Find the Equation of the Parabola (3,1) , (-3,-5) , (9,-5)

$(3, 1)$ , $(- 3, - 5)$ , $(9, - 5)$

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.

$1 = a {(3)}^{2} + b (3) + c$

$- 5 = a {(- 3)}^{2} + b (- 3) + c$

$- 5 = a {(9)}^{2} + b (9) + c$

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

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Simplify each equation.

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Remove parentheses.

$1 = a {(3)}^{2} + b (3) + c$

$- 5 = a {(- 3)}^{2} + b (- 3) + c$

$- 5 = a {(9)}^{2} + b (9) + c$

Simplify each term.

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Raise $3$ to the power of $2$ .

$1 = a \cdot 9 + b (3) + c$

$- 5 = a {(- 3)}^{2} + b (- 3) + c$

$- 5 = a {(9)}^{2} + b (9) + c$

Move $9$ to the left of $a$ .

$1 = 9 \cdot a + b (3) + c$

$- 5 = a {(- 3)}^{2} + b (- 3) + c$

$- 5 = a {(9)}^{2} + b (9) + c$

Move $3$ to the left of $b$ .

$1 = 9 a + 3 b + c$

$- 5 = a {(- 3)}^{2} + b (- 3) + c$

$- 5 = a {(9)}^{2} + b (9) + c$

$1 = 9 a + 3 b + c$

$- 5 = a {(- 3)}^{2} + b (- 3) + c$

$- 5 = a {(9)}^{2} + b (9) + c$

Remove parentheses.

$1 = 9 a + 3 b + c$

$- 5 = a {(- 3)}^{2} + b (- 3) + c$

$- 5 = a {(9)}^{2} + b (9) + c$

Simplify each term.

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Raise $- 3$ to the power of $2$ .

$1 = 9 a + 3 b + c$

$- 5 = a \cdot 9 + b (- 3) + c$

$- 5 = a {(9)}^{2} + b (9) + c$

Move $9$ to the left of $a$ .

$1 = 9 a + 3 b + c$

$- 5 = 9 \cdot a + b (- 3) + c$

$- 5 = a {(9)}^{2} + b (9) + c$

Move $- 3$ to the left of $b$ .

$1 = 9 a + 3 b + c$

$- 5 = 9 a - 3 b + c$

$- 5 = a {(9)}^{2} + b (9) + c$

$1 = 9 a + 3 b + c$

$- 5 = 9 a - 3 b + c$

$- 5 = a {(9)}^{2} + b (9) + c$

Remove parentheses.

$1 = 9 a + 3 b + c$

$- 5 = 9 a - 3 b + c$

$- 5 = a {(9)}^{2} + b (9) + c$

Simplify each term.

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Raise $9$ to the power of $2$ .

$1 = 9 a + 3 b + c$

$- 5 = 9 a - 3 b + c$

$- 5 = a \cdot 81 + b (9) + c$

Move $81$ to the left of $a$ .

$1 = 9 a + 3 b + c$

$- 5 = 9 a - 3 b + c$

$- 5 = 81 \cdot a + b (9) + c$

Move $9$ to the left of $b$ .

$1 = 9 a + 3 b + c$

$- 5 = 9 a - 3 b + c$

$- 5 = 81 a + 9 b + c$

$1 = 9 a + 3 b + c$

$- 5 = 9 a - 3 b + c$

$- 5 = 81 a + 9 b + c$

$1 = 9 a + 3 b + c$

$- 5 = 9 a - 3 b + c$

$- 5 = 81 a + 9 b + c$

Solve for $c$ in the first equation.

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Rewrite the equation as $9 a + 3 b + c = 1$ .

$9 a + 3 b + c = 1$

$- 5 = 9 a - 3 b + c$

$- 5 = 81 a + 9 b + c$

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

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Subtract $9 a$ from both sides of the equation.

$3 b + c = 1 - 9 a$

$- 5 = 9 a - 3 b + c$

$- 5 = 81 a + 9 b + c$

Subtract $3 b$ from both sides of the equation.

