Find the Equation of the Parabola (0,1) , (2,-3) , (4,-7)

$(0, 1)$ , $(2, - 3)$ , $(4, - 7)$

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.

$1 = a {(0)}^{2} + b (0) + c, - 3 = a {(2)}^{2} + b (2) + c, - 7 = a {(4)}^{2} + b (4) + c$

Step 3

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

Solve for $c$ in $1 = a \cdot 0^{2} + c$ .

Rewrite the equation as $a \cdot 0^{2} + c = 1$ .

$a \cdot 0^{2} + c = 1$

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

$- 7 = a \cdot 4^{2} + b (4) + c$

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

Simplify each term.

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

$a \cdot 0 + c = 1$

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

$- 7 = a \cdot 4^{2} + b (4) + c$

Multiply $a$ by $0$ .

$0 + c = 1$

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

$- 7 = a \cdot 4^{2} + b (4) + c$

$0 + c = 1$

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

$- 7 = a \cdot 4^{2} + b (4) + c$

Add $0$ and $c$ .

$c = 1$

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

$- 7 = a \cdot 4^{2} + b (4) + c$

$c = 1$

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

$- 7 = a \cdot 4^{2} + b (4) + c$

$c = 1$

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

$- 7 = a \cdot 4^{2} + b (4) + c$

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

Replace all occurrences of $c$ in $- 3 = a \cdot 2^{2} + b (2) + c$ with $1$ .

$- 3 = a \cdot 2^{2} + b (2) + 1$

$c = 1$

$- 7 = a \cdot 4^{2} + b (4) + c$

Simplify $- 3 = a \cdot 2^{2} + b (2) + 1$ .

Simplify the left side.

Remove parentheses.

$- 3 = a \cdot 2^{2} + b (2) + 1$

$c = 1$

$- 7 = a \cdot 4^{2} + b (4) + c$

$- 3 = a \cdot 2^{2} + b (2) + 1$

$c = 1$

$- 7 = a \cdot 4^{2} + b (4) + c$

Simplify the right side.

Simplify each term.

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

$- 3 = a \cdot 4 + b (2) + 1$

$c = 1$

$- 7 = a \cdot 4^{2} + b (4) + c$

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

$- 3 = 4 \cdot a + b (2) + 1$

$c = 1$

$- 7 = a \cdot 4^{2} + b (4) + c$

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

$- 3 = 4 a + 2 b + 1$

$c = 1$

$- 7 = a \cdot 4^{2} + b (4) + c$

$- 3 = 4 a + 2 b + 1$

$c = 1$

$- 7 = a \cdot 4^{2} + b (4) + c$

$- 3 = 4 a + 2 b + 1$

$c = 1$

$- 7 = a \cdot 4^{2} + b (4) + c$

$- 3 = 4 a + 2 b + 1$

$c = 1$

$- 7 = a \cdot 4^{2} + b (4) + c$

Replace all occurrences of $c$ in $- 7 = a \cdot 4^{2} + b (4) + c$ with $1$ .

$- 7 = a \cdot 4^{2} + b (4) + 1$

$- 3 = 4 a + 2 b + 1$

$c = 1$

Simplify $- 7 = a \cdot 4^{2} + b (4) + 1$ .

Simplify the left side.

Remove parentheses.

$- 7 = a \cdot 4^{2} + b (4) + 1$

$- 3 = 4 a + 2 b + 1$

$c = 1$

$- 7 = a \cdot 4^{2} + b (4) + 1$

$- 3 = 4 a + 2 b + 1$

$c = 1$

Simplify the right side.

Simplify each term.

