Precalculus Examples

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

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 $b$ .

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

Step 2.2

Remove parentheses.

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

Step 2.3

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

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Step 2.3.1

Cancel the common factor of $- 1$ and $1$ .

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Step 2.3.1.1

Rewrite $- 1$ as $- 1 (1)$ .

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

Step 2.3.1.2

Cancel the common factor.

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

Step 2.3.1.3

Rewrite the expression.

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

Step 2.3.2

Move the negative in front of the fraction.

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

Step 2.3.3

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

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Step 2.3.3.1

Multiply $- 1$ by $- 1$ .

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

Step 2.3.3.2

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

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

Step 3

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

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Step 3.1

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

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

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 $\frac{1}{2}$ .

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

Step 3.2.1.2

One to any power is one.

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

Step 3.2.1.3

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

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

Step 3.2.2

Find the common denominator.

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Step 3.2.2.1

Write $2$ as a fraction with denominator $1$ .

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

Step 3.2.2.2

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

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

Step 3.2.2.3

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

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

Step 3.2.2.4

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

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

Step 3.2.2.5

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

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

Step 3.2.2.6

Multiply $2$ by $2$ .

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

Step 3.2.3

Combine the numerators over the common denominator.

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

Step 3.2.4

Simplify the expression.

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Step 3.2.4.1

Multiply $2$ by $4$ .

$f (\frac{1}{2}) = \frac{8 - 2 + 1}{4}$

Step 3.2.4.2

Subtract $2$ from $8$ .

$f (\frac{1}{2}) = \frac{6 + 1}{4}$

Step 3.2.4.3

Add $6$ and $1$ .

$f (\frac{1}{2}) = \frac{7}{4}$

Step 3.2.5

The final answer is $\frac{7}{4}$ .

$\frac{7}{4}$

Step 4

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

$(\frac{1}{2}, \frac{7}{4})$

Step 5