a \cdot 2^{2} + b (2) + 5$

$c = 5$

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

Simplify $11 = a \cdot 2^{2} + b (2) + 5$ .

Simplify the left side.

Remove parentheses.

$11 = a \cdot 2^{2} + b (2) + 5$

$c = 5$

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

$11 = a \cdot 2^{2} + b (2) + 5$

$c = 5$

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

Simplify the right side.

Simplify each term.

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

$11 = a \cdot 4 + b (2) + 5$

$c = 5$

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

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

$11 = 4 \cdot a + b (2) + 5$

$c = 5$

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

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

$11 = 4 a + 2 b + 5$

$c = 5$

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

$11 = 4 a + 2 b + 5$

$c = 5$

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

$11 = 4 a + 2 b + 5$

$c = 5$

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

$11 = 4 a + 2 b + 5$

$c = 5$

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

Replace all occurrences of $c$ in $1 = a \cdot 4^{2} + b (4) + c$ with $5$ .

$1 = a \cdot 4^{2} + b (4) + 5$

$11 = 4 a + 2 b + 5$

$c = 5$

Simplify $1 = a \cdot 4^{2} + b (4) + 5$ .

Simplify the left side.

Remove parentheses.

$1 = a \cdot 4^{2} + b (4) + 5$

$11 = 4 a + 2 b + 5$

$c = 5$

$1 = a \cdot 4^{2} + b (4) + 5$

$11 = 4 a + 2 b + 5$

$c = 5$

Simplify the right side.

Simplify each term.

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

$1 = a \cdot 16 + b (4) + 5$

$11 = 4 a + 2 b + 5$

$c = 5$

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

$1 = 16 \cdot a + b (4) + 5$

$11 = 4 a + 2 b + 5$

$c = 5$

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

$1 = 16 a + 4 b + 5$

$11 = 4 a + 2 b + 5$

$c = 5$

$1 = 16 a + 4 b + 5$

$11 = 4 a + 2 b + 5$

$c = 5$

$1 = 16 a + 4 b + 5$

$11 = 4 a + 2 b + 5$

$c = 5$

$1 = 16 a + 4 b + 5$

$11 = 4 a + 2 b + 5$

$c = 5$

$1 = 16 a + 4 b + 5$

$11 = 4 a + 2 b + 5$

$c = 5$

Solve for $a$ in $1 = 16 a + 4 b + 5$ .

Rewrite the equation as $16 a + 4 b + 5 = 1$ .

$16 a + 4 b + 5 = 1$

$11 = 4 a + 2 b + 5$

$c = 5$

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

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

$16 a + 5 = 1 - 4 b$

$11 = 4 a + 2 b + 5$

$c = 5$

Subtract $5$ from both sides of the equation.

$16 a = 1 - 4 b - 5$

$11 = 4 a + 2 b + 5$

$c = 5$

Subtract $5$ from $1$ .

$16 a = - 4 b - 4$

$11 = 4 a + 2 b + 5$

$c = 5$

$16 a = - 4 b - 4$

$11 = 4 a + 2 b + 5$

$c = 5$

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

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

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

$11 = 4 a + 2 b + 5$

$c = 5$

Simplify the left side.

Cancel the common factor of $16$ .

Cancel the common factor.

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

$11 = 4 a + 2 b + 5$

$c = 5$

Divide $a$ by $1$ .

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

$11 = 4 a + 2 b + 5$

$c = 5$

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

$11 = 4 a + 2 b + 5$

$c = 5$

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

$11 = 4 a + 2 b + 5$

$c = 5$

Simplify the right side.

Simplify each term.

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

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

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

$11 = 4 a + 2 b + 5$

$c = 5$

Cancel the common factors.

Factor $4$ out of $16$ .

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

$11 = 4 a + 2 b + 5$

$c = 5$

Cancel the common factor.

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

$11 = 4 a + 2 b + 5$

$c = 5$

Rewrite the expression.

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

$11 = 4 a + 2 b + 5$

$c = 5$

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

$11 = 4 a + 2 b + 5$

$c = 5$

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

$11 = 4 a + 2 b + 5$

$c = 5$

Move the negative in front of the fraction.

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

$11 = 4 a + 2 b + 5$

$c = 5$

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

Factor $4$ out of $- 4$ .

$a = - \frac{b}{4} + \frac{4 (- 1)}{16}$

$11 = 4 a + 2 b + 5$

$c = 5$

Cancel the common factors.

Factor $4$ out of $16$ .

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

$11 = 4 a + 2 b + 5$

$c = 5$

Cancel the common factor.

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

$11 = 4 a + 2 b + 5$

$c = 5$

Rewrite the expression.

$a = - \frac{b}{4} + \frac{- 1}{4}$

$11 = 4 a + 2 b + 5$

$c = 5$

$a = - \frac{b}{4} + \frac{- 1}{4}$

$11 = 4 a + 2 b + 5$

$c = 5$

$a = - \frac{b}{4} + \frac{- 1}{4}$

$11 = 4 a + 2 b + 5$

$c = 5$

Move the negative in front of the fraction.

