Finite Math Examples

$(- 17, 15)$ , $(0, 22)$ , $(17, 15)$

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.

$15 = a {(- 17)}^{2} + b (- 17) + c, 22 = a {(0)}^{2} + b (0) + c, 15 = a {(17)}^{2} + b (17) + c$

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 $22 = a \cdot 0^{2} + c$ .

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Step 3.1.1

Rewrite the equation as $a \cdot 0^{2} + c = 22$ .

$a \cdot 0^{2} + c = 22$

$15 = a {(- 17)}^{2} + b (- 17) + c$

$15 = a \cdot 17^{2} + b (17) + c$

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 $0$ .

$a \cdot 0 + c = 22$

$15 = a {(- 17)}^{2} + b (- 17) + c$

$15 = a \cdot 17^{2} + b (17) + c$

Step 3.1.2.1.2

Multiply $a$ by $0$ .

$0 + c = 22$

$15 = a {(- 17)}^{2} + b (- 17) + c$

$15 = a \cdot 17^{2} + b (17) + c$

$0 + c = 22$

$15 = a {(- 17)}^{2} + b (- 17) + c$

$15 = a \cdot 17^{2} + b (17) + c$

Step 3.1.2.2

Add $0$ and $c$ .

$c = 22$

$15 = a {(- 17)}^{2} + b (- 17) + c$

$15 = a \cdot 17^{2} + b (17) + c$

$c = 22$

$15 = a {(- 17)}^{2} + b (- 17) + c$

$15 = a \cdot 17^{2} + b (17) + c$

$c = 22$

$15 = a {(- 17)}^{2} + b (- 17) + c$

$15 = a \cdot 17^{2} + b (17) + c$

Step 3.2

Replace all occurrences of $c$ with $22$ in each equation.

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Step 3.2.1

Replace all occurrences of $c$ in $15 = a {(- 17)}^{2} + b (- 17) + c$ with $22$ .

$15 = a {(- 17)}^{2} + b (- 17) + 22$

$c = 22$

$15 = a \cdot 17^{2} + b (17) + c$

Step 3.2.2

Simplify $15 = a {(- 17)}^{2} + b (- 17) + 22$ .

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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.

$15 = a {(- 17)}^{2} + b (- 17) + 22$

$c = 22$

$15 = a \cdot 17^{2} + b (17) + c$

$15 = a {(- 17)}^{2} + b (- 17) + 22$

$c = 22$

$15 = a \cdot 17^{2} + b (17) + c$

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 $- 17$ to the power of $2$ .

$15 = a \cdot 289 + b (- 17) + 22$

$c = 22$

$15 = a \cdot 17^{2} + b (17) + c$

Step 3.2.2.2.1.2

Move $289$ to the left of $a$ .

$15 = 289 \cdot a + b (- 17) + 22$

$c = 22$

$15 = a \cdot 17^{2} + b (17) + c$

Step 3.2.2.2.1.3

Move $- 17$ to the left of $b$ .

$15 = 289 a - 17 b + 22$

$c = 22$

$15 = a \cdot 17^{2} + b (17) + c$

$15 = 289 a - 17 b + 22$

$c = 22$

$15 = a \cdot 17^{2} + b (17) + c$

$15 = 289 a - 17 b + 22$

$c = 22$

$15 = a \cdot 17^{2} + b (17) + c$

$15 = 289 a - 17 b + 22$

$c = 22$

$15 = a \cdot 17^{2} + b (17) + c$

Step 3.2.3

Replace all occurrences of $c$ in $15 = a \cdot 17^{2} + b (17) + c$ with $22$ .

$15 = a \cdot 17^{2} + b (17) + 22$

$15 = 289 a - 17 b + 22$

$c = 22$

Step 3.2.4

Simplify $15 = a \cdot 17^{2} + b (17) + 22$ .

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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.

$15 = a \cdot 17^{2} + b (17) + 22$

$15 = 289 a - 17 b + 22$

$c = 22$

$15 = a \cdot 17^{2} + b (17) + 22$

$15 = 289 a - 17 b + 22$

$c = 22$

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 $17$ to the power of $2$ .

$15 = a \cdot 289 + b (17) + 22$

$15 = 289 a - 17 b + 22$

$c = 22$

Step 3.2.4.2.1.2

Move $289$ to the left of $a$ .

$15 = 289 \cdot a + b (17) + 22$

$15 = 289 a - 17 b + 22$

$c = 22$

Step 3.2.4.2.1.3

Move $17$ to the left of $b$ .

$15 = 289 a + 17 b + 22$

$15 = 289 a - 17 b + 22$

$c = 22$

$15 = 289 a + 17 b + 22$

$15 = 289 a - 17 b + 22$

$c = 22$

$15 = 289 a + 17 b + 22$

$15 = 289 a - 17 b + 22$

$c = 22$

$15 = 289 a + 17 b + 22$

$15 = 289 a - 17 b + 22$

$c = 22$

$15 = 289 a + 17 b + 22$

$15 = 289 a - 17 b + 22$

$c = 22$

Step 3.3

Solve for $a$ in $15 = 289 a + 17 b + 22$ .

