Find the Equation of the Parabola (-2,18) , (0,4) , (4,24)

$(- 2, 18)$ , $(0, 4)$ , $(4, 24)$

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.

$18 = a {(- 2)}^{2} + b (- 2) + c, 4 = a {(0)}^{2} + b (0) + c, 24 = a {(4)}^{2} + b (4) + c$

Step 3

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

Solve for $c$ in $4 = a \cdot 0^{2} + c$ .

Rewrite the equation as $a \cdot 0^{2} + c = 4$ .

$a \cdot 0^{2} + c = 4$

$18 = a {(- 2)}^{2} + b (- 2) + c$

$24 = a \cdot 4^{2} + b (4) + c$

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

Simplify each term.

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

$a \cdot 0 + c = 4$

$18 = a {(- 2)}^{2} + b (- 2) + c$

$24 = a \cdot 4^{2} + b (4) + c$

Multiply $a$ by $0$ .

$0 + c = 4$

$18 = a {(- 2)}^{2} + b (- 2) + c$

$24 = a \cdot 4^{2} + b (4) + c$

$0 + c = 4$

$18 = a {(- 2)}^{2} + b (- 2) + c$

$24 = a \cdot 4^{2} + b (4) + c$

Add $0$ and $c$ .

$c = 4$

$18 = a {(- 2)}^{2} + b (- 2) + c$

$24 = a \cdot 4^{2} + b (4) + c$

$c = 4$

$18 = a {(- 2)}^{2} + b (- 2) + c$

$24 = a \cdot 4^{2} + b (4) + c$

$c = 4$

$18 = a {(- 2)}^{2} + b (- 2) + c$

$24 = a \cdot 4^{2} + b (4) + c$

Replace all occurrences of $c$ with $4$ in each equation.

Replace all occurrences of $c$ in $18 = a {(- 2)}^{2} + b (- 2) + c$ with $4$ .

$18 = a {(- 2)}^{2} + b (- 2) + 4$

$c = 4$

$24 = a \cdot 4^{2} + b (4) + c$

Simplify $18 = a {(- 2)}^{2} + b (- 2) + 4$ .

Simplify the left side.

Remove parentheses.

$18 = a {(- 2)}^{2} + b (- 2) + 4$

$c = 4$

$24 = a \cdot 4^{2} + b (4) + c$

$18 = a {(- 2)}^{2} + b (- 2) + 4$

$c = 4$

$24 = a \cdot 4^{2} + b (4) + c$

Simplify the right side.

Simplify each term.

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

$18 = a \cdot 4 + b (- 2) + 4$

$c = 4$

$24 = a \cdot 4^{2} + b (4) + c$

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

$18 = 4 \cdot a + b (- 2) + 4$

$c = 4$

$24 = a \cdot 4^{2} + b (4) + c$

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

$18 = 4 a - 2 b + 4$

$c = 4$

$24 = a \cdot 4^{2} + b (4) + c$

$18 = 4 a - 2 b + 4$

$c = 4$

$24 = a \cdot 4^{2} + b (4) + c$

$18 = 4 a - 2 b + 4$

$c = 4$

$24 = a \cdot 4^{2} + b (4) + c$

$18 = 4 a - 2 b + 4$

$c = 4$

$24 = a \cdot 4^{2} + b (4) + c$

Replace all occurrences of $c$ in $24 = a \cdot 4^{2} + b (4) + c$ with $4$ .

$24 = a \cdot 4^{2} + b (4) + 4$

$18 = 4 a - 2 b + 4$

$c = 4$

Simplify $24 = a \cdot 4^{2} + b (4) + 4$ .

Simplify the left side.

Remove parentheses.

$24 = a \cdot 4^{2} + b (4) + 4$

$18 = 4 a - 2 b + 4$

$c = 4$

$24 = a \cdot 4^{2} + b (4) + 4$

$18 = 4 a - 2 b + 4$

$c = 4$

Simplify the right side.

Simplify each term.

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

$24 = a \cdot 16 + b (4) + 4$

$18 = 4 a - 2 b + 4$

$c = 4$

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

$24 = 16 \cdot a + b (4) + 4$

$18 = 4 a - 2 b + 4$

$c = 4$

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

$24 = 16 a + 4 b + 4$

$18 = 4 a - 2 b + 4$

$c = 4$

$24 = 16 a + 4 b + 4$

$18 = 4 a - 2 b + 4$

$c = 4$

$24 = 16 a + 4 b + 4$

$18 = 4 a - 2 b + 4$

$c = 4$

$24 = 16 a + 4 b + 4$

$18 = 4 a - 2 b + 4$

$c = 4$

$24 = 16 a + 4 b + 4$

$18 = 4 a - 2 b + 4$

$c = 4$

Solve for $a$ in $24 = 16 a + 4 b + 4$ .

Rewrite the equation as $16 a + 4 b + 4 = 24$ .

$16 a + 4 b + 4 = 24$

$18 = 4 a - 2 b + 4$

$c = 4$

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

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

$16 a + 4 = 24 - 4 b$

$18 = 4 a - 2 b + 4$

$c = 4$

Subtract $4$ from both sides of the equation.

