Algebra Examples

$(0, 4)$ , $(2, 14)$ , $(3, 13)$

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.

$4 = a {(0)}^{2} + b (0) + c, 14 = a {(2)}^{2} + b (2) + c, 13 = a {(3)}^{2} + b (3) + c$

Step 3

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

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Step 3.1

Solve for $c$ in $4 = a \cdot 0^{2} + c$ .

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Step 3.1.1

Rewrite the equation as $a \cdot 0^{2} + c = 4$ .

$a \cdot 0^{2} + c = 4$

$14 = a \cdot 2^{2} + b (2) + c$

$13 = a \cdot 3^{2} + b (3) + c$

Step 3.1.2

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

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Step 3.1.2.1

Simplify each term.

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Step 3.1.2.1.1

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

$a \cdot 0 + c = 4$

$14 = a \cdot 2^{2} + b (2) + c$

$13 = a \cdot 3^{2} + b (3) + c$

Step 3.1.2.1.2

Multiply $a$ by $0$ .

$0 + c = 4$

$14 = a \cdot 2^{2} + b (2) + c$

$13 = a \cdot 3^{2} + b (3) + c$

$0 + c = 4$

$14 = a \cdot 2^{2} + b (2) + c$

$13 = a \cdot 3^{2} + b (3) + c$

Step 3.1.2.2

Add $0$ and $c$ .

$c = 4$

$14 = a \cdot 2^{2} + b (2) + c$

$13 = a \cdot 3^{2} + b (3) + c$

$c = 4$

$14 = a \cdot 2^{2} + b (2) + c$

$13 = a \cdot 3^{2} + b (3) + c$

$c = 4$

$14 = a \cdot 2^{2} + b (2) + c$

$13 = a \cdot 3^{2} + b (3) + c$

Step 3.2

Replace all occurrences of $c$ with $4$ in each equation.

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Step 3.2.1

Replace all occurrences of $c$ in $14 = a \cdot 2^{2} + b (2) + c$ with $4$ .

$14 = a \cdot 2^{2} + b (2) + 4$

$c = 4$

$13 = a \cdot 3^{2} + b (3) + c$

Step 3.2.2

Simplify $14 = a \cdot 2^{2} + b (2) + 4$ .

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Step 3.2.2.1

Simplify the left side.

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Step 3.2.2.1.1

Remove parentheses.

$14 = a \cdot 2^{2} + b (2) + 4$

$c = 4$

$13 = a \cdot 3^{2} + b (3) + c$

$14 = a \cdot 2^{2} + b (2) + 4$

$c = 4$

$13 = a \cdot 3^{2} + b (3) + c$

Step 3.2.2.2

Simplify the right side.

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Step 3.2.2.2.1

Simplify each term.

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Step 3.2.2.2.1.1

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

$14 = a \cdot 4 + b (2) + 4$

$c = 4$

$13 = a \cdot 3^{2} + b (3) + c$

Step 3.2.2.2.1.2

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

$14 = 4 \cdot a + b (2) + 4$

$c = 4$

$13 = a \cdot 3^{2} + b (3) + c$

Step 3.2.2.2.1.3

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

$14 = 4 a + 2 b + 4$

$c = 4$

$13 = a \cdot 3^{2} + b (3) + c$

$14 = 4 a + 2 b + 4$

$c = 4$

$13 = a \cdot 3^{2} + b (3) + c$

$14 = 4 a + 2 b + 4$

$c = 4$

$13 = a \cdot 3^{2} + b (3) + c$

$14 = 4 a + 2 b + 4$

$c = 4$

$13 = a \cdot 3^{2} + b (3) + c$

Step 3.2.3

Replace all occurrences of $c$ in $13 = a \cdot 3^{2} + b (3) + c$ with $4$ .

$13 = a \cdot 3^{2} + b (3) + 4$

$14 = 4 a + 2 b + 4$

$c = 4$

Step 3.2.4

Simplify $13 = a \cdot 3^{2} + b (3) + 4$ .

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Step 3.2.4.1

Simplify the left side.

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Step 3.2.4.1.1

Remove parentheses.