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Step 3.3.1

Rewrite the equation as $289 a + 17 b + 22 = 15$ .

$289 a + 17 b + 22 = 15$

$15 = 289 a - 17 b + 22$

$c = 22$

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 $17 b$ from both sides of the equation.

$289 a + 22 = 15 - 17 b$

$15 = 289 a - 17 b + 22$

$c = 22$

Step 3.3.2.2

Subtract $22$ from both sides of the equation.

$289 a = 15 - 17 b - 22$

$15 = 289 a - 17 b + 22$

$c = 22$

Step 3.3.2.3

Subtract $22$ from $15$ .

$289 a = - 17 b - 7$

$15 = 289 a - 17 b + 22$

$c = 22$

$289 a = - 17 b - 7$

$15 = 289 a - 17 b + 22$

$c = 22$

Step 3.3.3

Divide each term in $289 a = - 17 b - 7$ by $289$ and simplify.

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Step 3.3.3.1

Divide each term in $289 a = - 17 b - 7$ by $289$ .

$\frac{289 a}{289} = \frac{- 17 b}{289} + \frac{- 7}{289}$

$15 = 289 a - 17 b + 22$

$c = 22$

Step 3.3.3.2

Simplify the left side.

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Step 3.3.3.2.1

Cancel the common factor of $289$ .

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Step 3.3.3.2.1.1

Cancel the common factor.

$\frac{289 a}{289} = \frac{- 17 b}{289} + \frac{- 7}{289}$

$15 = 289 a - 17 b + 22$

$c = 22$

Step 3.3.3.2.1.2

Divide $a$ by $1$ .

$a = \frac{- 17 b}{289} + \frac{- 7}{289}$

$15 = 289 a - 17 b + 22$

$c = 22$

$a = \frac{- 17 b}{289} + \frac{- 7}{289}$

$15 = 289 a - 17 b + 22$

$c = 22$

$a = \frac{- 17 b}{289} + \frac{- 7}{289}$

$15 = 289 a - 17 b + 22$

$c = 22$

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 $- 17$ and $289$ .

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Step 3.3.3.3.1.1.1

Factor $17$ out of $- 17 b$ .

$a = \frac{17 (- b)}{289} + \frac{- 7}{289}$

$15 = 289 a - 17 b + 22$

$c = 22$

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 $17$ out of $289$ .

$a = \frac{17 (- b)}{17 (17)} + \frac{- 7}{289}$

$15 = 289 a - 17 b + 22$

$c = 22$

Step 3.3.3.3.1.1.2.2

Cancel the common factor.

$a = \frac{17 (- b)}{17 \cdot 17} + \frac{- 7}{289}$

$15 = 289 a - 17 b + 22$

$c = 22$

Step 3.3.3.3.1.1.2.3

Rewrite the expression.

$a = \frac{- b}{17} + \frac{- 7}{289}$

$15 = 289 a - 17 b + 22$

$c = 22$

$a = \frac{- b}{17} + \frac{- 7}{289}$

$15 = 289 a - 17 b + 22$

$c = 22$

$a = \frac{- b}{17} + \frac{- 7}{289}$

$15 = 289 a - 17 b + 22$

$c = 22$

Step 3.3.3.3.1.2

Move the negative in front of the fraction.

$a = - \frac{b}{17} + \frac{- 7}{289}$

$15 = 289 a - 17 b + 22$

$c = 22$

Step 3.3.3.3.1.3

Move the negative in front of the fraction.

$a = - \frac{b}{17} - \frac{7}{289}$

$15 = 289 a - 17 b + 22$

$c = 22$

$a = - \frac{b}{17} - \frac{7}{289}$

$15 = 289 a - 17 b + 22$

$c = 22$

$a = - \frac{b}{17} - \frac{7}{289}$

$15 = 289 a - 17 b + 22$

$c = 22$

$a = - \frac{b}{17} - \frac{7}{289}$

$15 = 289 a - 17 b + 22$

$c = 22$

$a = - \frac{b}{17} - \frac{7}{289}$

$15 = 289 a - 17 b + 22$

$c = 22$

Step 3.4

Replace all occurrences of $a$ with $- \frac{b}{17} - \frac{7}{289}$ in each equation.

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Step 3.4.1

Replace all occurrences of $a$ in $15 = 289 a - 17 b + 22$ with $- \frac{b}{17} - \frac{7}{289}$ .

$15 = 289 (- \frac{b}{17} - \frac{7}{289}) - 17 b + 22$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify $289 (- \frac{b}{17} - \frac{7}{289}) - 17 b + 22$ .

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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.

$15 = 289 (- \frac{b}{17}) + 289 (- \frac{7}{289}) - 17 b + 22$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

Step 3.4.2.1.1.2

Cancel the common factor of $17$ .

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Step 3.4.2.1.1.2.1

Move the leading negative in $- \frac{b}{17}$ into the numerator.

