Algebra Examples

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

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.

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

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

$- 7 = a {(3)}^{2} + b (3) + c$

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

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Solve for $a$ in the first equation.

$a = - b - c 2 = 16 a + 4 b + c - 7 = 9 a + 3 b + c$

Replace all occurrences of $a$ with the solution found by solving the last equation for $a$ . In this case, the value substituted is $- b - c$ .

$a = - b - c$

$2 = 16 (- b - c) + 4 b + c$

$- 7 = 9 a + 3 b + c$

Replace all occurrences of $a$ with the solution found by solving the last equation for $a$ . In this case, the value substituted is $- b - c$ .

$a = - b - c$

$2 = 16 (- b - c) + 4 b + c$

$- 7 = 9 (- b - c) + 3 b + c$

Solve for $b$ in the second equation.

$a = - b - c b = - \frac{1}{6} - \frac{5 c}{4} - 7 = - 6 b - 8 c$

Replace all occurrences of $b$ with the solution found by solving the last equation for $b$ . In this case, the value substituted is $- \frac{1}{6} - \frac{5 c}{4}$ .

$a = - (- \frac{1}{6} - \frac{5 c}{4}) - c$

$b = - \frac{1}{6} - \frac{5 c}{4}$

$- 7 = - 6 b - 8 c$

Replace all occurrences of $b$ with the solution found by solving the last equation for $b$ . In this case, the value substituted is $- \frac{1}{6} - \frac{5 c}{4}$ .

$a = - (- \frac{1}{6} - \frac{5 c}{4}) - c$

$b = - \frac{1}{6} - \frac{5 c}{4}$

$- 7 = - 6 (- \frac{1}{6} - \frac{5 c}{4}) - 8 c$

Solve for $c$ in the third equation.

$a = \frac{1}{6} + \frac{c}{4} b = - \frac{1}{6} - \frac{5 c}{4} c = 16$

Replace all occurrences of $c$ with the solution found by solving the last equation for $c$ . In this case, the value substituted is $16$ .

$a = \frac{1}{6} + \frac{16}{4}$

$b = - \frac{1}{6} - \frac{5 c}{4}$

$c = 16$

Replace all occurrences of $c$ with the solution found by solving the last equation for $c$ . In this case, the value substituted is $16$ .

$a = \frac{1}{6} + \frac{16}{4}$

$b = - \frac{1}{6} - \frac{5 (16)}{4}$

$c = 16$

Simplify.

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Simplify $\frac{1}{6} + \frac{16}{4}$ .

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Divide $16$ by $4$ .

$a = \frac{1}{6} + 4$

$b = - \frac{1}{6} - \frac{5 (16)}{4}$

$c = 16$

To write $\frac{4}{1}$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$a = \frac{1}{6} + \frac{4}{1} \cdot \frac{6}{6}$

$b = - \frac{1}{6} - \frac{5 (16)}{4}$

$c = 16$

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$a = \frac{1}{6} + \frac{4 \cdot 6}{1 \cdot 6}$

$b = - \frac{1}{6} - \frac{5 (16)}{4}$

$c = 16$

Multiply $6$ by $1$ .

$a = \frac{1}{6} + \frac{4 \cdot 6}{6}$

$b = - \frac{1}{6} - \frac{5 (16)}{4}$

$c = 16$

$a = \frac{1}{6} + \frac{4 \cdot 6}{6}$

$b = - \frac{1}{6} - \frac{5 (16)}{4}$

$c = 16$

Combine the numerators over the common denominator.

$a = \frac{1 + 4 \cdot 6}{6}$

$b = - \frac{1}{6} - \frac{5 (16)}{4}$

$c = 16$

Simplify the numerator.

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Multiply $4$ by $6$ .

$a = \frac{1 + 24}{6}$

$b = - \frac{1}{6} - \frac{5 (16)}{4}$

$c = 16$

Add $1$ and $24$ .

$a = \frac{25}{6}$

$b = - \frac{1}{6} - \frac{5 (16)}{4}$

$c = 16$

$a = \frac{25}{6}$

$b = - \frac{1}{6} - \frac{5 (16)}{4}$

$c = 16$

$a = \frac{25}{6}$

$b = - \frac{1}{6} - \frac{5 (16)}{4}$

$c = 16$

Simplify $- \frac{1}{6} - \frac{5 (16)}{4}$ .

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Simplify each term.

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Reduce the expression by cancelling the common factors.

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Factor $4$ out of $5 (16)$ .

$a = \frac{25}{6}$

$b = - \frac{1}{6} - \frac{4 (5 \cdot (4))}{4}$

$c = 16$

Cancel the common factors.

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Factor $4$ out of $4$ .

$a = \frac{25}{6}$

$b = - \frac{1}{6} - \frac{4 (5 \cdot (4))}{4 (1)}$

$c = 16$

Cancel the common factor.

$a = \frac{25}{6}$

$b = - \frac{1}{6} - \frac{4 (5 \cdot (4))}{4 \cdot 1}$

$c = 16$

Rewrite the expression.

$a = \frac{25}{6}$

$b = - \frac{1}{6} - \frac{5 \cdot (4)}{1}$

$c = 16$

Divide $5 \cdot (4)$ by $1$ .

$a = \frac{25}{6}$

$b = - \frac{1}{6} - (5 \cdot (4))$

$c = 16$

$a = \frac{25}{6}$

$b = - \frac{1}{6} - (5 \cdot (4))$

$c = 16$

$a = \frac{25}{6}$

$b = - \frac{1}{6} - (5 \cdot (4))$

$c = 16$

Multiply $5$ by $4$ .

$a = \frac{25}{6}$

$b = - \frac{1}{6} - 1 \cdot 20$

$c = 16$

Multiply $- 1$ by $20$ .

$a = \frac{25}{6}$

$b = - \frac{1}{6} - 20$

$c = 16$

$a = \frac{25}{6}$

$b = - \frac{1}{6} - 20$

$c = 16$

To write $- \frac{- 20}{1}$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$a = \frac{25}{6}$

$b = - \frac{1}{6} + \frac{- 20}{1} \cdot \frac{6}{6}$

$c = 16$

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$a = \frac{25}{6}$

$b = - \frac{1}{6} + \frac{- 20 \cdot 6}{1 \cdot 6}$

$c = 16$

Multiply $6$ by $1$ .

$a = \frac{25}{6}$

$b = - \frac{1}{6} + \frac{- 20 \cdot 6}{6}$

$c = 16$

$a = \frac{25}{6}$

$b = - \frac{1}{6} + \frac{- 20 \cdot 6}{6}$

$c = 16$

Combine the numerators over the common denominator.

$a = \frac{25}{6}$

$b = \frac{- 1 - 20 \cdot 6}{6}$

$c = 16$

Simplify the numerator.

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Multiply $- 20$ by $6$ .

$a = \frac{25}{6}$

$b = \frac{- 1 - 120}{6}$

$c = 16$

Subtract $120$ from $- 1$ .

$a = \frac{25}{6}$

$b = \frac{- 121}{6}$

$c = 16$

$a = \frac{25}{6}$

$b = \frac{- 121}{6}$

$c = 16$

Move the negative in front of the fraction.

$a = \frac{25}{6}$

$b = - \frac{121}{6}$

$c = 16$

$a = \frac{25}{6}$

$b = - \frac{121}{6}$

$c = 16$

$a = \frac{25}{6}$

$b = - \frac{121}{6}$

$c = 16$

$a = \frac{25}{6}$

$b = - \frac{121}{6}$

$c = 16$

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

$y = \frac{25 x^{2}}{6} - \frac{121 x}{6} + 16$

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