Find the Equation of the Parabola (1,20) , (3,15) , (5,10)

$(1, 20)$ , $(3, 15)$ , $(5, 10)$

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.

$20 = a {(1)}^{2} + b (1) + c$

$15 = a {(3)}^{2} + b (3) + c$

$10 = a {(5)}^{2} + b (5) + c$

Step 3

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

Solve for $a$ in $20 = a + b + c$ .

Rewrite the equation as $a + b + c = 20$ .

$a + b + c = 20$

$15 = a \cdot 3^{2} + b (3) + c$

$10 = a \cdot 5^{2} + b (5) + c$

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

Subtract $b$ from both sides of the equation.

$a + c = 20 - b$

$15 = a \cdot 3^{2} + b (3) + c$

$10 = a \cdot 5^{2} + b (5) + c$

Subtract $c$ from both sides of the equation.

$a = 20 - b - c$

$15 = a \cdot 3^{2} + b (3) + c$

$10 = a \cdot 5^{2} + b (5) + c$

$a = 20 - b - c$

$15 = a \cdot 3^{2} + b (3) + c$

$10 = a \cdot 5^{2} + b (5) + c$

$a = 20 - b - c$

$15 = a \cdot 3^{2} + b (3) + c$

$10 = a \cdot 5^{2} + b (5) + c$

Replace all occurrences of $a$ with $20 - b - c$ in each equation.

Replace all occurrences of $a$ in $15 = a \cdot 3^{2} + b (3) + c$ with $20 - b - c$ .

$15 = (20 - b - c) \cdot 3^{2} + b (3) + c$

$a = 20 - b - c$

$10 = a \cdot 5^{2} + b (5) + c$

Simplify the right side.

Simplify $(20 - b - c) \cdot 3^{2} + b (3) + c$ .

Simplify each term.

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

$15 = (20 - b - c) \cdot 9 + b (3) + c$

$a = 20 - b - c$

$10 = a \cdot 5^{2} + b (5) + c$

Apply the distributive property.

$15 = 20 \cdot 9 - b \cdot 9 - c \cdot 9 + b (3) + c$

$a = 20 - b - c$

$10 = a \cdot 5^{2} + b (5) + c$

Simplify.

Multiply $20$ by $9$ .

$15 = 180 - b \cdot 9 - c \cdot 9 + b (3) + c$

$a = 20 - b - c$

$10 = a \cdot 5^{2} + b (5) + c$

Multiply $9$ by $- 1$ .

$15 = 180 - 9 b - c \cdot 9 + b (3) + c$

$a = 20 - b - c$

$10 = a \cdot 5^{2} + b (5) + c$

Multiply $9$ by $- 1$ .

$15 = 180 - 9 b - 9 c + b (3) + c$

$a = 20 - b - c$

$10 = a \cdot 5^{2} + b (5) + c$

$15 = 180 - 9 b - 9 c + b (3) + c$

$a = 20 - b - c$

$10 = a \cdot 5^{2} + b (5) + c$

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

$15 = 180 - 9 b - 9 c + 3 b + c$

$a = 20 - b - c$

$10 = a \cdot 5^{2} + b (5) + c$

$15 = 180 - 9 b - 9 c + 3 b + c$

$a = 20 - b - c$

$10 = a \cdot 5^{2} + b (5) + c$

Simplify by adding terms.

Add $- 9 b$ and $3 b$ .

$15 = 180 - 6 b - 9 c + c$

$a = 20 - b - c$

$10 = a \cdot 5^{2} + b (5) + c$

Add $- 9 c$ and $c$ .

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

$10 = a \cdot 5^{2} + b (5) + c$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

$10 = a \cdot 5^{2} + b (5) + c$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

$10 = a \cdot 5^{2} + b (5) + c$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

$10 = a \cdot 5^{2} + b (5) + c$

Replace all occurrences of $a$ in $10 = a \cdot 5^{2} + b (5) + c$ with $20 - b - c$ .

$10 = (20 - b - c) \cdot 5^{2} + b (5) + c$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

Simplify the right side.

Simplify $(20 - b - c) \cdot 5^{2} + b (5) + c$ .

Simplify each term.

Raise $5$ to the power of $2$ .

$10 = (20 - b - c) \cdot 25 + b (5) + c$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

Apply the distributive property.

