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

$(1, 0)$ , $(\frac{1}{2}, 0)$ , $(- \frac{1}{3}, 0)$

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.

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

Step 3

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

Solve for $a$ in $0 = a + b + c$ .

Rewrite the equation as $a + b + c = 0$ .

$a + b + c = 0$

$0 = a {(\frac{1}{2})}^{2} + \frac{b}{2} + c$

$0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

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

Subtract $b$ from both sides of the equation.

$a + c = - b$

$0 = a {(\frac{1}{2})}^{2} + \frac{b}{2} + c$

$0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

Subtract $c$ from both sides of the equation.

$a = - b - c$

$0 = a {(\frac{1}{2})}^{2} + \frac{b}{2} + c$

$0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

$a = - b - c$

$0 = a {(\frac{1}{2})}^{2} + \frac{b}{2} + c$

$0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

$a = - b - c$

$0 = a {(\frac{1}{2})}^{2} + \frac{b}{2} + c$

$0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

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

Replace all occurrences of $a$ in $0 = a {(\frac{1}{2})}^{2} + \frac{b}{2} + c$ with $- b - c$ .

$0 = (- b - c) {(\frac{1}{2})}^{2} + \frac{b}{2} + c$

$a = - b - c$

$0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

Simplify the right side.

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

Simplify each term.

Apply the product rule to $\frac{1}{2}$ .

$0 = (- b - c) (\frac{1^{2}}{2^{2}}) + \frac{b}{2} + c$

$a = - b - c$

$0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

One to any power is one.

$0 = (- b - c) (\frac{1}{2^{2}}) + \frac{b}{2} + c$

$a = - b - c$

$0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

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

$0 = (- b - c) (\frac{1}{4}) + \frac{b}{2} + c$

$a = - b - c$

$0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

Apply the distributive property.

$0 = - b \frac{1}{4} - c \frac{1}{4} + \frac{b}{2} + c$

$a = - b - c$

$0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

Combine $\frac{1}{4}$ and $b$ .

$0 = - \frac{b}{4} - c \frac{1}{4} + \frac{b}{2} + c$

$a = - b - c$

$0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

Combine $\frac{1}{4}$ and $c$ .

$0 = - \frac{b}{4} - \frac{c}{4} + \frac{b}{2} + c$

$a = - b - c$

$0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

$0 = - \frac{b}{4} - \frac{c}{4} + \frac{b}{2} + c$

$a = - b - c$

$0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

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

$0 = - \frac{c}{4} - \frac{b}{4} + \frac{b}{2} \cdot \frac{2}{2} + c$

$a = - b - c$

$0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

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

$0 = - \frac{c}{4} - \frac{b}{4} + \frac{b \cdot 2}{2 \cdot 2} + c$

$a = - b - c$

$0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

Multiply $2$ by $2$ .

$0 = - \frac{c}{4} - \frac{b}{4} + \frac{b \cdot 2}{4} + c$

$a = - b - c$

$0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

$0 = - \frac{c}{4} - \frac{b}{4} + \frac{b \cdot 2}{4} + c$

$a = - b - c$

$0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

Simplify terms.

Combine the numerators over the common denominator.

$0 = - \frac{c}{4} + \frac{- b + b \cdot 2}{4} + c$

$a = - b - c$

$0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

Combine the numerators over the common denominator.

$0 = c + \frac{- c - b + b \cdot 2}{4}$

$a = - b - c$

$0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

$0 = c + \frac{- c - b + b \cdot 2}{4}$

$a = - b - c$

$0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

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

$0 = c + \frac{- c - b + 2 b}{4}$

$a = - b - c$

$0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

Add $- b$ and $2 b$ .

$0 = c + \frac{- c + b}{4}$

$a = - b - c$

$0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

To write $c$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$0 = c \cdot \frac{4}{4} + \frac{- c + b}{4}$

$a = - b - c$

$0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

Simplify terms.

Combine $c$ and $\frac{4}{4}$ .

$0 = \frac{c \cdot 4}{4} + \frac{- c + b}{4}$

$a = - b - c$

$0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

Combine the numerators over the common denominator.

$0 = \frac{c \cdot 4 - c + b}{4}$

$a = - b - c$

$0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

$0 = \frac{c \cdot 4 - c + b}{4}$

$a = - b - c$

$0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

Simplify the numerator.

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

$0 = \frac{4 \cdot c - c + b}{4}$

$a = - b - c$

$0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

Subtract $c$ from $4 c$ .

