Find the Equation of the Parabola (0,43) , (15,77) , (23,85)

$(0, 43)$ , $(15, 77)$ , $(23, 85)$

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.

$43 = a {(0)}^{2} + b (0) + c$

$77 = a {(15)}^{2} + b (15) + c$

$85 = a {(23)}^{2} + b (23) + c$

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

Solve for $c$ in $43 = a \cdot 0^{2} + c$ .

Rewrite the equation as $a \cdot 0^{2} + c = 43$ .

$a \cdot 0^{2} + c = 43$

$77 = a \cdot 15^{2} + b (15) + c$

$85 = a \cdot 23^{2} + b (23) + c$

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

Simplify each term.

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

$a \cdot 0 + c = 43$

$77 = a \cdot 15^{2} + b (15) + c$

$85 = a \cdot 23^{2} + b (23) + c$

Multiply $a$ by $0$ .

$0 + c = 43$

$77 = a \cdot 15^{2} + b (15) + c$

$85 = a \cdot 23^{2} + b (23) + c$

$0 + c = 43$

$77 = a \cdot 15^{2} + b (15) + c$

$85 = a \cdot 23^{2} + b (23) + c$

Add $0$ and $c$ .

$c = 43$

$77 = a \cdot 15^{2} + b (15) + c$

$85 = a \cdot 23^{2} + b (23) + c$

$c = 43$

$77 = a \cdot 15^{2} + b (15) + c$

$85 = a \cdot 23^{2} + b (23) + c$

$c = 43$

$77 = a \cdot 15^{2} + b (15) + c$

$85 = a \cdot 23^{2} + b (23) + c$

Replace all occurrences of $c$ with $43$ in each equation.

Replace all occurrences of $c$ in $77 = a \cdot 15^{2} + b (15) + c$ with $43$ .

$77 = a \cdot 15^{2} + b (15) + 43$

$c = 43$

$85 = a \cdot 23^{2} + b (23) + c$

Simplify $77 = a \cdot 15^{2} + b (15) + 43$ .

Simplify the left side.

Remove parentheses.

$77 = a \cdot 15^{2} + b (15) + 43$

$c = 43$

$85 = a \cdot 23^{2} + b (23) + c$

$77 = a \cdot 15^{2} + b (15) + 43$

$c = 43$

$85 = a \cdot 23^{2} + b (23) + c$

Simplify the right side.

Simplify each term.

Raise $15$ to the power of $2$ .

$77 = a \cdot 225 + b (15) + 43$

$c = 43$

$85 = a \cdot 23^{2} + b (23) + c$

Move $225$ to the left of $a$ .

$77 = 225 \cdot a + b (15) + 43$

$c = 43$

$85 = a \cdot 23^{2} + b (23) + c$

Move $15$ to the left of $b$ .

$77 = 225 a + 15 b + 43$

$c = 43$

$85 = a \cdot 23^{2} + b (23) + c$

$77 = 225 a + 15 b + 43$

$c = 43$

$85 = a \cdot 23^{2} + b (23) + c$

$77 = 225 a + 15 b + 43$

$c = 43$

$85 = a \cdot 23^{2} + b (23) + c$

$77 = 225 a + 15 b + 43$

$c = 43$

$85 = a \cdot 23^{2} + b (23) + c$

Replace all occurrences of $c$ in $85 = a \cdot 23^{2} + b (23) + c$ with $43$ .

$85 = a \cdot 23^{2} + b (23) + 43$

$77 = 225 a + 15 b + 43$

$c = 43$

Simplify $85 = a \cdot 23^{2} + b (23) + 43$ .

Simplify the left side.

Remove parentheses.

$85 = a \cdot 23^{2} + b (23) + 43$

$77 = 225 a + 15 b + 43$

$c = 43$

$85 = a \cdot 23^{2} + b (23) + 43$

$77 = 225 a + 15 b + 43$

$c = 43$

Simplify the right side.

Simplify each term.

Raise $23$ to the power of $2$ .

$85 = a \cdot 529 + b (23) + 43$

$77 = 225 a + 15 b + 43$

$c = 43$

Move $529$ to the left of $a$ .