$10 = 20 \cdot 25 - b \cdot 25 - c \cdot 25 + b (5) + c$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

Simplify.

Multiply $20$ by $25$ .

$10 = 500 - b \cdot 25 - c \cdot 25 + b (5) + c$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

Multiply $25$ by $- 1$ .

$10 = 500 - 25 b - c \cdot 25 + b (5) + c$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

Multiply $25$ by $- 1$ .

$10 = 500 - 25 b - 25 c + b (5) + c$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

$10 = 500 - 25 b - 25 c + b (5) + c$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

Move $5$ to the left of $b$ .

$10 = 500 - 25 b - 25 c + 5 b + c$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

$10 = 500 - 25 b - 25 c + 5 b + c$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

Simplify by adding terms.

Add $- 25 b$ and $5 b$ .

$10 = 500 - 20 b - 25 c + c$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

Add $- 25 c$ and $c$ .

$10 = 500 - 20 b - 24 c$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

$10 = 500 - 20 b - 24 c$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

$10 = 500 - 20 b - 24 c$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

$10 = 500 - 20 b - 24 c$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

$10 = 500 - 20 b - 24 c$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

Solve for $b$ in $10 = 500 - 20 b - 24 c$ .

Rewrite the equation as $500 - 20 b - 24 c = 10$ .

$500 - 20 b - 24 c = 10$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

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

Subtract $500$ from both sides of the equation.

$- 20 b - 24 c = 10 - 500$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

Add $24 c$ to both sides of the equation.

$- 20 b = 10 - 500 + 24 c$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

Subtract $500$ from $10$ .

$- 20 b = - 490 + 24 c$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

$- 20 b = - 490 + 24 c$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

Divide each term in $- 20 b = - 490 + 24 c$ by $- 20$ and simplify.

Divide each term in $- 20 b = - 490 + 24 c$ by $- 20$ .

$\frac{- 20 b}{- 20} = \frac{- 490}{- 20} + \frac{24 c}{- 20}$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

Simplify the left side.

Cancel the common factor of $- 20$ .

Cancel the common factor.

$\frac{- 20 b}{- 20} = \frac{- 490}{- 20} + \frac{24 c}{- 20}$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

Divide $b$ by $1$ .

$b = \frac{- 490}{- 20} + \frac{24 c}{- 20}$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

$b = \frac{- 490}{- 20} + \frac{24 c}{- 20}$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

$b = \frac{- 490}{- 20} + \frac{24 c}{- 20}$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

Simplify the right side.

Simplify each term.

Cancel the common factor of $- 490$ and $- 20$ .

Factor $- 10$ out of $- 490$ .

$b = \frac{- 10 \cdot 49}{- 20} + \frac{24 c}{- 20}$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

Cancel the common factors.

Factor $- 10$ out of $- 20$ .

$b = \frac{- 10 \cdot 49}{- 10 \cdot 2} + \frac{24 c}{- 20}$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

Cancel the common factor.

$b = \frac{- 10 \cdot 49}{- 10 \cdot 2} + \frac{24 c}{- 20}$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

Rewrite the expression.

$b = \frac{49}{2} + \frac{24 c}{- 20}$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

$b = \frac{49}{2} + \frac{24 c}{- 20}$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

$b = \frac{49}{2} + \frac{24 c}{- 20}$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

Cancel the common factor of $24$ and $- 20$ .

Factor $4$ out of $24 c$ .

$b = \frac{49}{2} + \frac{4 (6 c)}{- 20}$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

Cancel the common factors.

Factor $4$ out of $- 20$ .

$b = \frac{49}{2} + \frac{4 (6 c)}{4 (- 5)}$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

Cancel the common factor.

$b = \frac{49}{2} + \frac{4 (6 c)}{4 \cdot - 5}$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

Rewrite the expression.

$b = \frac{49}{2} + \frac{6 c}{- 5}$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

$b = \frac{49}{2} + \frac{6 c}{- 5}$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

$b = \frac{49}{2} + \frac{6 c}{- 5}$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

Move the negative in front of the fraction.

