Finite Math Examples

$(- 1, 4)$ , $(3, 9)$ , $(0, 3)$

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.

$4 = a {(- 1)}^{2} + b (- 1) + c, 9 = a {(3)}^{2} + b (3) + c, 3 = a {(0)}^{2} + b (0) + c$

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 $3 = a \cdot 0^{2} + c$ .

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Step 3.1.1

Rewrite the equation as $a \cdot 0^{2} + c = 3$ .

$a \cdot 0^{2} + c = 3$

$4 = a {(- 1)}^{2} + b (- 1) + c$

$9 = a \cdot 3^{2} + b (3) + c$

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 $0$ .

$a \cdot 0 + c = 3$

$4 = a {(- 1)}^{2} + b (- 1) + c$

$9 = a \cdot 3^{2} + b (3) + c$

Step 3.1.2.1.2

Multiply $a$ by $0$ .

$0 + c = 3$

$4 = a {(- 1)}^{2} + b (- 1) + c$

$9 = a \cdot 3^{2} + b (3) + c$

$0 + c = 3$

$4 = a {(- 1)}^{2} + b (- 1) + c$

$9 = a \cdot 3^{2} + b (3) + c$

Step 3.1.2.2

Add $0$ and $c$ .

$c = 3$

$4 = a {(- 1)}^{2} + b (- 1) + c$

$9 = a \cdot 3^{2} + b (3) + c$

$c = 3$

$4 = a {(- 1)}^{2} + b (- 1) + c$

$9 = a \cdot 3^{2} + b (3) + c$

$c = 3$

$4 = a {(- 1)}^{2} + b (- 1) + c$

$9 = a \cdot 3^{2} + b (3) + c$

Step 3.2

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

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Step 3.2.1

Replace all occurrences of $c$ in $4 = a {(- 1)}^{2} + b (- 1) + c$ with $3$ .

$4 = a {(- 1)}^{2} + b (- 1) + 3$

$c = 3$

$9 = a \cdot 3^{2} + b (3) + c$

Step 3.2.2

Simplify $4 = a {(- 1)}^{2} + b (- 1) + 3$ .

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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.

$4 = a {(- 1)}^{2} + b (- 1) + 3$

$c = 3$

$9 = a \cdot 3^{2} + b (3) + c$

$4 = a {(- 1)}^{2} + b (- 1) + 3$

$c = 3$

$9 = a \cdot 3^{2} + b (3) + c$

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 $- 1$ to the power of $2$ .

$4 = a \cdot 1 + b (- 1) + 3$

$c = 3$

$9 = a \cdot 3^{2} + b (3) + c$

Step 3.2.2.2.1.2

Multiply $a$ by $1$ .

$4 = a + b (- 1) + 3$

$c = 3$

$9 = a \cdot 3^{2} + b (3) + c$

Step 3.2.2.2.1.3

Move $- 1$ to the left of $b$ .

$4 = a - 1 b + 3$

$c = 3$

$9 = a \cdot 3^{2} + b (3) + c$

Step 3.2.2.2.1.4

Rewrite $- 1 b$ as $- b$ .

$4 = a - b + 3$

$c = 3$

$9 = a \cdot 3^{2} + b (3) + c$

$4 = a - b + 3$

$c = 3$

$9 = a \cdot 3^{2} + b (3) + c$

$4 = a - b + 3$

$c = 3$

$9 = a \cdot 3^{2} + b (3) + c$

$4 = a - b + 3$

$c = 3$

$9 = a \cdot 3^{2} + b (3) + c$

Step 3.2.3

Replace all occurrences of $c$ in $9 = a \cdot 3^{2} + b (3) + c$ with $3$ .

$9 = a \cdot 3^{2} + b (3) + 3$

$4 = a - b + 3$

$c = 3$

Step 3.2.4

Simplify $9 = a \cdot 3^{2} + b (3) + 3$ .

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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.

$9 = a \cdot 3^{2} + b (3) + 3$

$4 = a - b + 3$

$c = 3$

$9 = a \cdot 3^{2} + b (3) + 3$

$4 = a - b + 3$

$c = 3$

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 $2$ .

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

$4 = a - b + 3$

$c = 3$

Step 3.2.4.2.1.2

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

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

$4 = a - b + 3$

$c = 3$

Step 3.2.4.2.1.3

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

$9 = 9 a + 3 b + 3$

$4 = a - b + 3$

$c = 3$

$9 = 9 a + 3 b + 3$

$4 = a - b + 3$

$c = 3$

$9 = 9 a + 3 b + 3$

$4 = a - b + 3$

$c = 3$

$9 = 9 a + 3 b + 3$

$4 = a - b + 3$

$c = 3$

$9 = 9 a + 3 b + 3$

$4 = a - b + 3$

$c = 3$

Step 3.3

Solve for $a$ in $4 = a - b + 3$ .

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Step 3.3.1

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

$a - b + 3 = 4$

$9 = 9 a + 3 b + 3$

$c = 3$

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

Add $b$ to both sides of the equation.

$a + 3 = 4 + b$

$9 = 9 a + 3 b + 3$

$c = 3$

Step 3.3.2.2

Subtract $3$ from both sides of the equation.

$a = 4 + b - 3$

$9 = 9 a + 3 b + 3$

$c = 3$

Step 3.3.2.3

Subtract $3$ from $4$ .