$16 a = 24 - 4 b - 4$

$18 = 4 a - 2 b + 4$

$c = 4$

Subtract $4$ from $24$ .

$16 a = - 4 b + 20$

$18 = 4 a - 2 b + 4$

$c = 4$

$16 a = - 4 b + 20$

$18 = 4 a - 2 b + 4$

$c = 4$

Divide each term in $16 a = - 4 b + 20$ by $16$ and simplify.

Divide each term in $16 a = - 4 b + 20$ by $16$ .

$\frac{16 a}{16} = \frac{- 4 b}{16} + \frac{20}{16}$

$18 = 4 a - 2 b + 4$

$c = 4$

Simplify the left side.

Cancel the common factor of $16$ .

Cancel the common factor.

$\frac{16 a}{16} = \frac{- 4 b}{16} + \frac{20}{16}$

$18 = 4 a - 2 b + 4$

$c = 4$

Divide $a$ by $1$ .

$a = \frac{- 4 b}{16} + \frac{20}{16}$

$18 = 4 a - 2 b + 4$

$c = 4$

$a = \frac{- 4 b}{16} + \frac{20}{16}$

$18 = 4 a - 2 b + 4$

$c = 4$

$a = \frac{- 4 b}{16} + \frac{20}{16}$

$18 = 4 a - 2 b + 4$

$c = 4$

Simplify the right side.

Simplify each term.

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

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

$a = \frac{4 (- b)}{16} + \frac{20}{16}$

$18 = 4 a - 2 b + 4$

$c = 4$

Cancel the common factors.

Factor $4$ out of $16$ .

$a = \frac{4 (- b)}{4 (4)} + \frac{20}{16}$

$18 = 4 a - 2 b + 4$

$c = 4$

Cancel the common factor.

$a = \frac{4 (- b)}{4 \cdot 4} + \frac{20}{16}$

$18 = 4 a - 2 b + 4$

$c = 4$

Rewrite the expression.

$a = \frac{- b}{4} + \frac{20}{16}$

$18 = 4 a - 2 b + 4$

$c = 4$

$a = \frac{- b}{4} + \frac{20}{16}$

$18 = 4 a - 2 b + 4$

$c = 4$

$a = \frac{- b}{4} + \frac{20}{16}$

$18 = 4 a - 2 b + 4$

$c = 4$

Move the negative in front of the fraction.

$a = - \frac{b}{4} + \frac{20}{16}$

$18 = 4 a - 2 b + 4$

$c = 4$

Cancel the common factor of $20$ and $16$ .

Factor $4$ out of $20$ .

$a = - \frac{b}{4} + \frac{4 (5)}{16}$

$18 = 4 a - 2 b + 4$

$c = 4$

Cancel the common factors.

Factor $4$ out of $16$ .

$a = - \frac{b}{4} + \frac{4 \cdot 5}{4 \cdot 4}$

$18 = 4 a - 2 b + 4$

$c = 4$

Cancel the common factor.

$a = - \frac{b}{4} + \frac{4 \cdot 5}{4 \cdot 4}$

$18 = 4 a - 2 b + 4$

$c = 4$

Rewrite the expression.

$a = - \frac{b}{4} + \frac{5}{4}$

$18 = 4 a - 2 b + 4$

$c = 4$

$a = - \frac{b}{4} + \frac{5}{4}$

$18 = 4 a - 2 b + 4$

$c = 4$

$a = - \frac{b}{4} + \frac{5}{4}$

$18 = 4 a - 2 b + 4$

$c = 4$

$a = - \frac{b}{4} + \frac{5}{4}$

$18 = 4 a - 2 b + 4$

$c = 4$

$a = - \frac{b}{4} + \frac{5}{4}$

$18 = 4 a - 2 b + 4$

$c = 4$

$a = - \frac{b}{4} + \frac{5}{4}$

$18 = 4 a - 2 b + 4$

$c = 4$

$a = - \frac{b}{4} + \frac{5}{4}$

$18 = 4 a - 2 b + 4$

$c = 4$

Replace all occurrences of $a$ with $- \frac{b}{4} + \frac{5}{4}$ in each equation.

Replace all occurrences of $a$ in $18 = 4 a - 2 b + 4$ with $- \frac{b}{4} + \frac{5}{4}$ .

$18 = 4 (- \frac{b}{4} + \frac{5}{4}) - 2 b + 4$

$a = - \frac{b}{4} + \frac{5}{4}$

$c = 4$

Simplify the right side.

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

Simplify each term.

Apply the distributive property.

$18 = 4 (- \frac{b}{4}) + 4 (\frac{5}{4}) - 2 b + 4$

$a = - \frac{b}{4} + \frac{5}{4}$

$c = 4$

Cancel the common factor of $4$ .

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

$18 = 4 (\frac{- b}{4}) + 4 (\frac{5}{4}) - 2 b + 4$

$a = - \frac{b}{4} + \frac{5}{4}$

$c = 4$

Cancel the common factor.