$b = \frac{49}{2} - \frac{6 c}{5}$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

$b = \frac{49}{2} - \frac{6 c}{5}$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

$b = \frac{49}{2} - \frac{6 c}{5}$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

$b = \frac{49}{2} - \frac{6 c}{5}$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

$b = \frac{49}{2} - \frac{6 c}{5}$

$15 = 180 - 6 b - 8 c$

$a = 20 - b - c$

Replace all occurrences of $b$ with $\frac{49}{2} - \frac{6 c}{5}$ in each equation.

Replace all occurrences of $b$ in $15 = 180 - 6 b - 8 c$ with $\frac{49}{2} - \frac{6 c}{5}$ .

$15 = 180 - 6 (\frac{49}{2} - \frac{6 c}{5}) - 8 c$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 20 - b - c$

Simplify the right side.

Simplify $180 - 6 (\frac{49}{2} - \frac{6 c}{5}) - 8 c$ .

Simplify each term.

Apply the distributive property.

$15 = 180 - 6 (\frac{49}{2}) - 6 (- \frac{6 c}{5}) - 8 c$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 20 - b - c$

Cancel the common factor of $2$ .

Factor $2$ out of $- 6$ .

$15 = 180 + 2 (- 3) (\frac{49}{2}) - 6 (- \frac{6 c}{5}) - 8 c$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 20 - b - c$

Cancel the common factor.

$15 = 180 + 2 \cdot (- 3 (\frac{49}{2})) - 6 (- \frac{6 c}{5}) - 8 c$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 20 - b - c$

Rewrite the expression.

$15 = 180 - 3 \cdot 49 - 6 (- \frac{6 c}{5}) - 8 c$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 20 - b - c$

$15 = 180 - 3 \cdot 49 - 6 (- \frac{6 c}{5}) - 8 c$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 20 - b - c$

Multiply $- 3$ by $49$ .

$15 = 180 - 147 - 6 (- \frac{6 c}{5}) - 8 c$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 20 - b - c$

Multiply $- 6 (- \frac{6 c}{5})$ .

Multiply $- 1$ by $- 6$ .

$15 = 180 - 147 + 6 (\frac{6 c}{5}) - 8 c$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 20 - b - c$

Combine $6$ and $\frac{6 c}{5}$ .

$15 = 180 - 147 + \frac{6 (6 c)}{5} - 8 c$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 20 - b - c$

Multiply $6$ by $6$ .

$15 = 180 - 147 + \frac{36 c}{5} - 8 c$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 20 - b - c$

$15 = 180 - 147 + \frac{36 c}{5} - 8 c$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 20 - b - c$

$15 = 180 - 147 + \frac{36 c}{5} - 8 c$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 20 - b - c$

Subtract $147$ from $180$ .

$15 = 33 + \frac{36 c}{5} - 8 c$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 20 - b - c$

To write $- 8 c$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$15 = 33 + \frac{36 c}{5} - 8 c \cdot \frac{5}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 20 - b - c$

Simplify terms.

Combine $- 8 c$ and $\frac{5}{5}$ .

$15 = 33 + \frac{36 c}{5} + \frac{- 8 c \cdot 5}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 20 - b - c$

Combine the numerators over the common denominator.

$15 = 33 + \frac{36 c - 8 c \cdot 5}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 20 - b - c$

$15 = 33 + \frac{36 c - 8 c \cdot 5}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 20 - b - c$

Simplify each term.

Simplify the numerator.

Factor $4 c$ out of $36 c - 8 c \cdot 5$ .

Factor $4 c$ out of $36 c$ .

$15 = 33 + \frac{4 c (9) - 8 c \cdot 5}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 20 - b - c$

Factor $4 c$ out of $- 8 c \cdot 5$ .

$15 = 33 + \frac{4 c (9) + 4 c (- 2 \cdot 5)}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 20 - b - c$

Factor $4 c$ out of $4 c (9) + 4 c (- 2 \cdot 5)$ .

$15 = 33 + \frac{4 c (9 - 2 \cdot 5)}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 20 - b - c$

$15 = 33 + \frac{4 c (9 - 2 \cdot 5)}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 20 - b - c$

Multiply $- 2$ by $5$ .

$15 = 33 + \frac{4 c (9 - 10)}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 20 - b - c$

Subtract $10$ from $9$ .

$15 = 33 + \frac{4 c \cdot - 1}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 20 - b - c$

Multiply $- 1$ by $4$ .