$0 = \frac{3 c + b}{4}$

$a = - b - c$

$0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

$0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

$0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

$0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

Replace all occurrences of $a$ in $0 = a {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$ with $- b - c$ .

$0 = (- b - c) {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

Simplify the right side.

Simplify $(- b - c) {(- \frac{1}{3})}^{2} - \frac{b}{3} + c$ .

Simplify each term.

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{1}{3}$ .

$0 = (- b - c) ({(- 1)}^{2} {(\frac{1}{3})}^{2}) - \frac{b}{3} + c$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

Apply the product rule to $\frac{1}{3}$ .

$0 = (- b - c) ({(- 1)}^{2} (\frac{1^{2}}{3^{2}})) - \frac{b}{3} + c$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

$0 = (- b - c) ({(- 1)}^{2} (\frac{1^{2}}{3^{2}})) - \frac{b}{3} + c$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

Raise $- 1$ to the power of $2$ .

$0 = (- b - c) (1 (\frac{1^{2}}{3^{2}})) - \frac{b}{3} + c$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

Multiply $\frac{1^{2}}{3^{2}}$ by $1$ .

$0 = (- b - c) (\frac{1^{2}}{3^{2}}) - \frac{b}{3} + c$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

One to any power is one.

$0 = (- b - c) (\frac{1}{3^{2}}) - \frac{b}{3} + c$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

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

$0 = (- b - c) (\frac{1}{9}) - \frac{b}{3} + c$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

Apply the distributive property.

$0 = - b \frac{1}{9} - c \frac{1}{9} - \frac{b}{3} + c$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

Combine $\frac{1}{9}$ and $b$ .

$0 = - \frac{b}{9} - c \frac{1}{9} - \frac{b}{3} + c$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

Combine $\frac{1}{9}$ and $c$ .

$0 = - \frac{b}{9} - \frac{c}{9} - \frac{b}{3} + c$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

$0 = - \frac{b}{9} - \frac{c}{9} - \frac{b}{3} + c$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

To write $- \frac{b}{3}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$0 = - \frac{c}{9} - \frac{b}{9} - \frac{b}{3} \cdot \frac{3}{3} + c$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

Write each expression with a common denominator of $9$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{b}{3}$ by $\frac{3}{3}$ .

$0 = - \frac{c}{9} - \frac{b}{9} - \frac{b \cdot 3}{3 \cdot 3} + c$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

Multiply $3$ by $3$ .

$0 = - \frac{c}{9} - \frac{b}{9} - \frac{b \cdot 3}{9} + c$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

$0 = - \frac{c}{9} - \frac{b}{9} - \frac{b \cdot 3}{9} + c$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

Combine the numerators over the common denominator.

$0 = - \frac{c}{9} + \frac{- b - b \cdot 3}{9} + c$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

Combine the numerators over the common denominator.

$0 = c + \frac{- c - b - b \cdot 3}{9}$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

Multiply $3$ by $- 1$ .

$0 = c + \frac{- c - b - 3 b}{9}$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

Subtract $3 b$ from $- b$ .

$0 = c + \frac{- c - 4 b}{9}$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

To write $c$ as a fraction with a common denominator, multiply by $\frac{9}{9}$ .

$0 = c \cdot \frac{9}{9} + \frac{- c - 4 b}{9}$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

Simplify terms.

Combine $c$ and $\frac{9}{9}$ .

$0 = \frac{c \cdot 9}{9} + \frac{- c - 4 b}{9}$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

Combine the numerators over the common denominator.

$0 = \frac{c \cdot 9 - c - 4 b}{9}$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

$0 = \frac{c \cdot 9 - c - 4 b}{9}$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

Simplify the numerator.

Subtract $c$ from $c \cdot 9$ .

Reorder $c$ and $9$ .

$0 = \frac{9 \cdot c - c - 4 b}{9}$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

Subtract $c$ from $9 \cdot c$ .

$0 = \frac{8 \cdot c - 4 b}{9}$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

$0 = \frac{8 \cdot c - 4 b}{9}$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

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

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

$0 = \frac{4 (2 \cdot c) - 4 b}{9}$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

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

$0 = \frac{4 (2 \cdot c) + 4 (- b)}{9}$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

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

$0 = \frac{4 (2 \cdot c - b)}{9}$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

$0 = \frac{4 (2 c - b)}{9}$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

$0 = \frac{4 (2 c - b)}{9}$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

$0 = \frac{4 (2 c - b)}{9}$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

$0 = \frac{4 (2 c - b)}{9}$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

$0 = \frac{4 (2 c - b)}{9}$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

Solve for $c$ in $0 = \frac{4 (2 c - b)}{9}$ .

