Find the Equation of the Parabola (0,9) , (2,19) , (4,13)

$(0, 9)$ , $(2, 19)$ , $(4, 13)$

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.

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

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.

$9 = a {(0)}^{2} + b (0) + c$

$19 = a {(2)}^{2} + b (2) + c$

$13 = a {(4)}^{2} + b (4) + c$

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

Solve for $c$ in $9 = a \cdot 0^{2} + c$ .

Rewrite the equation as $a \cdot 0^{2} + c = 9$ .

$a \cdot 0^{2} + c = 9$

$19 = a \cdot 2^{2} + b (2) + c$

$13 = a \cdot 4^{2} + b (4) + c$

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

Simplify each term.

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

$a \cdot 0 + c = 9$

$19 = a \cdot 2^{2} + b (2) + c$

$13 = a \cdot 4^{2} + b (4) + c$

Multiply $a$ by $0$ .

$0 + c = 9$

$19 = a \cdot 2^{2} + b (2) + c$

$13 = a \cdot 4^{2} + b (4) + c$

$0 + c = 9$

$19 = a \cdot 2^{2} + b (2) + c$

$13 = a \cdot 4^{2} + b (4) + c$

Add $0$ and $c$ .

$c = 9$

$19 = a \cdot 2^{2} + b (2) + c$

$13 = a \cdot 4^{2} + b (4) + c$

$c = 9$

$19 = a \cdot 2^{2} + b (2) + c$

$13 = a \cdot 4^{2} + b (4) + c$

$c = 9$

$19 = a \cdot 2^{2} + b (2) + c$

$13 = a \cdot 4^{2} + b (4) + c$

Replace all occurrences of $c$ with $9$ in each equation.

Replace all occurrences of $c$ in $19 = a \cdot 2^{2} + b (2) + c$ with $9$ .

$19 = a \cdot 2^{2} + b (2) + 9$

$c = 9$

$13 = a \cdot 4^{2} + b (4) + c$

Simplify $19 = a \cdot 2^{2} + b (2) + 9$ .

Simplify the left side.

Remove parentheses.

$19 = a \cdot 2^{2} + b (2) + 9$

$c = 9$

$13 = a \cdot 4^{2} + b (4) + c$

$19 = a \cdot 2^{2} + b (2) + 9$

$c = 9$

$13 = a \cdot 4^{2} + b (4) + c$

Simplify the right side.

Simplify each term.

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

$19 = a \cdot 4 + b (2) + 9$

$c = 9$

$13 = a \cdot 4^{2} + b (4) + c$

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

$19 = 4 \cdot a + b (2) + 9$

$c = 9$

$13 = a \cdot 4^{2} + b (4) + c$

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

$19 = 4 a + 2 b + 9$

$c = 9$

$13 = a \cdot 4^{2} + b (4) + c$

$19 = 4 a + 2 b + 9$

$c = 9$

$13 = a \cdot 4^{2} + b (4) + c$

$19 = 4 a + 2 b + 9$

$c = 9$

$13 = a \cdot 4^{2} + b (4) + c$

$19 = 4 a + 2 b + 9$

$c = 9$

$13 = a \cdot 4^{2} + b (4) + c$

Replace all occurrences of $c$ in $13 = a \cdot 4^{2} + b (4) + c$ with $9$ .

$13 = a \cdot 4^{2} + b (4) + 9$

$19 = 4 a + 2 b + 9$

$c = 9$

Simplify $13 = a \cdot 4^{2} + b (4) + 9$ .

Simplify the left side.

Remove parentheses.

$13 = a \cdot 4^{2} + b (4) + 9$

$19 = 4 a + 2 b + 9$

$c = 9$

$13 = a \cdot 4^{2} + b (4) + 9$

$19 = 4 a + 2 b + 9$

$c = 9$

Simplify the right side.

Simplify each term.

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

$13 = a \cdot 16 + b (4) + 9$

$19 = 4 a + 2 b + 9$

$c = 9$

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

$13 = 16 \cdot a + b (4) + 9$

$19 = 4 a + 2 b + 9$

$c = 9$

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

$13 = 16 a + 4 b + 9$

$19 = 4 a + 2 b + 9$

$c = 9$

$13 = 16 a + 4 b + 9$

$19 = 4 a + 2 b + 9$

$c = 9$

$13 = 16 a + 4 b + 9$

$19 = 4 a + 2 b + 9$

$c = 9$

$13 = 16 a + 4 b + 9$

$19 = 4 a + 2 b + 9$

$c = 9$

$13 = 16 a + 4 b + 9$

$19 = 4 a + 2 b + 9$

$c = 9$

Solve for $a$ in $13 = 16 a + 4 b + 9$ .

Rewrite the equation as $16 a + 4 b + 9 = 13$ .

$16 a + 4 b + 9 = 13$

$19 = 4 a + 2 b + 9$

$c = 9$

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

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

$16 a + 9 = 13 - 4 b$

$19 = 4 a + 2 b + 9$

$c = 9$

Subtract $9$ from both sides of the equation.

$16 a = 13 - 4 b - 9$

$19 = 4 a + 2 b + 9$

$c = 9$

Subtract $9$ from $13$ .

$16 a = - 4 b + 4$

$19 = 4 a + 2 b + 9$

$c = 9$

$16 a = - 4 b + 4$

$19 = 4 a + 2 b + 9$

$c = 9$

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

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

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

$19 = 4 a + 2 b + 9$

$c = 9$

Simplify the left side.