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

$- 7 = a \cdot 16 + b (4) + 1$

$- 3 = 4 a + 2 b + 1$

$c = 1$

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

$- 7 = 16 \cdot a + b (4) + 1$

$- 3 = 4 a + 2 b + 1$

$c = 1$

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

$- 7 = 16 a + 4 b + 1$

$- 3 = 4 a + 2 b + 1$

$c = 1$

$- 7 = 16 a + 4 b + 1$

$- 3 = 4 a + 2 b + 1$

$c = 1$

$- 7 = 16 a + 4 b + 1$

$- 3 = 4 a + 2 b + 1$

$c = 1$

$- 7 = 16 a + 4 b + 1$

$- 3 = 4 a + 2 b + 1$

$c = 1$

$- 7 = 16 a + 4 b + 1$

$- 3 = 4 a + 2 b + 1$

$c = 1$

Solve for $a$ in $- 7 = 16 a + 4 b + 1$ .

Rewrite the equation as $16 a + 4 b + 1 = - 7$ .

$16 a + 4 b + 1 = - 7$

$- 3 = 4 a + 2 b + 1$

$c = 1$

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

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

$16 a + 1 = - 7 - 4 b$

$- 3 = 4 a + 2 b + 1$

$c = 1$

Subtract $1$ from both sides of the equation.

$16 a = - 7 - 4 b - 1$

$- 3 = 4 a + 2 b + 1$

$c = 1$

Subtract $1$ from $- 7$ .

$16 a = - 4 b - 8$

$- 3 = 4 a + 2 b + 1$

$c = 1$

$16 a = - 4 b - 8$

$- 3 = 4 a + 2 b + 1$

$c = 1$

Divide each term in $16 a = - 4 b - 8$ by $16$ and simplify.

Divide each term in $16 a = - 4 b - 8$ by $16$ .

$\frac{16 a}{16} = \frac{- 4 b}{16} + \frac{- 8}{16}$

$- 3 = 4 a + 2 b + 1$

$c = 1$

Simplify the left side.

Cancel the common factor of $16$ .

Cancel the common factor.

$\frac{16 a}{16} = \frac{- 4 b}{16} + \frac{- 8}{16}$

$- 3 = 4 a + 2 b + 1$

$c = 1$

Divide $a$ by $1$ .

$a = \frac{- 4 b}{16} + \frac{- 8}{16}$

$- 3 = 4 a + 2 b + 1$

$c = 1$

$a = \frac{- 4 b}{16} + \frac{- 8}{16}$

$- 3 = 4 a + 2 b + 1$

$c = 1$

$a = \frac{- 4 b}{16} + \frac{- 8}{16}$

$- 3 = 4 a + 2 b + 1$

$c = 1$

Simplify the right side.

Simplify each term.

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

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

$a = \frac{4 (- b)}{16} + \frac{- 8}{16}$

$- 3 = 4 a + 2 b + 1$

$c = 1$

Cancel the common factors.

Factor $4$ out of $16$ .

$a = \frac{4 (- b)}{4 (4)} + \frac{- 8}{16}$

$- 3 = 4 a + 2 b + 1$

$c = 1$

Cancel the common factor.

$a = \frac{4 (- b)}{4 \cdot 4} + \frac{- 8}{16}$

$- 3 = 4 a + 2 b + 1$

$c = 1$

Rewrite the expression.

$a = \frac{- b}{4} + \frac{- 8}{16}$

$- 3 = 4 a + 2 b + 1$

$c = 1$

$a = \frac{- b}{4} + \frac{- 8}{16}$

$- 3 = 4 a + 2 b + 1$

$c = 1$

$a = \frac{- b}{4} + \frac{- 8}{16}$

$- 3 = 4 a + 2 b + 1$

$c = 1$

Move the negative in front of the fraction.

$a = - \frac{b}{4} + \frac{- 8}{16}$

$- 3 = 4 a + 2 b + 1$

$c = 1$

Cancel the common factor of $- 8$ and $16$ .

Factor $8$ out of $- 8$ .

$a = - \frac{b}{4} + \frac{8 (- 1)}{16}$

$- 3 = 4 a + 2 b + 1$

$c = 1$

Cancel the common factors.

Factor $8$ out of $16$ .

