Algebra Examples

$(- 1, \frac{1}{4})$ , $(0, 1)$ , $(1, 4)$

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.

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

Step 3

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

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Step 3.1

Solve for $c$ in $1 = a \cdot 0^{2} + c$ .

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Step 3.1.1

Rewrite the equation as $a \cdot 0^{2} + c = 1$ .

$a \cdot 0^{2} + c = 1$

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

$4 = a + b + c$

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 $0$ .

$a \cdot 0 + c = 1$

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

$4 = a + b + c$

Step 3.1.2.1.2

Multiply $a$ by $0$ .

$0 + c = 1$

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

$4 = a + b + c$

$0 + c = 1$

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

$4 = a + b + c$

Step 3.1.2.2

Add $0$ and $c$ .

$c = 1$

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

$4 = a + b + c$

$c = 1$

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

$4 = a + b + c$

$c = 1$

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

$4 = a + b + c$

Step 3.2

Replace all occurrences of $c$ with $1$ in each equation.

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Step 3.2.1

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

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

$c = 1$

$4 = a + b + c$

Step 3.2.2

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

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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.

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

$c = 1$

$4 = a + b + c$

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

$c = 1$

$4 = a + b + c$

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 $2$ .

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

$c = 1$

$4 = a + b + c$

Step 3.2.2.2.1.2

Multiply $a$ by $1$ .

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

$c = 1$

$4 = a + b + c$

Step 3.2.2.2.1.3

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

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

$c = 1$

$4 = a + b + c$

Step 3.2.2.2.1.4

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

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

$c = 1$

$4 = a + b + c$

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

$c = 1$

$4 = a + b + c$

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

$c = 1$

$4 = a + b + c$

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

$c = 1$

$4 = a + b + c$

Step 3.2.3

Replace all occurrences of $c$ in $4 = a + b + c$ with $1$ .

$4 = a + b + 1$

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

$c = 1$

Step 3.2.4

Simplify the left side.

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Step 3.2.4.1

Remove parentheses.

$4 = a + b + 1$

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

$c = 1$

$4 = a + b + 1$

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

$c = 1$

$4 = a + b + 1$

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

$c = 1$

Step 3.3

Solve for $a$ in $4 = a + b + 1$ .

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Step 3.3.1

Rewrite the equation as $a + b + 1 = 4$ .

$a + b + 1 = 4$

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

$c = 1$

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 $b$ from both sides of the equation.

$a + 1 = 4 - b$

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

$c = 1$

Step 3.3.2.2

Subtract $1$ from both sides of the equation.

$a = 4 - b - 1$

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

$c = 1$

Step 3.3.2.3

Subtract $1$ from $4$ .

$a = - b + 3$

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

$c = 1$

$a = - b + 3$

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

$c = 1$

$a = - b + 3$

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

$c = 1$

Step 3.4

Replace all occurrences of $a$ with $- b + 3$ in each equation.

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Step 3.4.1

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

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

$a = - b + 3$

$c = 1$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

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

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Step 3.4.2.1.1

Subtract $b$ from $- b$ .

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

$a = - b + 3$

$c = 1$

Step 3.4.2.1.2

Add $3$ and $1$ .

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

$a = - b + 3$

$c = 1$

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

$a = - b + 3$

$c = 1$

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

$a = - b + 3$

$c = 1$

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

$a = - b + 3$

$c = 1$

Step 3.5

Solve for $b$ in $\frac{1}{4} = - 2 b + 4$ .

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Step 3.5.1

Rewrite the equation as $- 2 b + 4 = \frac{1}{4}$ .

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

$a = - b + 3$

$c = 1$

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 $4$ from both sides of the equation.

$- 2 b = \frac{1}{4} - 4$

$a = - b + 3$

$c = 1$

Step 3.5.2.2

To write $- 4$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

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

$a = - b + 3$

$c = 1$

Step 3.5.2.3

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

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

$a = - b + 3$

$c = 1$

Step 3.5.2.4

Combine the numerators over the common denominator.

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

$a = - b + 3$

$c = 1$

Step 3.5.2.5

Simplify the numerator.

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Step 3.5.2.5.1

Multiply $- 4$ by $4$ .

$- 2 b = \frac{1 - 16}{4}$

$a = - b + 3$

$c = 1$

Step 3.5.2.5.2

Subtract $16$ from $1$ .

$- 2 b = \frac{- 15}{4}$

$a = - b + 3$

$c = 1$

$- 2 b = \frac{- 15}{4}$

$a = - b + 3$

$c = 1$

Step 3.5.2.6

Move the negative in front of the fraction.