$85 = 529 \cdot a + b (23) + 43$

$77 = 225 a + 15 b + 43$

$c = 43$

Move $23$ to the left of $b$ .

$85 = 529 a + 23 b + 43$

$77 = 225 a + 15 b + 43$

$c = 43$

$85 = 529 a + 23 b + 43$

$77 = 225 a + 15 b + 43$

$c = 43$

$85 = 529 a + 23 b + 43$

$77 = 225 a + 15 b + 43$

$c = 43$

$85 = 529 a + 23 b + 43$

$77 = 225 a + 15 b + 43$

$c = 43$

$85 = 529 a + 23 b + 43$

$77 = 225 a + 15 b + 43$

$c = 43$

Solve for $a$ in $85 = 529 a + 23 b + 43$ .

Rewrite the equation as $529 a + 23 b + 43 = 85$ .

$529 a + 23 b + 43 = 85$

$77 = 225 a + 15 b + 43$

$c = 43$

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

Subtract $23 b$ from both sides of the equation.

$529 a + 43 = 85 - 23 b$

$77 = 225 a + 15 b + 43$

$c = 43$

Subtract $43$ from both sides of the equation.

$529 a = 85 - 23 b - 43$

$77 = 225 a + 15 b + 43$

$c = 43$

Subtract $43$ from $85$ .

$529 a = - 23 b + 42$

$77 = 225 a + 15 b + 43$

$c = 43$

$529 a = - 23 b + 42$

$77 = 225 a + 15 b + 43$

$c = 43$

Divide each term in $529 a = - 23 b + 42$ by $529$ and simplify.

Divide each term in $529 a = - 23 b + 42$ by $529$ .

$\frac{529 a}{529} = \frac{- 23 b}{529} + \frac{42}{529}$

$77 = 225 a + 15 b + 43$

$c = 43$

Simplify the left side.

Cancel the common factor of $529$ .

Cancel the common factor.

$\frac{529 a}{529} = \frac{- 23 b}{529} + \frac{42}{529}$

$77 = 225 a + 15 b + 43$

$c = 43$

Divide $a$ by $1$ .

$a = \frac{- 23 b}{529} + \frac{42}{529}$

$77 = 225 a + 15 b + 43$

$c = 43$

$a = \frac{- 23 b}{529} + \frac{42}{529}$

$77 = 225 a + 15 b + 43$

$c = 43$

$a = \frac{- 23 b}{529} + \frac{42}{529}$

$77 = 225 a + 15 b + 43$

$c = 43$

Simplify the right side.

Simplify each term.

Cancel the common factor of $- 23$ and $529$ .

Factor $23$ out of $- 23 b$ .

$a = \frac{23 (- b)}{529} + \frac{42}{529}$

$77 = 225 a + 15 b + 43$

$c = 43$

Cancel the common factors.

Factor $23$ out of $529$ .

$a = \frac{23 (- b)}{23 (23)} + \frac{42}{529}$

$77 = 225 a + 15 b + 43$

$c = 43$

Cancel the common factor.

$a = \frac{23 (- b)}{23 \cdot 23} + \frac{42}{529}$

$77 = 225 a + 15 b + 43$

$c = 43$

Rewrite the expression.

$a = \frac{- b}{23} + \frac{42}{529}$

$77 = 225 a + 15 b + 43$

$c = 43$

$a = \frac{- b}{23} + \frac{42}{529}$

$77 = 225 a + 15 b + 43$

$c = 43$

$a = \frac{- b}{23} + \frac{42}{529}$

$77 = 225 a + 15 b + 43$

$c = 43$

Move the negative in front of the fraction.

$a = - \frac{b}{23} + \frac{42}{529}$

$77 = 225 a + 15 b + 43$

$c = 43$

$a = - \frac{b}{23} + \frac{42}{529}$

$77 = 225 a + 15 b + 43$

$c = 43$

$a = - \frac{b}{23} + \frac{42}{529}$

$77 = 225 a + 15 b + 43$

$c = 43$

$a = - \frac{b}{23} + \frac{42}{529}$

$77 = 225 a + 15 b + 43$

$c = 43$

$a = - \frac{b}{23} + \frac{42}{529}$

$77 = 225 a + 15 b + 43$

$c = 43$

Replace all occurrences of $a$ with $- \frac{b}{23} + \frac{42}{529}$ in each equation.