$13 = a \cdot 3^{2} + b (3) + 4$

$14 = 4 a + 2 b + 4$

$c = 4$

$13 = a \cdot 3^{2} + b (3) + 4$

$14 = 4 a + 2 b + 4$

$c = 4$

Step 3.2.4.2

Simplify the right side.

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Step 3.2.4.2.1

Simplify each term.

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Step 3.2.4.2.1.1

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

$13 = a \cdot 9 + b (3) + 4$

$14 = 4 a + 2 b + 4$

$c = 4$

Step 3.2.4.2.1.2

Move $9$ to the left of $a$ .

$13 = 9 \cdot a + b (3) + 4$

$14 = 4 a + 2 b + 4$

$c = 4$

Step 3.2.4.2.1.3

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

$13 = 9 a + 3 b + 4$

$14 = 4 a + 2 b + 4$

$c = 4$

$13 = 9 a + 3 b + 4$

$14 = 4 a + 2 b + 4$

$c = 4$

$13 = 9 a + 3 b + 4$

$14 = 4 a + 2 b + 4$

$c = 4$

$13 = 9 a + 3 b + 4$

$14 = 4 a + 2 b + 4$

$c = 4$

$13 = 9 a + 3 b + 4$

$14 = 4 a + 2 b + 4$

$c = 4$

Step 3.3

Solve for $a$ in $13 = 9 a + 3 b + 4$ .

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Step 3.3.1

Rewrite the equation as $9 a + 3 b + 4 = 13$ .

$9 a + 3 b + 4 = 13$

$14 = 4 a + 2 b + 4$

$c = 4$

Step 3.3.2

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

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Step 3.3.2.1

Subtract $3 b$ from both sides of the equation.

$9 a + 4 = 13 - 3 b$

$14 = 4 a + 2 b + 4$

$c = 4$

Step 3.3.2.2

Subtract $4$ from both sides of the equation.

$9 a = 13 - 3 b - 4$

$14 = 4 a + 2 b + 4$

$c = 4$

Step 3.3.2.3

Subtract $4$ from $13$ .

$9 a = - 3 b + 9$

$14 = 4 a + 2 b + 4$

$c = 4$

$9 a = - 3 b + 9$

$14 = 4 a + 2 b + 4$

$c = 4$

Step 3.3.3

Divide each term in $9 a = - 3 b + 9$ by $9$ and simplify.

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Step 3.3.3.1

Divide each term in $9 a = - 3 b + 9$ by $9$ .

$\frac{9 a}{9} = \frac{- 3 b}{9} + \frac{9}{9}$

$14 = 4 a + 2 b + 4$

$c = 4$

Step 3.3.3.2

Simplify the left side.

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Step 3.3.3.2.1

Cancel the common factor of $9$ .

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Step 3.3.3.2.1.1

Cancel the common factor.

$\frac{9 a}{9} = \frac{- 3 b}{9} + \frac{9}{9}$

$14 = 4 a + 2 b + 4$

$c = 4$

Step 3.3.3.2.1.2

Divide $a$ by $1$ .

$a = \frac{- 3 b}{9} + \frac{9}{9}$

$14 = 4 a + 2 b + 4$

$c = 4$

$a = \frac{- 3 b}{9} + \frac{9}{9}$

$14 = 4 a + 2 b + 4$

$c = 4$

$a = \frac{- 3 b}{9} + \frac{9}{9}$

$14 = 4 a + 2 b + 4$

$c = 4$

Step 3.3.3.3

Simplify the right side.

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Step 3.3.3.3.1

Simplify each term.

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Step 3.3.3.3.1.1

Cancel the common factor of $- 3$ and $9$ .

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Step 3.3.3.3.1.1.1

Factor $3$ out of $- 3 b$ .

$a = \frac{3 (- b)}{9} + \frac{9}{9}$

$14 = 4 a + 2 b + 4$

$c = 4$

Step 3.3.3.3.1.1.2

Cancel the common factors.

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Step 3.3.3.3.1.1.2.1

Factor $3$ out of $9$ .

