Algebra Examples

$(0, 9)$ , $(5, 4)$ , $(1, 4)$

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.

$9 = a {(0)}^{2} + b (0) + c$

$4 = a {(5)}^{2} + b (5) + c$

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

Step 3

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

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Solve for $c$ in $9 = a \cdot 0^{2} + c$ .

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Rewrite the equation as $a \cdot 0^{2} + c = 9$ .

$a \cdot 0^{2} + c = 9$

$4 = a \cdot 5^{2} + b (5) + c$

$4 = a + b + c$

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

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Simplify each term.

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

$a \cdot 0 + c = 9$

$4 = a \cdot 5^{2} + b (5) + c$

$4 = a + b + c$

Multiply $a$ by $0$ .

$0 + c = 9$

$4 = a \cdot 5^{2} + b (5) + c$

$4 = a + b + c$

$0 + c = 9$

$4 = a \cdot 5^{2} + b (5) + c$

$4 = a + b + c$

Add $0$ and $c$ .

$c = 9$

$4 = a \cdot 5^{2} + b (5) + c$

$4 = a + b + c$

$c = 9$

$4 = a \cdot 5^{2} + b (5) + c$

$4 = a + b + c$

$c = 9$

$4 = a \cdot 5^{2} + b (5) + c$

$4 = a + b + c$

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

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Replace all occurrences of $c$ in $4 = a \cdot 5^{2} + b (5) + c$ with $9$ .

$4 = a \cdot 5^{2} + b (5) + 9$

$c = 9$

$4 = a + b + c$

Simplify $4 = a \cdot 5^{2} + b (5) + 9$ .

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Simplify the left side.

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Remove parentheses.

$4 = a \cdot 5^{2} + b (5) + 9$

$c = 9$

$4 = a + b + c$

$4 = a \cdot 5^{2} + b (5) + 9$

$c = 9$

$4 = a + b + c$

Simplify the right side.

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Simplify each term.

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Raise $5$ to the power of $2$ .

$4 = a \cdot 25 + b (5) + 9$

$c = 9$

$4 = a + b + c$

Move $25$ to the left of $a$ .

$4 = 25 \cdot a + b (5) + 9$

$c = 9$

$4 = a + b + c$

Move $5$ to the left of $b$ .

$4 = 25 a + 5 b + 9$

$c = 9$

$4 = a + b + c$

$4 = 25 a + 5 b + 9$

$c = 9$

$4 = a + b + c$

$4 = 25 a + 5 b + 9$

$c = 9$

$4 = a + b + c$

$4 = 25 a + 5 b + 9$

$c = 9$

$4 = a + b + c$

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

$4 = a + b + 9$

$4 = 25 a + 5 b + 9$

$c = 9$

Simplify the left side.

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Remove parentheses.

$4 = a + b + 9$

$4 = 25 a + 5 b + 9$

$c = 9$

$4 = a + b + 9$

$4 = 25 a + 5 b + 9$

$c = 9$

$4 = a + b + 9$

$4 = 25 a + 5 b + 9$

$c = 9$

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

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

$a + b + 9 = 4$

$4 = 25 a + 5 b + 9$

$c = 9$

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

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Subtract $b$ from both sides of the equation.

$a + 9 = 4 - b$

$4 = 25 a + 5 b + 9$

$c = 9$

Subtract $9$ from both sides of the equation.

$a = 4 - b - 9$

$4 = 25 a + 5 b + 9$

$c = 9$

Subtract $9$ from $4$ .

$a = - b - 5$

$4 = 25 a + 5 b + 9$

$c = 9$

$a = - b - 5$

$4 = 25 a + 5 b + 9$

$c = 9$

$a = - b - 5$

$4 = 25 a + 5 b + 9$

$c = 9$

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

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Replace all occurrences of $a$ in $4 = 25 a + 5 b + 9$ with $- b - 5$ .

$4 = 25 (- b - 5) + 5 b + 9$

$a = - b - 5$

$c = 9$

Simplify the right side.

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Simplify $25 (- b - 5) + 5 b + 9$ .

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Simplify each term.

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Apply the distributive property.

$4 = 25 (- b) + 25 \cdot - 5 + 5 b + 9$

$a = - b - 5$

$c = 9$

Multiply $- 1$ by $25$ .

$4 = - 25 b + 25 \cdot - 5 + 5 b + 9$

$a = - b - 5$

$c = 9$

Multiply $25$ by $- 5$ .

$4 = - 25 b - 125 + 5 b + 9$

$a = - b - 5$

$c = 9$

$4 = - 25 b - 125 + 5 b + 9$

$a = - b - 5$

$c = 9$

Simplify by adding terms.

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Add $- 25 b$ and $5 b$ .

$4 = - 20 b - 125 + 9$

$a = - b - 5$

$c = 9$

Add $- 125$ and $9$ .

$4 = - 20 b - 116$

$a = - b - 5$

$c = 9$

$4 = - 20 b - 116$

$a = - b - 5$

$c = 9$

$4 = - 20 b - 116$

$a = - b - 5$

$c = 9$

$4 = - 20 b - 116$

$a = - b - 5$

$c = 9$

$4 = - 20 b - 116$

$a = - b - 5$

$c = 9$

Solve for $b$ in $4 = - 20 b - 116$ .

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Rewrite the equation as $- 20 b - 116 = 4$ .

$- 20 b - 116 = 4$

$a = - b - 5$

$c = 9$

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

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Add $116$ to both sides of the equation.

$- 20 b = 4 + 116$

$a = - b - 5$

$c = 9$

Add $4$ and $116$ .

$- 20 b = 120$

$a = - b - 5$

$c = 9$

$- 20 b = 120$

$a = - b - 5$

$c = 9$

Divide each term in $- 20 b = 120$ by $- 20$ and simplify.

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Divide each term in $- 20 b = 120$ by $- 20$ .

$\frac{- 20 b}{- 20} = \frac{120}{- 20}$

$a = - b - 5$

$c = 9$

Simplify the left side.

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Cancel the common factor of $- 20$ .

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Cancel the common factor.

$\frac{- 20 b}{- 20} = \frac{120}{- 20}$

$a = - b - 5$

$c = 9$

Divide $b$ by $1$ .

$b = \frac{120}{- 20}$

$a = - b - 5$

$c = 9$

$b = \frac{120}{- 20}$

$a = - b - 5$

$c = 9$

$b = \frac{120}{- 20}$

$a = - b - 5$

$c = 9$

Simplify the right side.

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Divide $120$ by $- 20$ .

$b = - 6$

$a = - b - 5$

$c = 9$

$b = - 6$

$a = - b - 5$

$c = 9$

$b = - 6$

$a = - b - 5$

$c = 9$

$b = - 6$

$a = - b - 5$

$c = 9$

Replace all occurrences of $b$ with $- 6$ in each equation.

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Replace all occurrences of $b$ in $a = - b - 5$ with $- 6$ .

$a = - (- 6) - 5$

$b = - 6$

$c = 9$

Simplify the right side.

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Simplify $- (- 6) - 5$ .

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Multiply $- 1$ by $- 6$ .

$a = 6 - 5$

$b = - 6$

$c = 9$

Subtract $5$ from $6$ .

$a = 1$

$b = - 6$

$c = 9$

$a = 1$

$b = - 6$

$c = 9$

$a = 1$

$b = - 6$

$c = 9$

$a = 1$

$b = - 6$

$c = 9$

List all of the solutions.

$a = 1, b = - 6, c = 9$

Step 4

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

$y = x^{2} - 6 x + 9$

Step 5

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