Replace all occurrences of $a$ in $77 = 225 a + 15 b + 43$ with $- \frac{b}{23} + \frac{42}{529}$ .

$77 = 225 (- \frac{b}{23} + \frac{42}{529}) + 15 b + 43$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Simplify the right side.

Simplify $225 (- \frac{b}{23} + \frac{42}{529}) + 15 b + 43$ .

Simplify each term.

Apply the distributive property.

$77 = 225 (- \frac{b}{23}) + 225 (\frac{42}{529}) + 15 b + 43$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Multiply $225 (- \frac{b}{23})$ .

Multiply $- 1$ by $225$ .

$77 = - 225 \frac{b}{23} + 225 (\frac{42}{529}) + 15 b + 43$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Combine $- 225$ and $\frac{b}{23}$ .

$77 = \frac{- 225 b}{23} + 225 (\frac{42}{529}) + 15 b + 43$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

$77 = \frac{- 225 b}{23} + 225 (\frac{42}{529}) + 15 b + 43$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Multiply $225 (\frac{42}{529})$ .

Combine $225$ and $\frac{42}{529}$ .

$77 = \frac{- 225 b}{23} + \frac{225 \cdot 42}{529} + 15 b + 43$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Multiply $225$ by $42$ .

$77 = \frac{- 225 b}{23} + \frac{9450}{529} + 15 b + 43$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

$77 = \frac{- 225 b}{23} + \frac{9450}{529} + 15 b + 43$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Move the negative in front of the fraction.

$77 = - \frac{225 b}{23} + \frac{9450}{529} + 15 b + 43$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

$77 = - \frac{225 b}{23} + \frac{9450}{529} + 15 b + 43$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

To write $15 b$ as a fraction with a common denominator, multiply by $\frac{23}{23}$ .

$77 = - \frac{225 b}{23} + 15 b \cdot \frac{23}{23} + \frac{9450}{529} + 43$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Combine $15 b$ and $\frac{23}{23}$ .

$77 = - \frac{225 b}{23} + \frac{15 b \cdot 23}{23} + \frac{9450}{529} + 43$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Combine the numerators over the common denominator.

$77 = \frac{- 225 b + 15 b \cdot 23}{23} + \frac{9450}{529} + 43$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Find the common denominator.

Multiply $\frac{- 225 b + 15 b \cdot 23}{23}$ by $\frac{23}{23}$ .

$77 = \frac{- 225 b + 15 b \cdot 23}{23} \cdot \frac{23}{23} + \frac{9450}{529} + 43$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Multiply $\frac{- 225 b + 15 b \cdot 23}{23}$ by $\frac{23}{23}$ .

$77 = \frac{(- 225 b + 15 b \cdot 23) \cdot 23}{23 \cdot 23} + \frac{9450}{529} + 43$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Write $43$ as a fraction with denominator $1$ .

$77 = \frac{(- 225 b + 15 b \cdot 23) \cdot 23}{23 \cdot 23} + \frac{9450}{529} + \frac{43}{1}$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Multiply $\frac{43}{1}$ by $\frac{529}{529}$ .

$77 = \frac{(- 225 b + 15 b \cdot 23) \cdot 23}{23 \cdot 23} + \frac{9450}{529} + \frac{43}{1} \cdot \frac{529}{529}$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Multiply $\frac{43}{1}$ by $\frac{529}{529}$ .

$77 = \frac{(- 225 b + 15 b \cdot 23) \cdot 23}{23 \cdot 23} + \frac{9450}{529} + \frac{43 \cdot 529}{529}$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Multiply $23$ by $23$ .