$15 = 33 + \frac{- 4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 20 - b - c$

$15 = 33 + \frac{- 4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 20 - b - c$

Move the negative in front of the fraction.

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 20 - b - c$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 20 - b - c$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 20 - b - c$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 20 - b - c$

Replace all occurrences of $b$ in $a = 20 - b - c$ with $\frac{49}{2} - \frac{6 c}{5}$ .

$a = 20 - (\frac{49}{2} - \frac{6 c}{5}) - c$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Simplify the right side.

Simplify $20 - (\frac{49}{2} - \frac{6 c}{5}) - c$ .

Simplify each term.

Apply the distributive property.

$a = 20 - \frac{49}{2} + \frac{6 c}{5} - c$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Multiply $- - \frac{6 c}{5}$ .

Multiply $- 1$ by $- 1$ .

$a = 20 - \frac{49}{2} + 1 (\frac{6 c}{5}) - c$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Multiply $\frac{6 c}{5}$ by $1$ .

$a = 20 - \frac{49}{2} + \frac{6 c}{5} - c$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 20 - \frac{49}{2} + \frac{6 c}{5} - c$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 20 - \frac{49}{2} + \frac{6 c}{5} - c$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

To write $20$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$a = 20 \cdot \frac{2}{2} - \frac{49}{2} + \frac{6 c}{5} - c$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Combine $20$ and $\frac{2}{2}$ .

$a = \frac{20 \cdot 2}{2} - \frac{49}{2} + \frac{6 c}{5} - c$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Combine the numerators over the common denominator.

$a = \frac{20 \cdot 2 - 49}{2} + \frac{6 c}{5} - c$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Simplify the numerator.

Multiply $20$ by $2$ .

$a = \frac{40 - 49}{2} + \frac{6 c}{5} - c$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Subtract $49$ from $40$ .

$a = \frac{- 9}{2} + \frac{6 c}{5} - c$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = \frac{- 9}{2} + \frac{6 c}{5} - c$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Move the negative in front of the fraction.

$a = - \frac{9}{2} + \frac{6 c}{5} - c$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

To write $- c$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$a = - \frac{9}{2} + \frac{6 c}{5} - c \cdot \frac{5}{5}$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Simplify terms.

Combine $- c$ and $\frac{5}{5}$ .

$a = - \frac{9}{2} + \frac{6 c}{5} + \frac{- c \cdot 5}{5}$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Combine the numerators over the common denominator.

$a = - \frac{9}{2} + \frac{6 c - c \cdot 5}{5}$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = - \frac{9}{2} + \frac{6 c - c \cdot 5}{5}$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Simplify each term.

Simplify the numerator.

Factor $c$ out of $6 c - c \cdot 5$ .

Factor $c$ out of $6 c$ .

$a = - \frac{9}{2} + \frac{c \cdot 6 - c \cdot 5}{5}$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Factor $c$ out of $- c \cdot 5$ .

$a = - \frac{9}{2} + \frac{c \cdot 6 + c (- 1 \cdot 5)}{5}$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Factor $c$ out of $c \cdot 6 + c (- 1 \cdot 5)$ .

$a = - \frac{9}{2} + \frac{c (6 - 1 \cdot 5)}{5}$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = - \frac{9}{2} + \frac{c (6 - 1 \cdot 5)}{5}$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Multiply $- 1$ by $5$ .

$a = - \frac{9}{2} + \frac{c (6 - 5)}{5}$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Subtract $5$ from $6$ .

$a = - \frac{9}{2} + \frac{c \cdot 1}{5}$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = - \frac{9}{2} + \frac{c \cdot 1}{5}$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Multiply $c$ by $1$ .

$a = - \frac{9}{2} + \frac{c}{5}$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = - \frac{9}{2} + \frac{c}{5}$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = - \frac{9}{2} + \frac{c}{5}$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = - \frac{9}{2} + \frac{c}{5}$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = - \frac{9}{2} + \frac{c}{5}$

$15 = 33 - \frac{4 c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Solve for $c$ in $15 = 33 - \frac{4 c}{5}$ .

Rewrite the equation as $33 - \frac{4 c}{5} = 15$ .