$c = 1 - 9 a - 3 b$

$- 5 = 9 a - 3 b + c$

$- 5 = 81 a + 9 b + c$

$c = 1 - 9 a - 3 b$

$- 5 = 9 a - 3 b + c$

$- 5 = 81 a + 9 b + c$

$c = 1 - 9 a - 3 b$

$- 5 = 9 a - 3 b + c$

$- 5 = 81 a + 9 b + c$

Replace all occurrences of $c$ with $1 - 9 a - 3 b$ in each equation.

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Replace all occurrences of $c$ in $- 5 = 9 a - 3 b + c$ with $1 - 9 a - 3 b$ .

$c = 1 - 9 a - 3 b$

$- 5 = 9 a - 3 b + 1 - 9 a - 3 b$

$- 5 = 81 a + 9 b + c$

Replace all occurrences of $c$ in $- 5 = 81 a + 9 b + c$ with $1 - 9 a - 3 b$ .

$c = 1 - 9 a - 3 b$

$- 5 = 9 a - 3 b + 1 - 9 a - 3 b$

$- 5 = 81 a + 9 b + 1 - 9 a - 3 b$

$c = 1 - 9 a - 3 b$

$- 5 = 9 a - 3 b + 1 - 9 a - 3 b$

$- 5 = 81 a + 9 b + 1 - 9 a - 3 b$

Simplify each equation.

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Remove parentheses.

$c = 1 - 9 a - 3 b$

$- 5 = 9 a - 3 b + 1 - 9 a - 3 b$

$- 5 = 81 a + 9 b + 1 - 9 a - 3 b$

Simplify $9 a - 3 b + 1 - 9 a - 3 b$ .

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Combine the opposite terms in $9 a - 3 b + 1 - 9 a - 3 b$ .

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Subtract $9 a$ from $9 a$ .

$c = 1 - 9 a - 3 b$

$- 5 = - 3 b + 1 + 0 - 3 b$

$- 5 = 81 a + 9 b + 1 - 9 a - 3 b$

Add $- 3 b + 1$ and $0$ .

$c = 1 - 9 a - 3 b$

$- 5 = - 3 b + 1 - 3 b$

$- 5 = 81 a + 9 b + 1 - 9 a - 3 b$

$c = 1 - 9 a - 3 b$

$- 5 = - 3 b + 1 - 3 b$

$- 5 = 81 a + 9 b + 1 - 9 a - 3 b$

Subtract $3 b$ from $- 3 b$ .

$c = 1 - 9 a - 3 b$

$- 5 = - 6 b + 1$

$- 5 = 81 a + 9 b + 1 - 9 a - 3 b$

$c = 1 - 9 a - 3 b$

$- 5 = - 6 b + 1$

$- 5 = 81 a + 9 b + 1 - 9 a - 3 b$

Remove parentheses.

$c = 1 - 9 a - 3 b$

$- 5 = - 6 b + 1$

$- 5 = 81 a + 9 b + 1 - 9 a - 3 b$

Simplify $81 a + 9 b + 1 - 9 a - 3 b$ .

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Subtract $9 a$ from $81 a$ .

$c = 1 - 9 a - 3 b$

$- 5 = - 6 b + 1$

$- 5 = 72 a + 9 b + 1 - 3 b$

Subtract $3 b$ from $9 b$ .

$c = 1 - 9 a - 3 b$

$- 5 = - 6 b + 1$

$- 5 = 72 a + 6 b + 1$

$c = 1 - 9 a - 3 b$

$- 5 = - 6 b + 1$

$- 5 = 72 a + 6 b + 1$

$c = 1 - 9 a - 3 b$

$- 5 = - 6 b + 1$

$- 5 = 72 a + 6 b + 1$

Solve for $b$ in the second equation.

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Rewrite the equation as $- 6 b + 1 = - 5$ .

$c = 1 - 9 a - 3 b$

$- 6 b + 1 = - 5$

$- 5 = 72 a + 6 b + 1$

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

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Subtract $1$ from both sides of the equation.

$c = 1 - 9 a - 3 b$

$- 6 b = - 5 - 1$

$- 5 = 72 a + 6 b + 1$

Subtract $1$ from $- 5$ .