Set the numerator equal to zero.

$4 (2 c - b) = 0$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

Solve the equation for $c$ .

Divide each term in $4 (2 c - b) = 0$ by $4$ and simplify.

Divide each term in $4 (2 c - b) = 0$ by $4$ .

$\frac{4 (2 c - b)}{4} = \frac{0}{4}$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 (2 c - b)}{4} = \frac{0}{4}$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

Divide $2 c - b$ by $1$ .

$2 c - b = \frac{0}{4}$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

$2 c - b = \frac{0}{4}$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

$2 c - b = \frac{0}{4}$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

Simplify the right side.

Divide $0$ by $4$ .

$2 c - b = 0$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

$2 c - b = 0$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

$2 c - b = 0$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

Add $b$ to both sides of the equation.

$2 c = b$

$0 = \frac{3 c + b}{4}$

$a = - b - c$

Divide each term in $2 c = b$ by $2$ and simplify.

Divide each term in $2 c = b$ by $2$ .

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

$0 = \frac{3 c + b}{4}$

$a = - b - c$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

$0 = \frac{3 c + b}{4}$

$a = - b - c$

Divide $c$ by $1$ .

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

$0 = \frac{3 c + b}{4}$

$a = - b - c$

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

$0 = \frac{3 c + b}{4}$

$a = - b - c$

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

$0 = \frac{3 c + b}{4}$

$a = - b - c$

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

$0 = \frac{3 c + b}{4}$

$a = - b - c$

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

$0 = \frac{3 c + b}{4}$

$a = - b - c$

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

$0 = \frac{3 c + b}{4}$

$a = - b - c$

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

Replace all occurrences of $c$ in $0 = \frac{3 c + b}{4}$ with $\frac{b}{2}$ .

$0 = \frac{3 (\frac{b}{2}) + b}{4}$

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

$a = - b - c$

Simplify the right side.

Simplify $\frac{3 (\frac{b}{2}) + b}{4}$ .

Simplify the numerator.

Combine $3$ and $\frac{b}{2}$ .

$0 = \frac{\frac{3 b}{2} + b}{4}$

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

$a = - b - c$

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

$0 = \frac{\frac{3 b}{2} + b \cdot \frac{2}{2}}{4}$

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

$a = - b - c$

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

$0 = \frac{\frac{3 b}{2} + \frac{b \cdot 2}{2}}{4}$

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

$a = - b - c$

Combine the numerators over the common denominator.

$0 = \frac{\frac{3 b + b \cdot 2}{2}}{4}$

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

$a = - b - c$

Rewrite $\frac{3 b + b \cdot 2}{2}$ in a factored form.

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

$0 = \frac{\frac{3 b + 2 \cdot b}{2}}{4}$

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

$a = - b - c$

Add $3 b$ and $2 b$ .

$0 = \frac{\frac{5 b}{2}}{4}$

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

$a = - b - c$

$0 = \frac{\frac{5 b}{2}}{4}$

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

$a = - b - c$

$0 = \frac{\frac{5 b}{2}}{4}$

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

$a = - b - c$

Multiply the numerator by the reciprocal of the denominator.

$0 = \frac{5 b}{2} \cdot \frac{1}{4}$

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

$a = - b - c$

Multiply $\frac{5 b}{2} \cdot \frac{1}{4}$ .

Multiply $\frac{5 b}{2}$ by $\frac{1}{4}$ .

$0 = \frac{5 b}{2 \cdot 4}$

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

$a = - b - c$

Multiply $2$ by $4$ .

$0 = \frac{5 b}{8}$

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

$a = - b - c$

$0 = \frac{5 b}{8}$

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

$a = - b - c$

$0 = \frac{5 b}{8}$

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

$a = - b - c$

$0 = \frac{5 b}{8}$

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

$a = - b - c$

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

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

$0 = \frac{5 b}{8}$

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

Simplify the right side.

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

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

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

$0 = \frac{5 b}{8}$

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

Simplify terms.

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

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

$0 = \frac{5 b}{8}$

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

Combine the numerators over the common denominator.

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

$0 = \frac{5 b}{8}$

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

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

$0 = \frac{5 b}{8}$

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

Simplify the numerator.