$a = \frac{3 (- b)}{3 (3)} + \frac{9}{9}$

$14 = 4 a + 2 b + 4$

$c = 4$

Step 3.3.3.3.1.1.2.2

Cancel the common factor.

$a = \frac{3 (- b)}{3 \cdot 3} + \frac{9}{9}$

$14 = 4 a + 2 b + 4$

$c = 4$

Step 3.3.3.3.1.1.2.3

Rewrite the expression.

$a = \frac{- b}{3} + \frac{9}{9}$

$14 = 4 a + 2 b + 4$

$c = 4$

$a = \frac{- b}{3} + \frac{9}{9}$

$14 = 4 a + 2 b + 4$

$c = 4$

$a = \frac{- b}{3} + \frac{9}{9}$

$14 = 4 a + 2 b + 4$

$c = 4$

Step 3.3.3.3.1.2

Move the negative in front of the fraction.

$a = - \frac{b}{3} + \frac{9}{9}$

$14 = 4 a + 2 b + 4$

$c = 4$

Step 3.3.3.3.1.3

Divide $9$ by $9$ .

$a = - \frac{b}{3} + 1$

$14 = 4 a + 2 b + 4$

$c = 4$

$a = - \frac{b}{3} + 1$

$14 = 4 a + 2 b + 4$

$c = 4$

$a = - \frac{b}{3} + 1$

$14 = 4 a + 2 b + 4$

$c = 4$

$a = - \frac{b}{3} + 1$

$14 = 4 a + 2 b + 4$

$c = 4$

$a = - \frac{b}{3} + 1$

$14 = 4 a + 2 b + 4$

$c = 4$

Step 3.4

Replace all occurrences of $a$ with $- \frac{b}{3} + 1$ in each equation.

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Step 3.4.1

Replace all occurrences of $a$ in $14 = 4 a + 2 b + 4$ with $- \frac{b}{3} + 1$ .

$14 = 4 (- \frac{b}{3} + 1) + 2 b + 4$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify $4 (- \frac{b}{3} + 1) + 2 b + 4$ .

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Step 3.4.2.1.1

Simplify each term.

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Step 3.4.2.1.1.1

Apply the distributive property.

$14 = 4 (- \frac{b}{3}) + 4 \cdot 1 + 2 b + 4$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.4.2.1.1.2

Multiply $4 (- \frac{b}{3})$ .

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Step 3.4.2.1.1.2.1

Multiply $- 1$ by $4$ .

$14 = - 4 \frac{b}{3} + 4 \cdot 1 + 2 b + 4$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.4.2.1.1.2.2

Combine $- 4$ and $\frac{b}{3}$ .

$14 = \frac{- 4 b}{3} + 4 \cdot 1 + 2 b + 4$

$a = - \frac{b}{3} + 1$

$c = 4$

$14 = \frac{- 4 b}{3} + 4 \cdot 1 + 2 b + 4$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.4.2.1.1.3

Multiply $4$ by $1$ .

$14 = \frac{- 4 b}{3} + 4 + 2 b + 4$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.4.2.1.1.4

Move the negative in front of the fraction.

$14 = - \frac{4 b}{3} + 4 + 2 b + 4$

$a = - \frac{b}{3} + 1$

$c = 4$

$14 = - \frac{4 b}{3} + 4 + 2 b + 4$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.4.2.1.2

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

$14 = - \frac{4 b}{3} + 2 b \cdot \frac{3}{3} + 4 + 4$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.4.2.1.3

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

$14 = - \frac{4 b}{3} + \frac{2 b \cdot 3}{3} + 4 + 4$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.4.2.1.4

Combine the numerators over the common denominator.

$14 = \frac{- 4 b + 2 b \cdot 3}{3} + 4 + 4$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.4.2.1.5

Find the common denominator.

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Step 3.4.2.1.5.1

Write $4$ as a fraction with denominator $1$ .