$a = - \frac{b}{4} + \frac{8 \cdot - 1}{8 \cdot 2}$

$- 3 = 4 a + 2 b + 1$

$c = 1$

Cancel the common factor.

$a = - \frac{b}{4} + \frac{8 \cdot - 1}{8 \cdot 2}$

$- 3 = 4 a + 2 b + 1$

$c = 1$

Rewrite the expression.

$a = - \frac{b}{4} + \frac{- 1}{2}$

$- 3 = 4 a + 2 b + 1$

$c = 1$

$a = - \frac{b}{4} + \frac{- 1}{2}$

$- 3 = 4 a + 2 b + 1$

$c = 1$

$a = - \frac{b}{4} + \frac{- 1}{2}$

$- 3 = 4 a + 2 b + 1$

$c = 1$

Move the negative in front of the fraction.

$a = - \frac{b}{4} - \frac{1}{2}$

$- 3 = 4 a + 2 b + 1$

$c = 1$

$a = - \frac{b}{4} - \frac{1}{2}$

$- 3 = 4 a + 2 b + 1$

$c = 1$

$a = - \frac{b}{4} - \frac{1}{2}$

$- 3 = 4 a + 2 b + 1$

$c = 1$

$a = - \frac{b}{4} - \frac{1}{2}$

$- 3 = 4 a + 2 b + 1$

$c = 1$

$a = - \frac{b}{4} - \frac{1}{2}$

$- 3 = 4 a + 2 b + 1$

$c = 1$

Replace all occurrences of $a$ with $- \frac{b}{4} - \frac{1}{2}$ in each equation.

Replace all occurrences of $a$ in $- 3 = 4 a + 2 b + 1$ with $- \frac{b}{4} - \frac{1}{2}$ .

$- 3 = 4 (- \frac{b}{4} - \frac{1}{2}) + 2 b + 1$

$a = - \frac{b}{4} - \frac{1}{2}$

$c = 1$

Simplify the right side.

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

Simplify each term.

Apply the distributive property.

$- 3 = 4 (- \frac{b}{4}) + 4 (- \frac{1}{2}) + 2 b + 1$

$a = - \frac{b}{4} - \frac{1}{2}$

$c = 1$

Cancel the common factor of $4$ .

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

$- 3 = 4 (\frac{- b}{4}) + 4 (- \frac{1}{2}) + 2 b + 1$

$a = - \frac{b}{4} - \frac{1}{2}$

$c = 1$

Cancel the common factor.

$- 3 = 4 (\frac{- b}{4}) + 4 (- \frac{1}{2}) + 2 b + 1$

$a = - \frac{b}{4} - \frac{1}{2}$

$c = 1$

Rewrite the expression.

$- 3 = - b + 4 (- \frac{1}{2}) + 2 b + 1$

$a = - \frac{b}{4} - \frac{1}{2}$

$c = 1$

$- 3 = - b + 4 (- \frac{1}{2}) + 2 b + 1$

$a = - \frac{b}{4} - \frac{1}{2}$

$c = 1$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$- 3 = - b + 4 (\frac{- 1}{2}) + 2 b + 1$

$a = - \frac{b}{4} - \frac{1}{2}$

$c = 1$

Factor $2$ out of $4$ .

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

$a = - \frac{b}{4} - \frac{1}{2}$

$c = 1$

Cancel the common factor.

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

$a = - \frac{b}{4} - \frac{1}{2}$

$c = 1$

Rewrite the expression.

$- 3 = - b + 2 \cdot - 1 + 2 b + 1$

$a = - \frac{b}{4} - \frac{1}{2}$

$c = 1$

$- 3 = - b + 2 \cdot - 1 + 2 b + 1$

$a = - \frac{b}{4} - \frac{1}{2}$

$c = 1$

Multiply $2$ by $- 1$ .

