Finite Math Examples

$(0, 3)$ , $(1, 4)$ , $(- 1, - 6)$

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.

$3 = a {(0)}^{2} + b (0) + c, 4 = a {(1)}^{2} + b (1) + c, - 6 = a {(- 1)}^{2} + b (- 1) + c$

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

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

$4 = a + b + c$

$- 6 = a {(- 1)}^{2} + b (- 1) + c$

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 $0$ .

$a \cdot 0 + c = 3$

$4 = a + b + c$

$- 6 = a {(- 1)}^{2} + b (- 1) + c$

Step 3.1.2.1.2

Multiply $a$ by $0$ .

$0 + c = 3$

$4 = a + b + c$

$- 6 = a {(- 1)}^{2} + b (- 1) + c$

$0 + c = 3$

$4 = a + b + c$

$- 6 = a {(- 1)}^{2} + b (- 1) + c$

Step 3.1.2.2

Add $0$ and $c$ .

$c = 3$

$4 = a + b + c$

$- 6 = a {(- 1)}^{2} + b (- 1) + c$

$c = 3$

$4 = a + b + c$

$- 6 = a {(- 1)}^{2} + b (- 1) + c$

$c = 3$

$4 = a + b + c$

$- 6 = a {(- 1)}^{2} + b (- 1) + c$

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 $4 = a + b + c$ with $3$ .

$4 = a + b + 3$

$c = 3$

$- 6 = a {(- 1)}^{2} + b (- 1) + c$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Remove parentheses.

$4 = a + b + 3$

$c = 3$

$- 6 = a {(- 1)}^{2} + b (- 1) + c$

$4 = a + b + 3$

$c = 3$

$- 6 = a {(- 1)}^{2} + b (- 1) + c$

Step 3.2.3

Replace all occurrences of $c$ in $- 6 = a {(- 1)}^{2} + b (- 1) + c$ with $3$ .

$- 6 = a {(- 1)}^{2} + b (- 1) + 3$

$4 = a + b + 3$

$c = 3$

Step 3.2.4

Simplify $- 6 = a {(- 1)}^{2} + b (- 1) + 3$ .

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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.

$- 6 = a {(- 1)}^{2} + b (- 1) + 3$

$4 = a + b + 3$

$c = 3$

$- 6 = a {(- 1)}^{2} + b (- 1) + 3$

$4 = a + b + 3$

$c = 3$

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

$- 6 = a \cdot 1 + b (- 1) + 3$

$4 = a + b + 3$

$c = 3$

Step 3.2.4.2.1.2

Multiply $a$ by $1$ .

$- 6 = a + b (- 1) + 3$

$4 = a + b + 3$

$c = 3$

Step 3.2.4.2.1.3

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

$- 6 = a - 1 b + 3$

$4 = a + b + 3$

$c = 3$

Step 3.2.4.2.1.4

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

$- 6 = a - b + 3$

$4 = a + b + 3$

$c = 3$

$- 6 = a - b + 3$

$4 = a + b + 3$

$c = 3$

$- 6 = a - b + 3$

$4 = a + b + 3$

$c = 3$

$- 6 = a - b + 3$

$4 = a + b + 3$

$c = 3$

$- 6 = a - b + 3$

$4 = a + b + 3$

$c = 3$

Step 3.3

Solve for $a$ in $- 6 = a - b + 3$ .

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Step 3.3.1

Rewrite the equation as $a - b + 3 = - 6$ .

$a - b + 3 = - 6$

$4 = a + b + 3$

$c = 3$

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.

$a + 3 = - 6 + b$

$4 = a + b + 3$

$c = 3$

Step 3.3.2.2

Subtract $3$ from both sides of the equation.

$a = - 6 + b - 3$

$4 = a + b + 3$

$c = 3$

Step 3.3.2.3

Subtract $3$ from $- 6$ .

$a = b - 9$

$4 = a + b + 3$

$c = 3$

$a = b - 9$

$4 = a + b + 3$

$c = 3$

$a = b - 9$

$4 = a + b + 3$

$c = 3$

Step 3.4

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

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Step 3.4.1

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

$4 = (b - 9) + b + 3$

$a = b - 9$

$c = 3$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

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

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Step 3.4.2.1.1

Add $b$ and $b$ .

$4 = 2 b - 9 + 3$

$a = b - 9$

$c = 3$

Step 3.4.2.1.2

Add $- 9$ and $3$ .

$4 = 2 b - 6$

$a = b - 9$

$c = 3$

$4 = 2 b - 6$

$a = b - 9$

$c = 3$

$4 = 2 b - 6$

$a = b - 9$

$c = 3$

$4 = 2 b - 6$

$a = b - 9$

$c = 3$

Step 3.5

Solve for $b$ in $4 = 2 b - 6$ .

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Step 3.5.1

Rewrite the equation as $2 b - 6 = 4$ .

$2 b - 6 = 4$

$a = b - 9$

$c = 3$

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

Add $6$ to both sides of the equation.

$2 b = 4 + 6$

$a = b - 9$

$c = 3$

Step 3.5.2.2

Add $4$ and $6$ .

$2 b = 10$

$a = b - 9$

$c = 3$

$2 b = 10$

$a = b - 9$

$c = 3$

Step 3.5.3

Divide each term in $2 b = 10$ by $2$ and simplify.

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Step 3.5.3.1

Divide each term in $2 b = 10$ by $2$ .

$\frac{2 b}{2} = \frac{10}{2}$

$a = b - 9$

$c = 3$

Step 3.5.3.2

Simplify the left side.

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Step 3.5.3.2.1

Cancel the common factor of $2$ .

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Step 3.5.3.2.1.1

Cancel the common factor.

$\frac{2 b}{2} = \frac{10}{2}$

$a = b - 9$

$c = 3$

Step 3.5.3.2.1.2

Divide $b$ by $1$ .

$b = \frac{10}{2}$

$a = b - 9$

$c = 3$

$b = \frac{10}{2}$

$a = b - 9$

$c = 3$

$b = \frac{10}{2}$

$a = b - 9$

$c = 3$

Step 3.5.3.3

Simplify the right side.

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Step 3.5.3.3.1

Divide $10$ by $2$ .

$b = 5$

$a = b - 9$

$c = 3$

$b = 5$

$a = b - 9$

$c = 3$

$b = 5$

$a = b - 9$

$c = 3$

$b = 5$

$a = b - 9$

$c = 3$

Step 3.6

Replace all occurrences of $b$ with $5$ in each equation.

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Step 3.6.1

Replace all occurrences of $b$ in $a = b - 9$ with $5$ .

$a = (5) - 9$

$b = 5$

$c = 3$

Step 3.6.2

Simplify the right side.

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Step 3.6.2.1

Subtract $9$ from $5$ .

$a = - 4$

$b = 5$

$c = 3$

$a = - 4$

$b = 5$

$c = 3$

$a = - 4$

$b = 5$

$c = 3$

Step 3.7

List all of the solutions.

$a = - 4, b = 5, c = 3$

Step 4

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

$y = - 4 x^{2} + 5 x + 3$

Step 5