$a = b + 1$

$9 = 9 a + 3 b + 3$

$c = 3$

$a = b + 1$

$9 = 9 a + 3 b + 3$

$c = 3$

$a = b + 1$

$9 = 9 a + 3 b + 3$

$c = 3$

Step 3.4

Replace all occurrences of $a$ with $b + 1$ in each equation.

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Step 3.4.1

Replace all occurrences of $a$ in $9 = 9 a + 3 b + 3$ with $b + 1$ .

$9 = 9 (b + 1) + 3 b + 3$

$a = b + 1$

$c = 3$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify $9 (b + 1) + 3 b + 3$ .

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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.

$9 = 9 b + 9 \cdot 1 + 3 b + 3$

$a = b + 1$

$c = 3$

Step 3.4.2.1.1.2

Multiply $9$ by $1$ .

$9 = 9 b + 9 + 3 b + 3$

$a = b + 1$

$c = 3$

$9 = 9 b + 9 + 3 b + 3$

$a = b + 1$

$c = 3$

Step 3.4.2.1.2

Simplify by adding terms.

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Step 3.4.2.1.2.1

Add $9 b$ and $3 b$ .

$9 = 12 b + 9 + 3$

$a = b + 1$

$c = 3$

Step 3.4.2.1.2.2

Add $9$ and $3$ .

$9 = 12 b + 12$

$a = b + 1$

$c = 3$

$9 = 12 b + 12$

$a = b + 1$

$c = 3$

$9 = 12 b + 12$

$a = b + 1$

$c = 3$

$9 = 12 b + 12$

$a = b + 1$

$c = 3$

$9 = 12 b + 12$

$a = b + 1$

$c = 3$

Step 3.5

Solve for $b$ in $9 = 12 b + 12$ .

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Step 3.5.1

Rewrite the equation as $12 b + 12 = 9$ .

$12 b + 12 = 9$

$a = b + 1$

$c = 3$

Step 3.5.2

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

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Step 3.5.2.1

Subtract $12$ from both sides of the equation.

$12 b = 9 - 12$

$a = b + 1$

$c = 3$

Step 3.5.2.2

Subtract $12$ from $9$ .

$12 b = - 3$

$a = b + 1$

$c = 3$

$12 b = - 3$

$a = b + 1$

$c = 3$

Step 3.5.3

Divide each term in $12 b = - 3$ by $12$ and simplify.

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Step 3.5.3.1

Divide each term in $12 b = - 3$ by $12$ .

$\frac{12 b}{12} = \frac{- 3}{12}$

$a = b + 1$

$c = 3$

Step 3.5.3.2

Simplify the left side.

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Step 3.5.3.2.1

Cancel the common factor of $12$ .

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Step 3.5.3.2.1.1

Cancel the common factor.

$\frac{12 b}{12} = \frac{- 3}{12}$

$a = b + 1$

$c = 3$

Step 3.5.3.2.1.2

Divide $b$ by $1$ .

$b = \frac{- 3}{12}$

$a = b + 1$

$c = 3$

$b = \frac{- 3}{12}$

$a = b + 1$

$c = 3$

$b = \frac{- 3}{12}$

$a = b + 1$

$c = 3$

Step 3.5.3.3

Simplify the right side.

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Step 3.5.3.3.1

Cancel the common factor of $- 3$ and $12$ .

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Step 3.5.3.3.1.1

Factor $3$ out of $- 3$ .

$b = \frac{3 (- 1)}{12}$

$a = b + 1$

$c = 3$

Step 3.5.3.3.1.2

Cancel the common factors.

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Step 3.5.3.3.1.2.1

Factor $3$ out of $12$ .

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

$a = b + 1$

$c = 3$

Step 3.5.3.3.1.2.2

Cancel the common factor.

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

$a = b + 1$

$c = 3$

Step 3.5.3.3.1.2.3

Rewrite the expression.

$b = \frac{- 1}{4}$

$a = b + 1$

$c = 3$

$b = \frac{- 1}{4}$

$a = b + 1$

$c = 3$

$b = \frac{- 1}{4}$

$a = b + 1$

$c = 3$

Step 3.5.3.3.2

Move the negative in front of the fraction.

$b = - \frac{1}{4}$

$a = b + 1$

$c = 3$

$b = - \frac{1}{4}$

$a = b + 1$

$c = 3$

$b = - \frac{1}{4}$

$a = b + 1$

$c = 3$

$b = - \frac{1}{4}$

$a = b + 1$

$c = 3$

Step 3.6

Replace all occurrences of $b$ with $- \frac{1}{4}$ in each equation.

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Step 3.6.1

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

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

$b = - \frac{1}{4}$

$c = 3$

Step 3.6.2

Simplify the right side.

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Step 3.6.2.1

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

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Step 3.6.2.1.1

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

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

$b = - \frac{1}{4}$

$c = 3$

Step 3.6.2.1.2

Combine the numerators over the common denominator.

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

$b = - \frac{1}{4}$

$c = 3$

Step 3.6.2.1.3

Add $- 1$ and $4$ .

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

$b = - \frac{1}{4}$

$c = 3$

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

$b = - \frac{1}{4}$

$c = 3$

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

$b = - \frac{1}{4}$

$c = 3$

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

$b = - \frac{1}{4}$

$c = 3$

Step 3.7

List all of the solutions.

$a = \frac{3}{4}, b = - \frac{1}{4}, c = 3$

Step 4

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

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

Step 5