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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Simplify each equation.

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Remove parentheses.

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

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

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

Simplify $a {(0)}^{2} + b (0) + c$ .

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Simplify each term.

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

$3 = a \cdot 0 + b (0) + c$

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

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

Multiply $a$ by $0$ .

$3 = 0 + b (0) + c$

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

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

Multiply $b$ by $0$ .

$3 = 0 + 0 + c$

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

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

$3 = 0 + 0 + c$

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

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

Combine the opposite terms in $0 + 0 + c$ .

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Add $0$ and $0$ .

$3 = 0 + c$

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

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

Add $0$ and $c$ .

$3 = c$

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

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

$3 = c$

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

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

$3 = c$

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

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

Remove parentheses.

$3 = c$

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

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

Simplify each term.

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Raise $4$ to the power of $2$ .

$3 = c$

$9 = a \cdot 16 + b (4) + c$

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

Move $16$ to the left of $a$ .

$3 = c$

$9 = 16 \cdot a + b (4) + c$

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

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

$3 = c$

$9 = 16 a + 4 b + c$

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

$3 = c$

$9 = 16 a + 4 b + c$

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

Remove parentheses.

$3 = c$

$9 = 16 a + 4 b + c$

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

Simplify each term.

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Raise $- 4$ to the power of $2$ .

$3 = c$

$9 = 16 a + 4 b + c$

$3 = a \cdot 16 + b (- 4) + c$

Move $16$ to the left of $a$ .

$3 = c$

$9 = 16 a + 4 b + c$

$3 = 16 \cdot a + b (- 4) + c$

Move $- 4$ to the left of $b$ .

$3 = c$

$9 = 16 a + 4 b + c$

$3 = 16 a - 4 b + c$

$3 = c$

$9 = 16 a + 4 b + c$

$3 = 16 a - 4 b + c$

$3 = c$

$9 = 16 a + 4 b + c$

$3 = 16 a - 4 b + c$

Rewrite the equation as $c = 3$ .

$c = 3$

$9 = 16 a + 4 b + c$

$3 = 16 a - 4 b + c$

Replace all occurrences of $c$ with $3$ in each equation.

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Replace all occurrences of $c$ in $9 = 16 a + 4 b + c$ with $3$ .

$c = 3$

$9 = 16 a + 4 b + 3$

$3 = 16 a - 4 b + c$

Replace all occurrences of $c$ in $3 = 16 a - 4 b + c$ with $3$ .

$c = 3$

$9 = 16 a + 4 b + 3$

$3 = 16 a - 4 b + 3$

$c = 3$

$9 = 16 a + 4 b + 3$

$3 = 16 a - 4 b + 3$

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.

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Rewrite the equation as $16 a + 4 b + 3 = 9$ .

$c = 3$

$16 a + 4 b + 3 = 9$

$3 = 16 a - 4 b + 3$

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

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

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Subtract $4 b$ from both sides of the equation.

$c = 3$

$16 a + 3 = 9 - 4 b$

$3 = 16 a - 4 b + 3$

Subtract $3$ from both sides of the equation.

$c = 3$

$16 a = 9 - 4 b - 3$

$3 = 16 a - 4 b + 3$

$c = 3$

$16 a = 9 - 4 b - 3$

$3 = 16 a - 4 b + 3$

Subtract $3$ from $9$ .

$c = 3$

$16 a = - 4 b + 6$

$3 = 16 a - 4 b + 3$

$c = 3$

$16 a = - 4 b + 6$

$3 = 16 a - 4 b + 3$

Divide each term by $16$ and simplify.

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Divide each term in $16 a = - 4 b + 6$ by $16$ .

$c = 3$

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

$3 = 16 a - 4 b + 3$

Cancel the common factor of $16$ .

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Cancel the common factor.

$c = 3$

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

$3 = 16 a - 4 b + 3$

Divide $a$ by $1$ .

$c = 3$

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

$3 = 16 a - 4 b + 3$

$c = 3$

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

$3 = 16 a - 4 b + 3$

Simplify each term.

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Cancel the common factor of $- 4$ and $16$ .

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

$c = 3$

$a = \frac{4 (- b)}{16} + \frac{6}{16}$

$3 = 16 a - 4 b + 3$

Cancel the common factors.

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Factor $4$ out of $16$ .

$c = 3$

$a = \frac{4 (- b)}{4 (4)} + \frac{6}{16}$

$3 = 16 a - 4 b + 3$

Cancel the common factor.

$c = 3$

$a = \frac{4 (- b)}{4 \cdot 4} + \frac{6}{16}$

$3 = 16 a - 4 b + 3$

Rewrite the expression.

$c = 3$

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

$3 = 16 a - 4 b + 3$

$c = 3$

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

$3 = 16 a - 4 b + 3$

$c = 3$

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

$3 = 16 a - 4 b + 3$

Move the negative in front of the fraction.

