Find the Maximum/Minimum Value 2x^2-2x+2

$2 x^{2} - 2 x + 2$

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

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

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

Remove parentheses.

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

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

Cancel the common factor of $- 2$ and $2$ .

Factor $2$ out of $- 2$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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

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

Multiply $- 1$ by $- 1$ .

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

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

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

Step 3

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

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

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

Simplify the result.

Simplify each term.

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

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

One to any power is one.

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

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

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

Cancel the common factor of $2$ .

Factor $2$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $2$ .

Factor $2$ out of $- 2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Find the common denominator.

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

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

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

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

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

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

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

Simplify each term.

Multiply $- 1$ by $2$ .

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

Multiply $2$ by $2$ .

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

Simplify by adding and subtracting.

Subtract $2$ from $1$ .

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

Add $- 1$ and $4$ .

$f (\frac{1}{2}) = \frac{3}{2}$

The final answer is $\frac{3}{2}$ .

$\frac{3}{2}$

Step 4

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

$(\frac{1}{2}, \frac{3}{2})$

Step 5

Find the Maximum/Minimum Value 2x^2-2x+2

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