$- 2 b = - \frac{15}{4}$

$a = - b + 3$

$c = 1$

$- 2 b = - \frac{15}{4}$

$a = - b + 3$

$c = 1$

Step 3.5.3

Divide each term in $- 2 b = - \frac{15}{4}$ by $- 2$ and simplify.

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Step 3.5.3.1

Divide each term in $- 2 b = - \frac{15}{4}$ by $- 2$ .

$\frac{- 2 b}{- 2} = \frac{- \frac{15}{4}}{- 2}$

$a = - b + 3$

$c = 1$

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.

$\frac{- 2 b}{- 2} = \frac{- \frac{15}{4}}{- 2}$

$a = - b + 3$

$c = 1$

Step 3.5.3.2.1.2

Divide $b$ by $1$ .

$b = \frac{- \frac{15}{4}}{- 2}$

$a = - b + 3$

$c = 1$

$b = \frac{- \frac{15}{4}}{- 2}$

$a = - b + 3$

$c = 1$

$b = \frac{- \frac{15}{4}}{- 2}$

$a = - b + 3$

$c = 1$

Step 3.5.3.3

Simplify the right side.

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Step 3.5.3.3.1

Multiply the numerator by the reciprocal of the denominator.

$b = - \frac{15}{4} \cdot \frac{1}{- 2}$

$a = - b + 3$

$c = 1$

Step 3.5.3.3.2

Move the negative in front of the fraction.

$b = - \frac{15}{4} \cdot (- \frac{1}{2})$

$a = - b + 3$

$c = 1$

Step 3.5.3.3.3

Multiply $- \frac{15}{4} (- \frac{1}{2})$ .

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Step 3.5.3.3.3.1

Multiply $- 1$ by $- 1$ .

$b = 1 (\frac{15}{4}) \cdot \frac{1}{2}$

$a = - b + 3$

$c = 1$

Step 3.5.3.3.3.2

Multiply $\frac{15}{4}$ by $1$ .

$b = \frac{15}{4} \cdot \frac{1}{2}$

$a = - b + 3$

$c = 1$

Step 3.5.3.3.3.3

Multiply $\frac{15}{4}$ by $\frac{1}{2}$ .

$b = \frac{15}{4 \cdot 2}$

$a = - b + 3$

$c = 1$

Step 3.5.3.3.3.4

Multiply $4$ by $2$ .

$b = \frac{15}{8}$

$a = - b + 3$

$c = 1$

$b = \frac{15}{8}$

$a = - b + 3$

$c = 1$

$b = \frac{15}{8}$

$a = - b + 3$

$c = 1$

$b = \frac{15}{8}$

$a = - b + 3$

$c = 1$

$b = \frac{15}{8}$

$a = - b + 3$

$c = 1$

Step 3.6

Replace all occurrences of $b$ with $\frac{15}{8}$ in each equation.

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Step 3.6.1

Replace all occurrences of $b$ in $a = - b + 3$ with $\frac{15}{8}$ .

$a = - (\frac{15}{8}) + 3$

$b = \frac{15}{8}$

$c = 1$

Step 3.6.2

Simplify the right side.

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Step 3.6.2.1

Simplify $- (\frac{15}{8}) + 3$ .

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Step 3.6.2.1.1

To write $3$ as a fraction with a common denominator, multiply by $\frac{8}{8}$ .

$a = - \frac{15}{8} + 3 \cdot \frac{8}{8}$

$b = \frac{15}{8}$

$c = 1$

Step 3.6.2.1.2

Combine $3$ and $\frac{8}{8}$ .

$a = - \frac{15}{8} + \frac{3 \cdot 8}{8}$

$b = \frac{15}{8}$

$c = 1$

Step 3.6.2.1.3

Combine the numerators over the common denominator.

$a = \frac{- 15 + 3 \cdot 8}{8}$

$b = \frac{15}{8}$

$c = 1$

Step 3.6.2.1.4

Simplify the numerator.

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Step 3.6.2.1.4.1

Multiply $3$ by $8$ .

$a = \frac{- 15 + 24}{8}$

$b = \frac{15}{8}$

$c = 1$

Step 3.6.2.1.4.2

Add $- 15$ and $24$ .

$a = \frac{9}{8}$

$b = \frac{15}{8}$

$c = 1$

$a = \frac{9}{8}$

$b = \frac{15}{8}$

$c = 1$

$a = \frac{9}{8}$

$b = \frac{15}{8}$

$c = 1$

$a = \frac{9}{8}$

$b = \frac{15}{8}$

$c = 1$

$a = \frac{9}{8}$

$b = \frac{15}{8}$

$c = 1$

Step 3.7

List all of the solutions.

$a = \frac{9}{8}, b = \frac{15}{8}, c = 1$

Step 4

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

$y = \frac{9 x^{2}}{8} + \frac{15 x}{8} + 1$

Step 5