Factor $b$ out of $- b \cdot 2 - b$ .

Factor $b$ out of $- b \cdot 2$ .

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

$0 = \frac{5 b}{8}$

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

Factor $b$ out of $- b$ .

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

$0 = \frac{5 b}{8}$

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

Factor $b$ out of $b (- 1 \cdot 2) + b \cdot - 1$ .

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

$0 = \frac{5 b}{8}$

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

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

$0 = \frac{5 b}{8}$

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

Multiply $- 1$ by $2$ .

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

$0 = \frac{5 b}{8}$

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

Subtract $1$ from $- 2$ .

$a = \frac{b \cdot - 3}{2}$

$0 = \frac{5 b}{8}$

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

$a = \frac{b \cdot - 3}{2}$

$0 = \frac{5 b}{8}$

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

Simplify the expression.

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

$a = \frac{- 3 \cdot b}{2}$

$0 = \frac{5 b}{8}$

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

Move the negative in front of the fraction.

$a = - \frac{3 b}{2}$

$0 = \frac{5 b}{8}$

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

$a = - \frac{3 b}{2}$

$0 = \frac{5 b}{8}$

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

$a = - \frac{3 b}{2}$

$0 = \frac{5 b}{8}$

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

$a = - \frac{3 b}{2}$

$0 = \frac{5 b}{8}$

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

$a = - \frac{3 b}{2}$

$0 = \frac{5 b}{8}$

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

Solve for $b$ in $0 = \frac{5 b}{8}$ .

Set the numerator equal to zero.

$5 b = 0$

$a = - \frac{3 b}{2}$

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

Divide each term in $5 b = 0$ by $5$ and simplify.

Divide each term in $5 b = 0$ by $5$ .

$\frac{5 b}{5} = \frac{0}{5}$

$a = - \frac{3 b}{2}$

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

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 b}{5} = \frac{0}{5}$

$a = - \frac{3 b}{2}$

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

Divide $b$ by $1$ .

$b = \frac{0}{5}$

$a = - \frac{3 b}{2}$

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

$b = \frac{0}{5}$

$a = - \frac{3 b}{2}$

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

$b = \frac{0}{5}$

$a = - \frac{3 b}{2}$

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

Simplify the right side.

Divide $0$ by $5$ .

$b = 0$

$a = - \frac{3 b}{2}$

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

$b = 0$

$a = - \frac{3 b}{2}$

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

$b = 0$

$a = - \frac{3 b}{2}$

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

$b = 0$

$a = - \frac{3 b}{2}$

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

Replace all occurrences of $b$ with $0$ in each equation.

Replace all occurrences of $b$ in $a = - \frac{3 b}{2}$ with $0$ .

$a = - \frac{3 (0)}{2}$

$b = 0$

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

Simplify the right side.

Simplify $- \frac{3 (0)}{2}$ .

Cancel the common factor of $0$ and $2$ .

Factor $2$ out of $3 (0)$ .

$a = - \frac{2 (3 \cdot (0))}{2}$

$b = 0$

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

Cancel the common factors.

Factor $2$ out of $2$ .

$a = - \frac{2 (3 \cdot (0))}{2 (1)}$

$b = 0$

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

Cancel the common factor.

$a = - \frac{2 (3 \cdot (0))}{2 \cdot 1}$

$b = 0$

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

Rewrite the expression.

$a = - \frac{3 \cdot (0)}{1}$

$b = 0$

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

Divide $3 \cdot (0)$ by $1$ .

$a = - (3 \cdot (0))$

$b = 0$

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

$a = - (3 \cdot (0))$

$b = 0$

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

$a = - (3 \cdot (0))$

$b = 0$

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

Multiply $- (3 \cdot (0))$ .

Multiply $3$ by $0$ .

$a = - 0$

$b = 0$

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

Multiply $- 1$ by $0$ .

$a = 0$

$b = 0$

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

$a = 0$

$b = 0$

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

$a = 0$

$b = 0$

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

$a = 0$

$b = 0$

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

Replace all occurrences of $b$ in $c = \frac{b}{2}$ with $0$ .

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

$a = 0$

$b = 0$

Simplify the right side.

Divide $0$ by $2$ .

$c = 0$

$a = 0$

$b = 0$

$c = 0$

$a = 0$

$b = 0$

$c = 0$

$a = 0$

$b = 0$

List all of the solutions.

$c = 0, a = 0, b = 0$

Step 4

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

$y = 0$

Step 5

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

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