Cancel the common factor of $16$ .

Cancel the common factor.

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

$19 = 4 a + 2 b + 9$

$c = 9$

Divide $a$ by $1$ .

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

$19 = 4 a + 2 b + 9$

$c = 9$

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

$19 = 4 a + 2 b + 9$

$c = 9$

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

$19 = 4 a + 2 b + 9$

$c = 9$

Simplify the right side.

Simplify each term.

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

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

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

$19 = 4 a + 2 b + 9$

$c = 9$

Cancel the common factors.

Factor $4$ out of $16$ .

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

$19 = 4 a + 2 b + 9$

$c = 9$

Cancel the common factor.

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

$19 = 4 a + 2 b + 9$

$c = 9$

Rewrite the expression.

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

$19 = 4 a + 2 b + 9$

$c = 9$

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

$19 = 4 a + 2 b + 9$

$c = 9$

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

$19 = 4 a + 2 b + 9$

$c = 9$

Move the negative in front of the fraction.

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

$19 = 4 a + 2 b + 9$

$c = 9$

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

Factor $4$ out of $4$ .

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

$19 = 4 a + 2 b + 9$

$c = 9$

Cancel the common factors.

Factor $4$ out of $16$ .

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

$19 = 4 a + 2 b + 9$

$c = 9$

Cancel the common factor.

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

$19 = 4 a + 2 b + 9$

$c = 9$

Rewrite the expression.

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

$19 = 4 a + 2 b + 9$

$c = 9$

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

$19 = 4 a + 2 b + 9$

$c = 9$

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

$19 = 4 a + 2 b + 9$

$c = 9$

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

$19 = 4 a + 2 b + 9$

$c = 9$

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

$19 = 4 a + 2 b + 9$

$c = 9$

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

$19 = 4 a + 2 b + 9$

$c = 9$

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

$19 = 4 a + 2 b + 9$

$c = 9$

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

Replace all occurrences of $a$ in $19 = 4 a + 2 b + 9$ with $- \frac{b}{4} + \frac{1}{4}$ .

$19 = 4 (- \frac{b}{4} + \frac{1}{4}) + 2 b + 9$

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

$c = 9$

Simplify the right side.

Simplify $4 (- \frac{b}{4} + \frac{1}{4}) + 2 b + 9$ .

Simplify each term.

Apply the distributive property.

$19 = 4 (- \frac{b}{4}) + 4 (\frac{1}{4}) + 2 b + 9$

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

$c = 9$

Cancel the common factor of $4$ .

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

$19 = 4 (\frac{- b}{4}) + 4 (\frac{1}{4}) + 2 b + 9$

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

$c = 9$

Cancel the common factor.

$19 = 4 (\frac{- b}{4}) + 4 (\frac{1}{4}) + 2 b + 9$

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

$c = 9$

Rewrite the expression.

$19 = - b + 4 (\frac{1}{4}) + 2 b + 9$

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

$c = 9$

$19 = - b + 4 (\frac{1}{4}) + 2 b + 9$

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

$c = 9$

Cancel the common factor of $4$ .

Cancel the common factor.

$19 = - b + 4 (\frac{1}{4}) + 2 b + 9$

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

$c = 9$

Rewrite the expression.

$19 = - b + 1 + 2 b + 9$

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

$c = 9$

$19 = - b + 1 + 2 b + 9$

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

$c = 9$

$19 = - b + 1 + 2 b + 9$

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

$c = 9$

Simplify by adding terms.

Add $- b$ and $2 b$ .

$19 = b + 1 + 9$

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

$c = 9$

Add $1$ and $9$ .

$19 = b + 10$

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

$c = 9$

$19 = b + 10$

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

$c = 9$

$19 = b + 10$

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

$c = 9$

$19 = b + 10$

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

$c = 9$

$19 = b + 10$

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

$c = 9$

Solve for $b$ in $19 = b + 10$ .

Rewrite the equation as $b + 10 = 19$ .

$b + 10 = 19$

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

$c = 9$

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

Subtract $10$ from both sides of the equation.

$b = 19 - 10$

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

$c = 9$

Subtract $10$ from $19$ .

$b = 9$

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

$c = 9$

$b = 9$

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

$c = 9$

$b = 9$

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

$c = 9$

Replace all occurrences of $b$ with $9$ in each equation.

Replace all occurrences of $b$ in $a = - \frac{b}{4} + \frac{1}{4}$ with $9$ .

$a = - \frac{9}{4} + \frac{1}{4}$

$b = 9$

$c = 9$

Simplify the right side.

Simplify $- \frac{9}{4} + \frac{1}{4}$ .

Combine the numerators over the common denominator.

$a = \frac{- 9 + 1}{4}$

$b = 9$

$c = 9$

Simplify the expression.

Add $- 9$ and $1$ .

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

$b = 9$

$c = 9$

Divide $- 8$ by $4$ .

$a = - 2$

$b = 9$

$c = 9$

$a = - 2$

$b = 9$

$c = 9$

$a = - 2$

$b = 9$

$c = 9$

$a = - 2$

$b = 9$

$c = 9$

$a = - 2$

$b = 9$

$c = 9$

List all of the solutions.

$a = - 2, b = 9, c = 9$

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

$y = - 2 x^{2} + 9 x + 9$

Find the Equation of the Parabola (0,9) , (2,19) , (4,13)

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