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

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

Step 1

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

$f_{max}$ $x = a x^{2} + b x + c$ 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 $2$ .

Cancel the common factor.

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

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 $- 2$ .

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

Factor $2$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Rewrite $- 1 (\frac{1}{2})$ as $- (\frac{1}{2})$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

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 $\frac{2}{2}$ .

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

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

Simplify the expression.

Multiply $- 2$ by $2$ .

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

Add $- 1$ and $2$ .

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

Subtract $4$ from $1$ .

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

Move the negative in front of the fraction.

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

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

$- \frac{3}{2}$

Step 4

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

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

Step 5

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

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