$14 = \frac{- 4 b + 2 b \cdot 3}{3} + \frac{4}{1} + 4$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.4.2.1.5.2

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

$14 = \frac{- 4 b + 2 b \cdot 3}{3} + \frac{4}{1} \cdot \frac{3}{3} + 4$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.4.2.1.5.3

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

$14 = \frac{- 4 b + 2 b \cdot 3}{3} + \frac{4 \cdot 3}{3} + 4$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.4.2.1.5.4

Write $4$ as a fraction with denominator $1$ .

$14 = \frac{- 4 b + 2 b \cdot 3}{3} + \frac{4 \cdot 3}{3} + \frac{4}{1}$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.4.2.1.5.5

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

$14 = \frac{- 4 b + 2 b \cdot 3}{3} + \frac{4 \cdot 3}{3} + \frac{4}{1} \cdot \frac{3}{3}$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.4.2.1.5.6

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

$14 = \frac{- 4 b + 2 b \cdot 3}{3} + \frac{4 \cdot 3}{3} + \frac{4 \cdot 3}{3}$

$a = - \frac{b}{3} + 1$

$c = 4$

$14 = \frac{- 4 b + 2 b \cdot 3}{3} + \frac{4 \cdot 3}{3} + \frac{4 \cdot 3}{3}$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.4.2.1.6

Combine the numerators over the common denominator.

$14 = \frac{- 4 b + 2 b \cdot 3 + 4 \cdot 3 + 4 \cdot 3}{3}$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.4.2.1.7

Simplify each term.

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Step 3.4.2.1.7.1

Multiply $3$ by $2$ .

$14 = \frac{- 4 b + 6 b + 4 \cdot 3 + 4 \cdot 3}{3}$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.4.2.1.7.2

Multiply $4$ by $3$ .

$14 = \frac{- 4 b + 6 b + 12 + 4 \cdot 3}{3}$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.4.2.1.7.3

Multiply $4$ by $3$ .

$14 = \frac{- 4 b + 6 b + 12 + 12}{3}$

$a = - \frac{b}{3} + 1$

$c = 4$

$14 = \frac{- 4 b + 6 b + 12 + 12}{3}$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.4.2.1.8

Simplify terms.

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Step 3.4.2.1.8.1

Add $- 4 b$ and $6 b$ .

$14 = \frac{2 b + 12 + 12}{3}$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.4.2.1.8.2

Add $12$ and $12$ .

$14 = \frac{2 b + 24}{3}$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.4.2.1.8.3

Factor $2$ out of $2 b + 24$ .

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Step 3.4.2.1.8.3.1

Factor $2$ out of $2 b$ .

$14 = \frac{2 (b) + 24}{3}$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.4.2.1.8.3.2

Factor $2$ out of $24$ .

$14 = \frac{2 b + 2 \cdot 12}{3}$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.4.2.1.8.3.3

Factor $2$ out of $2 b + 2 \cdot 12$ .

$14 = \frac{2 (b + 12)}{3}$

$a = - \frac{b}{3} + 1$

$c = 4$

$14 = \frac{2 (b + 12)}{3}$

$a = - \frac{b}{3} + 1$

$c = 4$

$14 = \frac{2 (b + 12)}{3}$

$a = - \frac{b}{3} + 1$

$c = 4$

$14 = \frac{2 (b + 12)}{3}$

$a = - \frac{b}{3} + 1$

$c = 4$

$14 = \frac{2 (b + 12)}{3}$

$a = - \frac{b}{3} + 1$

$c = 4$

$14 = \frac{2 (b + 12)}{3}$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.5

Solve for $b$ in $14 = \frac{2 (b + 12)}{3}$ .

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Step 3.5.1

Rewrite the equation as $\frac{2 (b + 12)}{3} = 14$ .

$\frac{2 (b + 12)}{3} = 14$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.5.2

Multiply both sides of the equation by $\frac{3}{2}$ .

$\frac{3}{2} \cdot \frac{2 (b + 12)}{3} = \frac{3}{2} \cdot 14$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.5.3

Simplify both sides of the equation.

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Step 3.5.3.1

Simplify the left side.

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Step 3.5.3.1.1

Simplify $\frac{3}{2} \cdot \frac{2 (b + 12)}{3}$ .

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Step 3.5.3.1.1.1

Cancel the common factor of $3$ .