$77 = \frac{(- 225 b + 15 b \cdot 23) \cdot 23}{529} + \frac{9450}{529} + \frac{43 \cdot 529}{529}$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

$77 = \frac{(- 225 b + 15 b \cdot 23) \cdot 23}{529} + \frac{9450}{529} + \frac{43 \cdot 529}{529}$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Combine the numerators over the common denominator.

$77 = \frac{(- 225 b + 15 b \cdot 23) \cdot 23 + 9450 + 43 \cdot 529}{529}$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Simplify each term.

Multiply $23$ by $15$ .

$77 = \frac{(- 225 b + 345 b) \cdot 23 + 9450 + 43 \cdot 529}{529}$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Add $- 225 b$ and $345 b$ .

$77 = \frac{120 b \cdot 23 + 9450 + 43 \cdot 529}{529}$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Multiply $23$ by $120$ .

$77 = \frac{2760 b + 9450 + 43 \cdot 529}{529}$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Multiply $43$ by $529$ .

$77 = \frac{2760 b + 9450 + 22747}{529}$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

$77 = \frac{2760 b + 9450 + 22747}{529}$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Add $9450$ and $22747$ .

$77 = \frac{2760 b + 32197}{529}$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

$77 = \frac{2760 b + 32197}{529}$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

$77 = \frac{2760 b + 32197}{529}$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

$77 = \frac{2760 b + 32197}{529}$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Solve for $b$ in $77 = \frac{2760 b + 32197}{529}$ .

Rewrite the equation as $\frac{2760 b + 32197}{529} = 77$ .

$\frac{2760 b + 32197}{529} = 77$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Multiply both sides by $529$ .

$\frac{2760 b + 32197}{529} \cdot 529 = 77 \cdot 529$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Simplify.

Simplify the left side.

Cancel the common factor of $529$ .

Cancel the common factor.

$\frac{2760 b + 32197}{529} \cdot 529 = 77 \cdot 529$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Rewrite the expression.

$2760 b + 32197 = 77 \cdot 529$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

$2760 b + 32197 = 77 \cdot 529$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

$2760 b + 32197 = 77 \cdot 529$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Simplify the right side.

Multiply $77$ by $529$ .

$2760 b + 32197 = 40733$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

$2760 b + 32197 = 40733$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

$2760 b + 32197 = 40733$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Solve for $b$ .

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

Subtract $32197$ from both sides of the equation.

$2760 b = 40733 - 32197$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Subtract $32197$ from $40733$ .

$2760 b = 8536$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

$2760 b = 8536$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Divide each term in $2760 b = 8536$ by $2760$ and simplify.

Divide each term in $2760 b = 8536$ by $2760$ .

$\frac{2760 b}{2760} = \frac{8536}{2760}$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Simplify the left side.

Cancel the common factor of $2760$ .

Cancel the common factor.

$\frac{2760 b}{2760} = \frac{8536}{2760}$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Divide $b$ by $1$ .

$b = \frac{8536}{2760}$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

$b = \frac{8536}{2760}$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

$b = \frac{8536}{2760}$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Simplify the right side.

Cancel the common factor of $8536$ and $2760$ .

Factor $8$ out of $8536$ .

$b = \frac{8 (1067)}{2760}$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Cancel the common factors.

Factor $8$ out of $2760$ .

$b = \frac{8 \cdot 1067}{8 \cdot 345}$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Cancel the common factor.

$b = \frac{8 \cdot 1067}{8 \cdot 345}$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Rewrite the expression.

$b = \frac{1067}{345}$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

$b = \frac{1067}{345}$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

$b = \frac{1067}{345}$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

$b = \frac{1067}{345}$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

$b = \frac{1067}{345}$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

$b = \frac{1067}{345}$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

$b = \frac{1067}{345}$

$a = - \frac{b}{23} + \frac{42}{529}$

$c = 43$

Replace all occurrences of $b$ with $\frac{1067}{345}$ in each equation.

Replace all occurrences of $b$ in $a = - \frac{b}{23} + \frac{42}{529}$ with $\frac{1067}{345}$ .

$a = - \frac{\frac{1067}{345}}{23} + \frac{42}{529}$

$b = \frac{1067}{345}$

$c = 43$

Simplify the right side.