$15 = 289 (\frac{- b}{17}) + 289 (- \frac{7}{289}) - 17 b + 22$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

Step 3.4.2.1.1.2.2

Factor $17$ out of $289$ .

$15 = 17 (17) (\frac{- b}{17}) + 289 (- \frac{7}{289}) - 17 b + 22$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

Step 3.4.2.1.1.2.3

Cancel the common factor.

$15 = 17 \cdot (17 (\frac{- b}{17})) + 289 (- \frac{7}{289}) - 17 b + 22$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

Step 3.4.2.1.1.2.4

Rewrite the expression.

$15 = 17 (- b) + 289 (- \frac{7}{289}) - 17 b + 22$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

$15 = 17 (- b) + 289 (- \frac{7}{289}) - 17 b + 22$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

Step 3.4.2.1.1.3

Multiply $- 1$ by $17$ .

$15 = - 17 b + 289 (- \frac{7}{289}) - 17 b + 22$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

Step 3.4.2.1.1.4

Cancel the common factor of $289$ .

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Step 3.4.2.1.1.4.1

Move the leading negative in $- \frac{7}{289}$ into the numerator.

$15 = - 17 b + 289 (\frac{- 7}{289}) - 17 b + 22$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

Step 3.4.2.1.1.4.2

Cancel the common factor.

$15 = - 17 b + 289 (\frac{- 7}{289}) - 17 b + 22$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

Step 3.4.2.1.1.4.3

Rewrite the expression.

$15 = - 17 b - 7 - 17 b + 22$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

$15 = - 17 b - 7 - 17 b + 22$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

$15 = - 17 b - 7 - 17 b + 22$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

Step 3.4.2.1.2

Simplify by adding terms.

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Step 3.4.2.1.2.1

Subtract $17 b$ from $- 17 b$ .

$15 = - 34 b - 7 + 22$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

Step 3.4.2.1.2.2

Add $- 7$ and $22$ .

$15 = - 34 b + 15$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

$15 = - 34 b + 15$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

$15 = - 34 b + 15$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

$15 = - 34 b + 15$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

$15 = - 34 b + 15$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

Step 3.5

Solve for $b$ in $15 = - 34 b + 15$ .

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Step 3.5.1

Rewrite the equation as $- 34 b + 15 = 15$ .

$- 34 b + 15 = 15$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

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 $15$ from both sides of the equation.

$- 34 b = 15 - 15$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

Step 3.5.2.2

Subtract $15$ from $15$ .

$- 34 b = 0$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

$- 34 b = 0$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

Step 3.5.3

Divide each term in $- 34 b = 0$ by $- 34$ and simplify.

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Step 3.5.3.1

Divide each term in $- 34 b = 0$ by $- 34$ .

$\frac{- 34 b}{- 34} = \frac{0}{- 34}$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

Step 3.5.3.2

Simplify the left side.

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Step 3.5.3.2.1

Cancel the common factor of $- 34$ .

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Step 3.5.3.2.1.1

Cancel the common factor.

$\frac{- 34 b}{- 34} = \frac{0}{- 34}$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

Step 3.5.3.2.1.2

Divide $b$ by $1$ .

$b = \frac{0}{- 34}$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

$b = \frac{0}{- 34}$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

$b = \frac{0}{- 34}$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

Step 3.5.3.3

Simplify the right side.

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Step 3.5.3.3.1

Divide $0$ by $- 34$ .

$b = 0$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

$b = 0$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

$b = 0$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

$b = 0$

$a = - \frac{b}{17} - \frac{7}{289}$

$c = 22$

Step 3.6

Replace all occurrences of $b$ with $0$ in each equation.

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Step 3.6.1

Replace all occurrences of $b$ in $a = - \frac{b}{17} - \frac{7}{289}$ with $0$ .

$a = - \frac{0}{17} - \frac{7}{289}$

$b = 0$

$c = 22$

Step 3.6.2

Simplify the right side.

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Step 3.6.2.1

Simplify $- \frac{0}{17} - \frac{7}{289}$ .

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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 $0$ by $17$ .

$a = - 0 - \frac{7}{289}$

$b = 0$

$c = 22$

Step 3.6.2.1.1.2

Multiply $- 1$ by $0$ .

$a = 0 - \frac{7}{289}$

$b = 0$

$c = 22$

$a = 0 - \frac{7}{289}$

$b = 0$

$c = 22$

Step 3.6.2.1.2

Subtract $\frac{7}{289}$ from $0$ .

$a = - \frac{7}{289}$

$b = 0$

$c = 22$

$a = - \frac{7}{289}$

$b = 0$

$c = 22$

$a = - \frac{7}{289}$

$b = 0$

$c = 22$

$a = - \frac{7}{289}$

$b = 0$

$c = 22$

Step 3.7

List all of the solutions.

$a = - \frac{7}{289}, b = 0, c = 22$

Step 4

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

$y = - \frac{7 x^{2}}{289} + 22$

Step 5