$33 - \frac{4 c}{5} = 15$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

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

Subtract $33$ from both sides of the equation.

$- \frac{4 c}{5} = 15 - 33$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Subtract $33$ from $15$ .

$- \frac{4 c}{5} = - 18$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$- \frac{4 c}{5} = - 18$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Multiply both sides of the equation by $- \frac{5}{4}$ .

$- \frac{5}{4} \cdot (- \frac{4 c}{5}) = - \frac{5}{4} \cdot - 18$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Simplify both sides of the equation.

Simplify the left side.

Simplify $- \frac{5}{4} (- \frac{4 c}{5})$ .

Cancel the common factor of $5$ .

Move the leading negative in $- \frac{5}{4}$ into the numerator.

$\frac{- 5}{4} \cdot (- \frac{4 c}{5}) = - \frac{5}{4} \cdot - 18$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Move the leading negative in $- \frac{4 c}{5}$ into the numerator.

$\frac{- 5}{4} \cdot \frac{- 4 c}{5} = - \frac{5}{4} \cdot - 18$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Factor $5$ out of $- 5$ .

$\frac{5 (- 1)}{4} \cdot \frac{- 4 c}{5} = - \frac{5}{4} \cdot - 18$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Cancel the common factor.

$\frac{5 \cdot - 1}{4} \cdot \frac{- 4 c}{5} = - \frac{5}{4} \cdot - 18$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Rewrite the expression.

$\frac{- 1}{4} \cdot (- 4 c) = - \frac{5}{4} \cdot - 18$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$\frac{- 1}{4} \cdot (- 4 c) = - \frac{5}{4} \cdot - 18$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Cancel the common factor of $4$ .

Factor $4$ out of $- 4 c$ .

$\frac{- 1}{4} \cdot (4 (- c)) = - \frac{5}{4} \cdot - 18$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Cancel the common factor.

$\frac{- 1}{4} \cdot (4 (- c)) = - \frac{5}{4} \cdot - 18$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Rewrite the expression.

$c = - \frac{5}{4} \cdot - 18$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$c = - \frac{5}{4} \cdot - 18$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Multiply.

Multiply $- 1$ by $- 1$ .

$1 c = - \frac{5}{4} \cdot - 18$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Multiply $c$ by $1$ .

$c = - \frac{5}{4} \cdot - 18$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$c = - \frac{5}{4} \cdot - 18$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$c = - \frac{5}{4} \cdot - 18$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$c = - \frac{5}{4} \cdot - 18$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Simplify the right side.

Simplify $- \frac{5}{4} \cdot - 18$ .

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{5}{4}$ into the numerator.

$c = \frac{- 5}{4} \cdot - 18$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Factor $2$ out of $4$ .

$c = \frac{- 5}{2 (2)} \cdot - 18$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Factor $2$ out of $- 18$ .

$c = \frac{- 5}{2 \cdot 2} \cdot (2 \cdot - 9)$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Cancel the common factor.

$c = \frac{- 5}{2 \cdot 2} \cdot (2 \cdot - 9)$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Rewrite the expression.

$c = \frac{- 5}{2} \cdot - 9$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$c = \frac{- 5}{2} \cdot - 9$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Combine $\frac{- 5}{2}$ and $- 9$ .

$c = \frac{- 5 \cdot - 9}{2}$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Multiply $- 5$ by $- 9$ .

$c = \frac{45}{2}$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$c = \frac{45}{2}$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$c = \frac{45}{2}$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$c = \frac{45}{2}$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$c = \frac{45}{2}$

$a = - \frac{9}{2} + \frac{c}{5}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Replace all occurrences of $c$ with $\frac{45}{2}$ in each equation.

Replace all occurrences of $c$ in $a = - \frac{9}{2} + \frac{c}{5}$ with $\frac{45}{2}$ .

$a = - \frac{9}{2} + \frac{\frac{45}{2}}{5}$

$c = \frac{45}{2}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Simplify the right side.

Simplify $- \frac{9}{2} + \frac{\frac{45}{2}}{5}$ .

Simplify each term.

Multiply the numerator by the reciprocal of the denominator.

$a = - \frac{9}{2} + \frac{45}{2} \cdot \frac{1}{5}$

$c = \frac{45}{2}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Cancel the common factor of $5$ .