Simplify $- \frac{\frac{1067}{345}}{23} + \frac{42}{529}$ .

Simplify each term.

Multiply the numerator by the reciprocal of the denominator.

$a = - (\frac{1067}{345} \cdot \frac{1}{23}) + \frac{42}{529}$

$b = \frac{1067}{345}$

$c = 43$

Multiply $\frac{1067}{345} \cdot \frac{1}{23}$ .

Multiply $\frac{1067}{345}$ by $\frac{1}{23}$ .

$a = - \frac{1067}{345 \cdot 23} + \frac{42}{529}$

$b = \frac{1067}{345}$

$c = 43$

Multiply $345$ by $23$ .

$a = - \frac{1067}{7935} + \frac{42}{529}$

$b = \frac{1067}{345}$

$c = 43$

$a = - \frac{1067}{7935} + \frac{42}{529}$

$b = \frac{1067}{345}$

$c = 43$

$a = - \frac{1067}{7935} + \frac{42}{529}$

$b = \frac{1067}{345}$

$c = 43$

To write $\frac{42}{529}$ as a fraction with a common denominator, multiply by $\frac{15}{15}$ .

$a = - \frac{1067}{7935} + \frac{42}{529} \cdot \frac{15}{15}$

$b = \frac{1067}{345}$

$c = 43$

Write each expression with a common denominator of $7935$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{42}{529}$ by $\frac{15}{15}$ .

$a = - \frac{1067}{7935} + \frac{42 \cdot 15}{529 \cdot 15}$

$b = \frac{1067}{345}$

$c = 43$

Multiply $529$ by $15$ .

$a = - \frac{1067}{7935} + \frac{42 \cdot 15}{7935}$

$b = \frac{1067}{345}$

$c = 43$

$a = - \frac{1067}{7935} + \frac{42 \cdot 15}{7935}$

$b = \frac{1067}{345}$

$c = 43$

Combine the numerators over the common denominator.

$a = \frac{- 1067 + 42 \cdot 15}{7935}$

$b = \frac{1067}{345}$

$c = 43$

Simplify the numerator.

Multiply $42$ by $15$ .

$a = \frac{- 1067 + 630}{7935}$

$b = \frac{1067}{345}$

$c = 43$

Add $- 1067$ and $630$ .

$a = \frac{- 437}{7935}$

$b = \frac{1067}{345}$

$c = 43$

$a = \frac{- 437}{7935}$

$b = \frac{1067}{345}$

$c = 43$

Cancel the common factor of $- 437$ and $7935$ .

Factor $23$ out of $- 437$ .

$a = \frac{23 (- 19)}{7935}$

$b = \frac{1067}{345}$

$c = 43$

Cancel the common factors.

Factor $23$ out of $7935$ .

$a = \frac{23 \cdot - 19}{23 \cdot 345}$

$b = \frac{1067}{345}$

$c = 43$

Cancel the common factor.

$a = \frac{23 \cdot - 19}{23 \cdot 345}$

$b = \frac{1067}{345}$

$c = 43$

Rewrite the expression.

$a = \frac{- 19}{345}$

$b = \frac{1067}{345}$

$c = 43$

$a = \frac{- 19}{345}$

$b = \frac{1067}{345}$

$c = 43$

$a = \frac{- 19}{345}$

$b = \frac{1067}{345}$

$c = 43$

Move the negative in front of the fraction.

$a = - \frac{19}{345}$

$b = \frac{1067}{345}$

$c = 43$

$a = - \frac{19}{345}$

$b = \frac{1067}{345}$

$c = 43$

$a = - \frac{19}{345}$

$b = \frac{1067}{345}$

$c = 43$

$a = - \frac{19}{345}$

$b = \frac{1067}{345}$

$c = 43$

List all of the solutions.

$a = - \frac{19}{345}, b = \frac{1067}{345}, c = 43$

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

$y = - \frac{19 x^{2}}{345} + \frac{1067 x}{345} + 43$

Find the Equation of the Parabola (0,43) , (15,77) , (23,85)

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