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Step 3.5.3.1.1.1.1

Cancel the common factor.

$\frac{3}{2} \cdot \frac{2 (b + 12)}{3} = \frac{3}{2} \cdot 14$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.5.3.1.1.1.2

Rewrite the expression.

$\frac{1}{2} \cdot (2 (b + 12)) = \frac{3}{2} \cdot 14$

$a = - \frac{b}{3} + 1$

$c = 4$

$\frac{1}{2} \cdot (2 (b + 12)) = \frac{3}{2} \cdot 14$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.5.3.1.1.2

Cancel the common factor of $2$ .

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Step 3.5.3.1.1.2.1

Cancel the common factor.

$\frac{1}{2} \cdot (2 (b + 12)) = \frac{3}{2} \cdot 14$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.5.3.1.1.2.2

Rewrite the expression.

$b + 12 = \frac{3}{2} \cdot 14$

$a = - \frac{b}{3} + 1$

$c = 4$

$b + 12 = \frac{3}{2} \cdot 14$

$a = - \frac{b}{3} + 1$

$c = 4$

$b + 12 = \frac{3}{2} \cdot 14$

$a = - \frac{b}{3} + 1$

$c = 4$

$b + 12 = \frac{3}{2} \cdot 14$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.5.3.2

Simplify the right side.

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Step 3.5.3.2.1

Simplify $\frac{3}{2} \cdot 14$ .

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Step 3.5.3.2.1.1

Cancel the common factor of $2$ .

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Step 3.5.3.2.1.1.1

Factor $2$ out of $14$ .

$b + 12 = \frac{3}{2} \cdot (2 (7))$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.5.3.2.1.1.2

Cancel the common factor.

$b + 12 = \frac{3}{2} \cdot (2 \cdot 7)$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.5.3.2.1.1.3

Rewrite the expression.

$b + 12 = 3 \cdot 7$

$a = - \frac{b}{3} + 1$

$c = 4$

$b + 12 = 3 \cdot 7$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.5.3.2.1.2

Multiply $3$ by $7$ .

$b + 12 = 21$

$a = - \frac{b}{3} + 1$

$c = 4$

$b + 12 = 21$

$a = - \frac{b}{3} + 1$

$c = 4$

$b + 12 = 21$

$a = - \frac{b}{3} + 1$

$c = 4$

$b + 12 = 21$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.5.4

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

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Step 3.5.4.1

Subtract $12$ from both sides of the equation.

$b = 21 - 12$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.5.4.2

Subtract $12$ from $21$ .

$b = 9$

$a = - \frac{b}{3} + 1$

$c = 4$

$b = 9$

$a = - \frac{b}{3} + 1$

$c = 4$

$b = 9$

$a = - \frac{b}{3} + 1$

$c = 4$

Step 3.6

Replace all occurrences of $b$ with $9$ in each equation.

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Step 3.6.1

Replace all occurrences of $b$ in $a = - \frac{b}{3} + 1$ with $9$ .

$a = - \frac{9}{3} + 1$

$b = 9$

$c = 4$

Step 3.6.2

Simplify the right side.

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Step 3.6.2.1

Simplify $- \frac{9}{3} + 1$ .

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Step 3.6.2.1.1

Simplify each term.

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Step 3.6.2.1.1.1

Divide $9$ by $3$ .

$a = - 1 \cdot 3 + 1$

$b = 9$

$c = 4$

Step 3.6.2.1.1.2

Multiply $- 1$ by $3$ .

$a = - 3 + 1$

$b = 9$

$c = 4$

$a = - 3 + 1$

$b = 9$

$c = 4$

Step 3.6.2.1.2

Add $- 3$ and $1$ .

$a = - 2$

$b = 9$

$c = 4$

$a = - 2$

$b = 9$

$c = 4$

$a = - 2$

$b = 9$

$c = 4$

$a = - 2$

$b = 9$

$c = 4$

Step 3.7

List all of the solutions.

$a = - 2, b = 9, c = 4$

Step 4

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

$y = - 2 x^{2} + 9 x + 4$

Step 5