Factor $5$ out of $45$ .

$a = - \frac{9}{2} + \frac{5 (9)}{2} \cdot \frac{1}{5}$

$c = \frac{45}{2}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Cancel the common factor.

$a = - \frac{9}{2} + \frac{5 \cdot 9}{2} \cdot \frac{1}{5}$

$c = \frac{45}{2}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Rewrite the expression.

$a = - \frac{9}{2} + \frac{9}{2}$

$c = \frac{45}{2}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = - \frac{9}{2} + \frac{9}{2}$

$c = \frac{45}{2}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = - \frac{9}{2} + \frac{9}{2}$

$c = \frac{45}{2}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Combine fractions.

Combine the numerators over the common denominator.

$a = \frac{- 9 + 9}{2}$

$c = \frac{45}{2}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Simplify the expression.

Add $- 9$ and $9$ .

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

$c = \frac{45}{2}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Divide $0$ by $2$ .

$a = 0$

$c = \frac{45}{2}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 0$

$c = \frac{45}{2}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 0$

$c = \frac{45}{2}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 0$

$c = \frac{45}{2}$

$b = \frac{49}{2} - \frac{6 c}{5}$

$a = 0$

$c = \frac{45}{2}$

$b = \frac{49}{2} - \frac{6 c}{5}$

Replace all occurrences of $c$ in $b = \frac{49}{2} - \frac{6 c}{5}$ with $\frac{45}{2}$ .

$b = \frac{49}{2} - \frac{6 (\frac{45}{2})}{5}$

$a = 0$

$c = \frac{45}{2}$

Simplify the right side.

Simplify $\frac{49}{2} - \frac{6 (\frac{45}{2})}{5}$ .

Simplify each term.

Combine $6$ and $\frac{45}{2}$ .

$b = \frac{49}{2} - \frac{\frac{6 \cdot 45}{2}}{5}$

$a = 0$

$c = \frac{45}{2}$

Multiply $6$ by $45$ .

$b = \frac{49}{2} - \frac{\frac{270}{2}}{5}$

$a = 0$

$c = \frac{45}{2}$

Divide $270$ by $2$ .

$b = \frac{49}{2} - \frac{135}{5}$

$a = 0$

$c = \frac{45}{2}$

Divide $135$ by $5$ .

$b = \frac{49}{2} - 1 \cdot 27$

$a = 0$

$c = \frac{45}{2}$

Multiply $- 1$ by $27$ .

$b = \frac{49}{2} - 27$

$a = 0$

$c = \frac{45}{2}$

$b = \frac{49}{2} - 27$

$a = 0$

$c = \frac{45}{2}$

To write $- 27$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$b = \frac{49}{2} - 27 \cdot \frac{2}{2}$

$a = 0$

$c = \frac{45}{2}$

Combine $- 27$ and $\frac{2}{2}$ .

$b = \frac{49}{2} + \frac{- 27 \cdot 2}{2}$

$a = 0$

$c = \frac{45}{2}$

Combine the numerators over the common denominator.

$b = \frac{49 - 27 \cdot 2}{2}$

$a = 0$

$c = \frac{45}{2}$

Simplify the numerator.

Multiply $- 27$ by $2$ .

$b = \frac{49 - 54}{2}$

$a = 0$

$c = \frac{45}{2}$

Subtract $54$ from $49$ .

$b = \frac{- 5}{2}$

$a = 0$

$c = \frac{45}{2}$

$b = \frac{- 5}{2}$

$a = 0$

$c = \frac{45}{2}$

Move the negative in front of the fraction.

$b = - \frac{5}{2}$

$a = 0$

$c = \frac{45}{2}$

$b = - \frac{5}{2}$

$a = 0$

$c = \frac{45}{2}$

$b = - \frac{5}{2}$

$a = 0$

$c = \frac{45}{2}$

$b = - \frac{5}{2}$

$a = 0$

$c = \frac{45}{2}$

List all of the solutions.

$b = - \frac{5}{2}, a = 0, c = \frac{45}{2}$

Step 4

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

$y = - \frac{5 x}{2} + \frac{45}{2}$

Step 5

Find the Equation of the Parabola (1,20) , (3,15) , (5,10)

Related Questions

Support LCMGCF.com by disabling your adblocker.