Precalculus Examples

$f (x) = x^{2} - 5 x + 6$

Step 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 $x = - \frac{b}{2 a}$

Step 2

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

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Step 2.1

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

$x = - \frac{- 5}{2 (1)}$

Step 2.2

Remove parentheses.

$x = - \frac{- 5}{2 (1)}$

Step 2.3

Simplify $- \frac{- 5}{2 (1)}$ .

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Step 2.3.1

Multiply $2$ by $1$ .

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

Step 2.3.2

Move the negative in front of the fraction.

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

Step 2.3.3

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

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Step 2.3.3.1

Multiply $- 1$ by $- 1$ .

$x = 1 (\frac{5}{2})$

Step 2.3.3.2

Multiply $\frac{5}{2}$ by $1$ .

$x = \frac{5}{2}$

Step 3

Evaluate $f (\frac{5}{2})$ .

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Step 3.1

Replace the variable $x$ with $\frac{5}{2}$ in the expression.

$f (\frac{5}{2}) = {(\frac{5}{2})}^{2} - 5 (\frac{5}{2}) + 6$

Step 3.2

Simplify the result.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Apply the product rule to $\frac{5}{2}$ .

$f (\frac{5}{2}) = \frac{5^{2}}{2^{2}} - 5 (\frac{5}{2}) + 6$

Step 3.2.1.2

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

$f (\frac{5}{2}) = \frac{25}{2^{2}} - 5 (\frac{5}{2}) + 6$

Step 3.2.1.3

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

$f (\frac{5}{2}) = \frac{25}{4} - 5 (\frac{5}{2}) + 6$

Step 3.2.1.4

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

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Step 3.2.1.4.1

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

$f (\frac{5}{2}) = \frac{25}{4} + \frac{- 5 \cdot 5}{2} + 6$

Step 3.2.1.4.2

Multiply $- 5$ by $5$ .

$f (\frac{5}{2}) = \frac{25}{4} + \frac{- 25}{2} + 6$

Step 3.2.1.5

Move the negative in front of the fraction.

$f (\frac{5}{2}) = \frac{25}{4} - \frac{25}{2} + 6$

Step 3.2.2

Find the common denominator.

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Step 3.2.2.1

Multiply $\frac{25}{2}$ by $\frac{2}{2}$ .

$f (\frac{5}{2}) = \frac{25}{4} - (\frac{25}{2} \cdot \frac{2}{2}) + 6$

Step 3.2.2.2

Multiply $\frac{25}{2}$ by $\frac{2}{2}$ .

$f (\frac{5}{2}) = \frac{25}{4} - \frac{25 \cdot 2}{2 \cdot 2} + 6$

Step 3.2.2.3

Write $6$ as a fraction with denominator $1$ .

$f (\frac{5}{2}) = \frac{25}{4} - \frac{25 \cdot 2}{2 \cdot 2} + \frac{6}{1}$

Step 3.2.2.4

Multiply $\frac{6}{1}$ by $\frac{4}{4}$ .

$f (\frac{5}{2}) = \frac{25}{4} - \frac{25 \cdot 2}{2 \cdot 2} + \frac{6}{1} \cdot \frac{4}{4}$

Step 3.2.2.5

Multiply $\frac{6}{1}$ by $\frac{4}{4}$ .

$f (\frac{5}{2}) = \frac{25}{4} - \frac{25 \cdot 2}{2 \cdot 2} + \frac{6 \cdot 4}{4}$

Step 3.2.2.6

Multiply $2$ by $2$ .

$f (\frac{5}{2}) = \frac{25}{4} - \frac{25 \cdot 2}{4} + \frac{6 \cdot 4}{4}$

Step 3.2.3

Combine the numerators over the common denominator.

$f (\frac{5}{2}) = \frac{25 - 25 \cdot 2 + 6 \cdot 4}{4}$

Step 3.2.4

Simplify each term.

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Step 3.2.4.1

Multiply $- 25$ by $2$ .

$f (\frac{5}{2}) = \frac{25 - 50 + 6 \cdot 4}{4}$

Step 3.2.4.2

Multiply $6$ by $4$ .

$f (\frac{5}{2}) = \frac{25 - 50 + 24}{4}$

Step 3.2.5

Simplify the expression.

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Step 3.2.5.1

Subtract $50$ from $25$ .

$f (\frac{5}{2}) = \frac{- 25 + 24}{4}$

Step 3.2.5.2

Add $- 25$ and $24$ .

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

Step 3.2.5.3

Move the negative in front of the fraction.

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

Step 3.2.6

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

$- \frac{1}{4}$

Step 4

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

$(\frac{5}{2}, - \frac{1}{4})$

Step 5

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