$c = 1 - 9 a - 3 b$

$- 6 b = - 6$

$- 5 = 72 a + 6 b + 1$

$c = 1 - 9 a - 3 b$

$- 6 b = - 6$

$- 5 = 72 a + 6 b + 1$

Divide each term by $- 6$ and simplify.

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

$c = 1 - 9 a - 3 b$

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

$- 5 = 72 a + 6 b + 1$

Cancel the common factor of $- 6$ .

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Cancel the common factor.

$c = 1 - 9 a - 3 b$

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

$- 5 = 72 a + 6 b + 1$

Divide $b$ by $1$ .

$c = 1 - 9 a - 3 b$

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

$- 5 = 72 a + 6 b + 1$

$c = 1 - 9 a - 3 b$

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

$- 5 = 72 a + 6 b + 1$

Divide $- 6$ by $- 6$ .

$c = 1 - 9 a - 3 b$

$b = 1$

$- 5 = 72 a + 6 b + 1$

$c = 1 - 9 a - 3 b$

$b = 1$

$- 5 = 72 a + 6 b + 1$

$c = 1 - 9 a - 3 b$

$b = 1$

$- 5 = 72 a + 6 b + 1$

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

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Replace all occurrences of $b$ in $c = 1 - 9 a - 3 b$ with $1$ .

$c = 1 - 9 a - 3 \cdot 1$

$b = 1$

$- 5 = 72 a + 6 b + 1$

Replace all occurrences of $b$ in $- 5 = 72 a + 6 b + 1$ with $1$ .

$c = 1 - 9 a - 3 \cdot 1$

$b = 1$

$- 5 = 72 a + 6 (1) + 1$

$c = 1 - 9 a - 3 \cdot 1$

$b = 1$

$- 5 = 72 a + 6 (1) + 1$

Simplify.

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Simplify $1 - 9 a - 3 \cdot 1$ .

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Multiply $- 3$ by $1$ .

$c = 1 - 9 a - 3$

$b = 1$

$- 5 = 72 a + 6 (1) + 1$

Subtract $3$ from $1$ .

$c = - 9 a - 2$

$b = 1$

$- 5 = 72 a + 6 (1) + 1$

$c = - 9 a - 2$

$b = 1$

$- 5 = 72 a + 6 (1) + 1$

Simplify $72 a + 6 (1) + 1$ .

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Multiply $6$ by $1$ .

$c = - 9 a - 2$

$b = 1$

$- 5 = 72 a + 6 + 1$

Add $6$ and $1$ .

$c = - 9 a - 2$

$b = 1$

$- 5 = 72 a + 7$

$c = - 9 a - 2$

$b = 1$

$- 5 = 72 a + 7$

$c = - 9 a - 2$

$b = 1$

$- 5 = 72 a + 7$

Solve for $a$ in the third equation.

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Rewrite the equation as $72 a + 7 = - 5$ .

$c = - 9 a - 2$

$b = 1$

$72 a + 7 = - 5$

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

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Subtract $7$ from both sides of the equation.

$c = - 9 a - 2$

$b = 1$

$72 a = - 5 - 7$

Subtract $7$ from $- 5$ .

$c = - 9 a - 2$

$b = 1$

$72 a = - 12$

$c = - 9 a - 2$

$b = 1$

$72 a = - 12$

Divide each term by $72$ and simplify.

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Divide each term in $72 a = - 12$ by $72$ .

$c = - 9 a - 2$

$b = 1$

$\frac{72 a}{72} = \frac{- 12}{72}$

Cancel the common factor of $72$ .

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Cancel the common factor.

$c = - 9 a - 2$

$b = 1$

$\frac{72 a}{72} = \frac{- 12}{72}$

Divide $a$ by $1$ .

$c = - 9 a - 2$

$b = 1$

$a = \frac{- 12}{72}$

$c = - 9 a - 2$

$b = 1$

$a = \frac{- 12}{72}$

Simplify $\frac{- 12}{72}$ .

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Cancel the common factor of $- 12$ and $72$ .

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Factor $12$ out of $- 12$ .