$- 3 = - b - 2 + 2 b + 1$

$a = - \frac{b}{4} - \frac{1}{2}$

$c = 1$

$- 3 = - b - 2 + 2 b + 1$

$a = - \frac{b}{4} - \frac{1}{2}$

$c = 1$

Simplify by adding terms.

Add $- b$ and $2 b$ .

$- 3 = b - 2 + 1$

$a = - \frac{b}{4} - \frac{1}{2}$

$c = 1$

Add $- 2$ and $1$ .

$- 3 = b - 1$

$a = - \frac{b}{4} - \frac{1}{2}$

$c = 1$

$- 3 = b - 1$

$a = - \frac{b}{4} - \frac{1}{2}$

$c = 1$

$- 3 = b - 1$

$a = - \frac{b}{4} - \frac{1}{2}$

$c = 1$

$- 3 = b - 1$

$a = - \frac{b}{4} - \frac{1}{2}$

$c = 1$

$- 3 = b - 1$

$a = - \frac{b}{4} - \frac{1}{2}$

$c = 1$

Solve for $b$ in $- 3 = b - 1$ .

Rewrite the equation as $b - 1 = - 3$ .

$b - 1 = - 3$

$a = - \frac{b}{4} - \frac{1}{2}$

$c = 1$

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

Add $1$ to both sides of the equation.

$b = - 3 + 1$

$a = - \frac{b}{4} - \frac{1}{2}$

$c = 1$

Add $- 3$ and $1$ .

$b = - 2$

$a = - \frac{b}{4} - \frac{1}{2}$

$c = 1$

$b = - 2$

$a = - \frac{b}{4} - \frac{1}{2}$

$c = 1$

$b = - 2$

$a = - \frac{b}{4} - \frac{1}{2}$

$c = 1$

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

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

$a = - \frac{- 2}{4} - \frac{1}{2}$

$b = - 2$

$c = 1$

Simplify the right side.

Simplify $- \frac{- 2}{4} - \frac{1}{2}$ .

Simplify each term.

Cancel the common factor of $- 2$ and $4$ .

Factor $2$ out of $- 2$ .

$a = - \frac{2 (- 1)}{4} - \frac{1}{2}$

$b = - 2$

$c = 1$

Cancel the common factors.

Factor $2$ out of $4$ .

$a = - \frac{2 \cdot - 1}{2 \cdot 2} - \frac{1}{2}$

$b = - 2$

$c = 1$

Cancel the common factor.

$a = - \frac{2 \cdot - 1}{2 \cdot 2} - \frac{1}{2}$

$b = - 2$

$c = 1$

Rewrite the expression.

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

$b = - 2$

$c = 1$

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

$b = - 2$

$c = 1$

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

$b = - 2$

$c = 1$

Move the negative in front of the fraction.

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

$b = - 2$

$c = 1$

Multiply $- - \frac{1}{2}$ .

Multiply $- 1$ by $- 1$ .

$a = 1 (\frac{1}{2}) - \frac{1}{2}$

$b = - 2$

$c = 1$

Multiply $\frac{1}{2}$ by $1$ .

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

$b = - 2$

$c = 1$

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

$b = - 2$

$c = 1$

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

$b = - 2$

$c = 1$

Combine fractions.

Combine the numerators over the common denominator.

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

$b = - 2$

$c = 1$

Simplify the expression.

Subtract $1$ from $1$ .

$a = \frac{0}{2}$

$b = - 2$

$c = 1$

Divide $0$ by $2$ .

$a = 0$

$b = - 2$

$c = 1$

$a = 0$

$b = - 2$

$c = 1$

$a = 0$

$b = - 2$

$c = 1$

$a = 0$

$b = - 2$

$c = 1$

$a = 0$

$b = - 2$

$c = 1$

$a = 0$

$b = - 2$

$c = 1$

List all of the solutions.

$a = 0, b = - 2, c = 1$

Step 4

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

$y = - 2 x + 1$

Step 5

Find the Equation of the Parabola (0,1) , (2,-3) , (4,-7)

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