$18 = 4 (\frac{- b}{4}) + 4 (\frac{5}{4}) - 2 b + 4$

$a = - \frac{b}{4} + \frac{5}{4}$

$c = 4$

Rewrite the expression.

$18 = - b + 4 (\frac{5}{4}) - 2 b + 4$

$a = - \frac{b}{4} + \frac{5}{4}$

$c = 4$

$18 = - b + 4 (\frac{5}{4}) - 2 b + 4$

$a = - \frac{b}{4} + \frac{5}{4}$

$c = 4$

Cancel the common factor of $4$ .

Cancel the common factor.

$18 = - b + 4 (\frac{5}{4}) - 2 b + 4$

$a = - \frac{b}{4} + \frac{5}{4}$

$c = 4$

Rewrite the expression.

$18 = - b + 5 - 2 b + 4$

$a = - \frac{b}{4} + \frac{5}{4}$

$c = 4$

$18 = - b + 5 - 2 b + 4$

$a = - \frac{b}{4} + \frac{5}{4}$

$c = 4$

$18 = - b + 5 - 2 b + 4$

$a = - \frac{b}{4} + \frac{5}{4}$

$c = 4$

Simplify by adding terms.

Subtract $2 b$ from $- b$ .

$18 = - 3 b + 5 + 4$

$a = - \frac{b}{4} + \frac{5}{4}$

$c = 4$

Add $5$ and $4$ .

$18 = - 3 b + 9$

$a = - \frac{b}{4} + \frac{5}{4}$

$c = 4$

$18 = - 3 b + 9$

$a = - \frac{b}{4} + \frac{5}{4}$

$c = 4$

$18 = - 3 b + 9$

$a = - \frac{b}{4} + \frac{5}{4}$

$c = 4$

$18 = - 3 b + 9$

$a = - \frac{b}{4} + \frac{5}{4}$

$c = 4$

$18 = - 3 b + 9$

$a = - \frac{b}{4} + \frac{5}{4}$

$c = 4$

Solve for $b$ in $18 = - 3 b + 9$ .

Rewrite the equation as $- 3 b + 9 = 18$ .

$- 3 b + 9 = 18$

$a = - \frac{b}{4} + \frac{5}{4}$

$c = 4$

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

Subtract $9$ from both sides of the equation.

$- 3 b = 18 - 9$

$a = - \frac{b}{4} + \frac{5}{4}$

$c = 4$

Subtract $9$ from $18$ .

$- 3 b = 9$

$a = - \frac{b}{4} + \frac{5}{4}$

$c = 4$

$- 3 b = 9$

$a = - \frac{b}{4} + \frac{5}{4}$

$c = 4$

Divide each term in $- 3 b = 9$ by $- 3$ and simplify.

Divide each term in $- 3 b = 9$ by $- 3$ .

$\frac{- 3 b}{- 3} = \frac{9}{- 3}$

$a = - \frac{b}{4} + \frac{5}{4}$

$c = 4$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 b}{- 3} = \frac{9}{- 3}$

$a = - \frac{b}{4} + \frac{5}{4}$

$c = 4$

Divide $b$ by $1$ .

$b = \frac{9}{- 3}$

$a = - \frac{b}{4} + \frac{5}{4}$

$c = 4$

$b = \frac{9}{- 3}$

$a = - \frac{b}{4} + \frac{5}{4}$

$c = 4$

$b = \frac{9}{- 3}$

$a = - \frac{b}{4} + \frac{5}{4}$

$c = 4$

Simplify the right side.

Divide $9$ by $- 3$ .

$b = - 3$

$a = - \frac{b}{4} + \frac{5}{4}$

$c = 4$

$b = - 3$

$a = - \frac{b}{4} + \frac{5}{4}$

$c = 4$

$b = - 3$

$a = - \frac{b}{4} + \frac{5}{4}$

$c = 4$

$b = - 3$

$a = - \frac{b}{4} + \frac{5}{4}$

$c = 4$

Replace all occurrences of $b$ with $- 3$ in each equation.

Replace all occurrences of $b$ in $a = - \frac{b}{4} + \frac{5}{4}$ with $- 3$ .

$a = - \frac{- 3}{4} + \frac{5}{4}$

$b = - 3$

$c = 4$

Simplify the right side.

Simplify $- \frac{- 3}{4} + \frac{5}{4}$ .

Combine the numerators over the common denominator.

$a = \frac{3 + 5}{4}$

$b = - 3$

$c = 4$

Simplify the expression.

Add $3$ and $5$ .

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

$b = - 3$

$c = 4$

Divide $8$ by $4$ .

$a = 2$

$b = - 3$

$c = 4$

$a = 2$

$b = - 3$

$c = 4$

$a = 2$

$b = - 3$

$c = 4$

$a = 2$

$b = - 3$

$c = 4$

$a = 2$

$b = - 3$

$c = 4$

List all of the solutions.

$a = 2, b = - 3, c = 4$

Step 4

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

$y = 2 x^{2} - 3 x + 4$

Step 5

Find the Equation of the Parabola (-2,18) , (0,4) , (4,24)

Related Questions

Support LCMGCF.com by disabling your adblocker.