$c = - 9 a - 2$

$b = 1$

$a = \frac{12 (- 1)}{72}$

Cancel the common factors.

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Factor $12$ out of $72$ .

$c = - 9 a - 2$

$b = 1$

$a = \frac{12 \cdot - 1}{12 \cdot 6}$

Cancel the common factor.

$c = - 9 a - 2$

$b = 1$

$a = \frac{12 \cdot - 1}{12 \cdot 6}$

Rewrite the expression.

$c = - 9 a - 2$

$b = 1$

$a = \frac{- 1}{6}$

$c = - 9 a - 2$

$b = 1$

$a = \frac{- 1}{6}$

$c = - 9 a - 2$

$b = 1$

$a = \frac{- 1}{6}$

Move the negative in front of the fraction.

$c = - 9 a - 2$

$b = 1$

$a = - \frac{1}{6}$

$c = - 9 a - 2$

$b = 1$

$a = - \frac{1}{6}$

$c = - 9 a - 2$

$b = 1$

$a = - \frac{1}{6}$

$c = - 9 a - 2$

$b = 1$

$a = - \frac{1}{6}$

Replace all occurrences of $a$ in $c = - 9 a - 2$ with $- \frac{1}{6}$ .

$c = - 9 (- \frac{1}{6}) - 2$

$b = 1$

$a = - \frac{1}{6}$

Simplify $- 9 (- \frac{1}{6}) - 2$ .

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Simplify each term.

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Cancel the common factor of $3$ .

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Move the leading negative in $- \frac{1}{6}$ into the numerator.

$c = - 9 (\frac{- 1}{6}) - 2$

$b = 1$

$a = - \frac{1}{6}$

Factor $3$ out of $- 9$ .

$c = 3 (- 3) (\frac{- 1}{6}) - 2$

$b = 1$

$a = - \frac{1}{6}$

Factor $3$ out of $6$ .

$c = 3 \cdot (- 3 \frac{- 1}{3 \cdot 2}) - 2$

$b = 1$

$a = - \frac{1}{6}$

Cancel the common factor.

$c = 3 \cdot (- 3 \frac{- 1}{3 \cdot 2}) - 2$

$b = 1$

$a = - \frac{1}{6}$

Rewrite the expression.

$c = - 3 (\frac{- 1}{2}) - 2$

$b = 1$

$a = - \frac{1}{6}$

$c = - 3 (\frac{- 1}{2}) - 2$

$b = 1$

$a = - \frac{1}{6}$

Combine $- 3$ and $\frac{- 1}{2}$ .

$c = \frac{- 3 \cdot - 1}{2} - 2$

$b = 1$

$a = - \frac{1}{6}$

Multiply $- 3$ by $- 1$ .

$c = \frac{3}{2} - 2$

$b = 1$

$a = - \frac{1}{6}$

$c = \frac{3}{2} - 2$

$b = 1$

$a = - \frac{1}{6}$

To write $- 2$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$c = \frac{3}{2} - 2 \cdot \frac{2}{2}$

$b = 1$

$a = - \frac{1}{6}$

Combine $- 2$ and $\frac{2}{2}$ .

$c = \frac{3}{2} + \frac{- 2 \cdot 2}{2}$

$b = 1$

$a = - \frac{1}{6}$

Combine the numerators over the common denominator.

$c = \frac{3 - 2 \cdot 2}{2}$

$b = 1$

$a = - \frac{1}{6}$

Simplify the numerator.

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Multiply $- 2$ by $2$ .

$c = \frac{3 - 4}{2}$

$b = 1$

$a = - \frac{1}{6}$

Subtract $4$ from $3$ .

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

$b = 1$

$a = - \frac{1}{6}$

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

$b = 1$

$a = - \frac{1}{6}$

Move the negative in front of the fraction.

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

$b = 1$

$a = - \frac{1}{6}$

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

$b = 1$

$a = - \frac{1}{6}$

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

$b = 1$

$a = - \frac{1}{6}$

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

$y = - \frac{x^{2}}{6} + x - \frac{1}{2}$

Find the Equation of the Parabola (3,1) , (-3,-5) , (9,-5)

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