$a = - \frac{b}{4} - \frac{1}{4}$

$11 = 4 a + 2 b + 5$

$c = 5$

$a = - \frac{b}{4} - \frac{1}{4}$

$11 = 4 a + 2 b + 5$

$c = 5$

$a = - \frac{b}{4} - \frac{1}{4}$

$11 = 4 a + 2 b + 5$

$c = 5$

$a = - \frac{b}{4} - \frac{1}{4}$

$11 = 4 a + 2 b + 5$

$c = 5$

$a = - \frac{b}{4} - \frac{1}{4}$

$11 = 4 a + 2 b + 5$

$c = 5$

Replace all occurrences of $a$ with $- \frac{b}{4} - \frac{1}{4}$ in each equation.

Replace all occurrences of $a$ in $11 = 4 a + 2 b + 5$ with $- \frac{b}{4} - \frac{1}{4}$ .

$11 = 4 (- \frac{b}{4} - \frac{1}{4}) + 2 b + 5$

$a = - \frac{b}{4} - \frac{1}{4}$

$c = 5$

Simplify the right side.

Simplify $4 (- \frac{b}{4} - \frac{1}{4}) + 2 b + 5$ .

Simplify each term.

Apply the distributive property.

$11 = 4 (- \frac{b}{4}) + 4 (- \frac{1}{4}) + 2 b + 5$

$a = - \frac{b}{4} - \frac{1}{4}$

$c = 5$

Cancel the common factor of $4$ .

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

$11 = 4 (\frac{- b}{4}) + 4 (- \frac{1}{4}) + 2 b + 5$

$a = - \frac{b}{4} - \frac{1}{4}$

$c = 5$

Cancel the common factor.

$11 = 4 (\frac{- b}{4}) + 4 (- \frac{1}{4}) + 2 b + 5$

$a = - \frac{b}{4} - \frac{1}{4}$

$c = 5$

Rewrite the expression.

$11 = - b + 4 (- \frac{1}{4}) + 2 b + 5$

$a = - \frac{b}{4} - \frac{1}{4}$

$c = 5$

$11 = - b + 4 (- \frac{1}{4}) + 2 b + 5$

$a = - \frac{b}{4} - \frac{1}{4}$

$c = 5$

Cancel the common factor of $4$ .

Move the leading negative in $- \frac{1}{4}$ into the numerator.

$11 = - b + 4 (\frac{- 1}{4}) + 2 b + 5$

$a = - \frac{b}{4} - \frac{1}{4}$

$c = 5$

Cancel the common factor.

$11 = - b + 4 (\frac{- 1}{4}) + 2 b + 5$

$a = - \frac{b}{4} - \frac{1}{4}$

$c = 5$

Rewrite the expression.

$11 = - b - 1 + 2 b + 5$

$a = - \frac{b}{4} - \frac{1}{4}$

$c = 5$

$11 = - b - 1 + 2 b + 5$

$a = - \frac{b}{4} - \frac{1}{4}$

$c = 5$

$11 = - b - 1 + 2 b + 5$

$a = - \frac{b}{4} - \frac{1}{4}$

$c = 5$

Simplify by adding terms.

Add $- b$ and $2 b$ .

$11 = b - 1 + 5$

$a = - \frac{b}{4} - \frac{1}{4}$

$c = 5$

Add $- 1$ and $5$ .

$11 = b + 4$

$a = - \frac{b}{4} - \frac{1}{4}$

$c = 5$

$11 = b + 4$

$a = - \frac{b}{4} - \frac{1}{4}$

$c = 5$

$11 = b + 4$

$a = - \frac{b}{4} - \frac{1}{4}$

$c = 5$

$11 = b + 4$

$a = - \frac{b}{4} - \frac{1}{4}$

$c = 5$

$11 = b + 4$

$a = - \frac{b}{4} - \frac{1}{4}$

$c = 5$

Solve for $b$ in $11 = b + 4$ .

Rewrite the equation as $b + 4 = 11$ .

$b + 4 = 11$

$a = - \frac{b}{4} - \frac{1}{4}$

$c = 5$

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

Subtract $4$ from both sides of the equation.

$b = 11 - 4$

$a = - \frac{b}{4} - \frac{1}{4}$

$c = 5$

Subtract $4$ from $11$ .

$b = 7$

$a = - \frac{b}{4} - \frac{1}{4}$

$c = 5$

$b = 7$

$a = - \frac{b}{4} - \frac{1}{4}$

$c = 5$

$b = 7$

$a = - \frac{b}{4} - \frac{1}{4}$

$c = 5$

Replace all occurrences of $b$ with $7$ in each equation.

Replace all occurrences of $b$ in $a = - \frac{b}{4} - \frac{1}{4}$ with $7$ .

$a = - \frac{7}{4} - \frac{1}{4}$

$b = 7$

$c = 5$

Simplify the right side.

Simplify $- \frac{7}{4} - \frac{1}{4}$ .

Combine the numerators over the common denominator.

$a = \frac{- 7 - 1}{4}$

$b = 7$

$c = 5$

Simplify the expression.

Subtract $1$ from $- 7$ .

$a = \frac{- 8}{4}$

$b = 7$

$c = 5$

Divide $- 8$ by $4$ .

$a = - 2$

$b = 7$

$c = 5$

$a = - 2$

$b = 7$

$c = 5$

$a = - 2$

$b = 7$

$c = 5$

$a = - 2$

$b = 7$

$c = 5$

$a = - 2$

$b = 7$

$c = 5$

List all of the solutions.

$a = - 2, b = 7, c = 5$

Step 4

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

$y = - 2 x^{2} + 7 x + 5$

Step 5

Find the Equation of the Parabola (0,5) , (2,11) , (4,1)

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