$c = 3$

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

$3 = 16 a - 4 b + 3$

Cancel the common factor of $6$ and $16$ .

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Factor $2$ out of $6$ .

$c = 3$

$a = - \frac{b}{4} + \frac{2 (3)}{16}$

$3 = 16 a - 4 b + 3$

Cancel the common factors.

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Factor $2$ out of $16$ .

$c = 3$

$a = - \frac{b}{4} + \frac{2 \cdot 3}{2 \cdot 8}$

$3 = 16 a - 4 b + 3$

Cancel the common factor.

$c = 3$

$a = - \frac{b}{4} + \frac{2 \cdot 3}{2 \cdot 8}$

$3 = 16 a - 4 b + 3$

Rewrite the expression.

$c = 3$

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

$3 = 16 a - 4 b + 3$

$c = 3$

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

$3 = 16 a - 4 b + 3$

$c = 3$

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

$3 = 16 a - 4 b + 3$

$c = 3$

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

$3 = 16 a - 4 b + 3$

$c = 3$

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

$3 = 16 a - 4 b + 3$

$c = 3$

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

$3 = 16 a - 4 b + 3$

Replace all occurrences of $a$ in $3 = 16 a - 4 b + 3$ with $- \frac{b}{4} + \frac{3}{8}$ .

$c = 3$

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

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

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

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Simplify each term.

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Apply the distributive property.

$c = 3$

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

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

Cancel the common factor of $4$ .

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

$c = 3$

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

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

Factor $4$ out of $16$ .

$c = 3$

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

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

Cancel the common factor.

$c = 3$

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

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

Rewrite the expression.

$c = 3$

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

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

$c = 3$

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

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

Multiply $- 1$ by $4$ .

$c = 3$

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

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

Cancel the common factor of $8$ .

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Factor $8$ out of $16$ .

$c = 3$

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

$3 = - 4 b + 8 (2) (\frac{3}{8}) - 4 b + 3$

Cancel the common factor.

$c = 3$

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

$3 = - 4 b + 8 \cdot (2 (\frac{3}{8})) - 4 b + 3$

Rewrite the expression.

$c = 3$

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

$3 = - 4 b + 2 \cdot 3 - 4 b + 3$

$c = 3$

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

$3 = - 4 b + 2 \cdot 3 - 4 b + 3$

Multiply $2$ by $3$ .

$c = 3$

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

$3 = - 4 b + 6 - 4 b + 3$

$c = 3$

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

$3 = - 4 b + 6 - 4 b + 3$

Simplify by adding terms.

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Subtract $4 b$ from $- 4 b$ .

$c = 3$

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

$3 = - 8 b + 6 + 3$

Add $6$ and $3$ .

$c = 3$

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

$3 = - 8 b + 9$

$c = 3$

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

$3 = - 8 b + 9$

$c = 3$

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

$3 = - 8 b + 9$

Solve for $b$ in the third equation.

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Rewrite the equation as $- 8 b + 9 = 3$ .

$c = 3$

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

$- 8 b + 9 = 3$

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

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Subtract $9$ from both sides of the equation.

$c = 3$

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

$- 8 b = 3 - 9$

Subtract $9$ from $3$ .

$c = 3$

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

$- 8 b = - 6$

$c = 3$

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

$- 8 b = - 6$

Divide each term by $- 8$ and simplify.

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Divide each term in $- 8 b = - 6$ by $- 8$ .

$c = 3$

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

$\frac{- 8 b}{- 8} = \frac{- 6}{- 8}$

Cancel the common factor of $- 8$ .

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Cancel the common factor.

$c = 3$

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

$\frac{- 8 b}{- 8} = \frac{- 6}{- 8}$

Divide $b$ by $1$ .

$c = 3$

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

$b = \frac{- 6}{- 8}$

$c = 3$

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

$b = \frac{- 6}{- 8}$

Cancel the common factor of $- 6$ and $- 8$ .

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Factor $- 2$ out of $- 6$ .

$c = 3$

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

$b = \frac{- 2 \cdot 3}{- 8}$

Cancel the common factors.

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Factor $- 2$ out of $- 8$ .

$c = 3$

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

$b = \frac{- 2 \cdot 3}{- 2 \cdot 4}$

Cancel the common factor.

$c = 3$

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

$b = \frac{- 2 \cdot 3}{- 2 \cdot 4}$

Rewrite the expression.

$c = 3$

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

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

$c = 3$

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

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

$c = 3$

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

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

$c = 3$

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

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

$c = 3$

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

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

Replace all occurrences of $b$ in $a = - \frac{b}{4} + \frac{3}{8}$ with $\frac{3}{4}$ .

$c = 3$

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

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

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.

$c = 3$

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

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

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

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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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Multiply $\frac{3}{8}$ and $\frac{2}{2}$ .

$c = 3$

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

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

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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