Algebra Examples

$(0, 3)$ , $(4, 9)$ , $(- 4, 3)$

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.

$3 = a {(0)}^{2} + b (0) + c$

$9 = a {(4)}^{2} + b (4) + c$

$3 = a {(- 4)}^{2} + b (- 4) + c$

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

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Solve for $c$ in the first equation.

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

Replace all occurrences of $c$ with the solution found by solving the last equation for $c$ . In this case, the value substituted is $3$ .

$c = 3$

$9 = 16 a + 4 b + 3$

$3 = 16 a - 4 b + c$

Replace all occurrences of $c$ with the solution found by solving the last equation for $c$ . In this case, the value substituted is $3$ .

$c = 3$

$9 = 16 a + 4 b + 3$

$3 = 16 a - 4 b + 3$

Simplify.

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Remove parentheses.

$c = 3$

$9 = 16 a + 4 b + 3$

$3 = 16 a - 4 b + 3$

Remove parentheses.

$c = 3$

$9 = 16 a + 4 b + 3$

$3 = 16 a - 4 b + 3$

$c = 3$

$9 = 16 a + 4 b + 3$

$3 = 16 a - 4 b + 3$

Solve for $a$ in the second equation.

$c = 3 a = - \frac{b}{4} + \frac{3}{8} 3 = 16 a - 4 b + 3$

Replace all occurrences of $a$ with the solution found by solving the last equation for $a$ . In this case, the value substituted is $- \frac{b}{4} + \frac{3}{8}$ .

$c = 3$

$a = - \frac{b}{4} + \frac{3}{8}$

$3 = 16 (- \frac{b}{4} + \frac{3}{8}) - 4 b + 3$

Solve for $b$ in the third equation.

$c = 3 a = - \frac{b}{4} + \frac{3}{8} b = \frac{3}{4}$

Replace all occurrences of $b$ with the solution found by solving the last equation for $b$ . In this case, the value substituted is $\frac{3}{4}$ .

$c = 3$

$a = - \frac{\frac{3}{4}}{4} + \frac{3}{8}$

$b = \frac{3}{4}$

Simplify $- \frac{\frac{3}{4}}{4} + \frac{3}{8}$ .

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Simplify each term.

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Multiply the numerator by the reciprocal of the denominator.

$c = 3$

$a = - (\frac{3}{4} \cdot \frac{1}{4}) + \frac{3}{8}$

$b = \frac{3}{4}$

Simplify $\frac{3}{4} \cdot \frac{1}{4}$ .

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Multiply $\frac{3}{4}$ and $\frac{1}{4}$ .

$c = 3$

$a = - \frac{3}{4 \cdot 4} + \frac{3}{8}$

$b = \frac{3}{4}$

Multiply $4$ by $4$ .

$c = 3$

$a = - \frac{3}{16} + \frac{3}{8}$

$b = \frac{3}{4}$

$c = 3$

$a = - \frac{3}{16} + \frac{3}{8}$

$b = \frac{3}{4}$

$c = 3$

$a = - \frac{3}{16} + \frac{3}{8}$

$b = \frac{3}{4}$

To write $\frac{3}{8}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$c = 3$

$a = - \frac{3}{16} + \frac{3}{8} \cdot \frac{2}{2}$

$b = \frac{3}{4}$

Write each expression with a common denominator of $16$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$c = 3$

$a = - \frac{3}{16} + \frac{3 \cdot 2}{8 \cdot 2}$

$b = \frac{3}{4}$

Multiply $8$ by $2$ .

$c = 3$

$a = - \frac{3}{16} + \frac{3 \cdot 2}{16}$

$b = \frac{3}{4}$

$c = 3$

$a = - \frac{3}{16} + \frac{3 \cdot 2}{16}$

$b = \frac{3}{4}$

Combine the numerators over the common denominator.

$c = 3$

$a = \frac{- 3 + 3 \cdot 2}{16}$

$b = \frac{3}{4}$

Simplify the numerator.

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Multiply $3$ by $2$ .

$c = 3$

$a = \frac{- 3 + 6}{16}$

$b = \frac{3}{4}$

Add $- 3$ and $6$ .

$c = 3$

$a = \frac{3}{16}$

$b = \frac{3}{4}$

$c = 3$

$a = \frac{3}{16}$

$b = \frac{3}{4}$

$c = 3$

$a = \frac{3}{16}$

$b = \frac{3}{4}$

$c = 3$

$a = \frac{3}{16}$

$b = \frac{3}{4}$

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

$y = \frac{3 x^{2}}{16} + \frac{3 x}{4} + 3$

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