Finite Math Examples

$f (x) = 5 x^{2} - 5 x + 1$

The minimum of a quadratic function occurs at $x = - \frac{b}{2 a}$ . If $a$ is positive, the minimum value of the function is $f (- \frac{b}{2 a})$ .

$f_{min}$ $x = a x^{2} + b x + c$ occurs at $x = - \frac{b}{2 a}$

Find the value of $x = - \frac{b}{2 a}$ .

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Substitute in the values of $a$ and $b$ .

$x = - \frac{- 5}{2 (5)}$

Remove parentheses.

$x = - \frac{- 5}{2 (5)}$

Simplify $- \frac{- 5}{2 (5)}$ .

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Cancel the common factor of $- 5$ and $5$ .

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Factor $5$ out of $- 5$ .

$x = - \frac{5 \cdot - 1}{2 \cdot 5}$

Cancel the common factors.

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Factor $5$ out of $2 \cdot 5$ .

$x = - \frac{5 \cdot - 1}{5 \cdot 2}$

Cancel the common factor.

$x = - \frac{5 \cdot - 1}{5 \cdot 2}$

Rewrite the expression.

$x = - \frac{- 1}{2}$

Move the negative in front of the fraction.

$x = - - \frac{1}{2}$

Multiply $- - \frac{1}{2}$ .

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Multiply $- 1$ by $- 1$ .

$x = 1 (\frac{1}{2})$

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

$x = \frac{1}{2}$

Evaluate $f (\frac{1}{2})$ .

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Replace the variable $x$ with $\frac{1}{2}$ in the expression.

$f (\frac{1}{2}) = 5 {(\frac{1}{2})}^{2} - 5 (\frac{1}{2}) + 1$

Simplify the result.

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Simplify each term.

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Apply the product rule to $\frac{1}{2}$ .

$f (\frac{1}{2}) = 5 (\frac{1^{2}}{2^{2}}) - 5 (\frac{1}{2}) + 1$

One to any power is one.

$f (\frac{1}{2}) = 5 (\frac{1}{2^{2}}) - 5 (\frac{1}{2}) + 1$

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

$f (\frac{1}{2}) = 5 (\frac{1}{4}) - 5 (\frac{1}{2}) + 1$

Combine $5$ and $\frac{1}{4}$ .

$f (\frac{1}{2}) = \frac{5}{4} - 5 (\frac{1}{2}) + 1$

Combine $- 5$ and $\frac{1}{2}$ .

$f (\frac{1}{2}) = \frac{5}{4} + \frac{- 5}{2} + 1$

Move the negative in front of the fraction.

$f (\frac{1}{2}) = \frac{5}{4} - \frac{5}{2} + 1$

Find the common denominator.

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Multiply $\frac{5}{2}$ by $\frac{2}{2}$ .

$f (\frac{1}{2}) = \frac{5}{4} - (\frac{5}{2} \cdot \frac{2}{2}) + 1$

Multiply $\frac{5}{2}$ and $\frac{2}{2}$ .

$f (\frac{1}{2}) = \frac{5}{4} - \frac{5 \cdot 2}{2 \cdot 2} + 1$

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

$f (\frac{1}{2}) = \frac{5}{4} - \frac{5 \cdot 2}{2 \cdot 2} + \frac{1}{1}$

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

$f (\frac{1}{2}) = \frac{5}{4} - \frac{5 \cdot 2}{2 \cdot 2} + \frac{1}{1} \cdot \frac{4}{4}$

Multiply $\frac{1}{1}$ and $\frac{4}{4}$ .

$f (\frac{1}{2}) = \frac{5}{4} - \frac{5 \cdot 2}{2 \cdot 2} + \frac{4}{4}$

Multiply $2$ by $2$ .

$f (\frac{1}{2}) = \frac{5}{4} - \frac{5 \cdot 2}{4} + \frac{4}{4}$

Combine fractions.

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Combine fractions with similar denominators.

$f (\frac{1}{2}) = \frac{5 - 5 \cdot 2 + 4}{4}$

Multiply $- 5$ by $2$ .

$f (\frac{1}{2}) = \frac{5 - 10 + 4}{4}$

Simplify the numerator.

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Subtract $10$ from $5$ .

$f (\frac{1}{2}) = \frac{- 5 + 4}{4}$

Add $- 5$ and $4$ .

$f (\frac{1}{2}) = \frac{- 1}{4}$

Move the negative in front of the fraction.

$f (\frac{1}{2}) = - \frac{1}{4}$

The final answer is $- \frac{1}{4}$ .

$- \frac{1}{4}$

Use the $x$ and $y$ values to find where the minimum occurs.

$(\frac{1}{2}, - \frac{1}{4})$

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