Find the Maximum/Minimum Value f(x)=-2x^2-2x+3

$f (x) = - 2 x^{2} - 2 x + 3$

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 $- 2$ 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.

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

Dividing two negative values results in a positive value.

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

Step 3

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

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

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

Simplify the result.

Simplify each term.

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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

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

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

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

Raise $- 1$ to the power of $2$ .

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

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

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

One to any power is one.

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

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

$f (- \frac{1}{2}) = - 2 (\frac{1}{4}) - 2 (- \frac{1}{2}) + 3$

Cancel the common factor of $2$ .

Factor $2$ out of $- 2$ .

$f (- \frac{1}{2}) = 2 (- 1) (\frac{1}{4}) - 2 (- \frac{1}{2}) + 3$

Factor $2$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{2}$ into the numerator.

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

Factor $2$ out of $- 2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 1$ by $- 1$ .

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

Find the common denominator.

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

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

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

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

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

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

Write $3$ as a fraction with denominator $1$ .

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

Multiply $\frac{3}{1}$ by $\frac{2}{2}$ .

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

Multiply $\frac{3}{1}$ by $\frac{2}{2}$ .

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

Combine the numerators over the common denominator.

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

Simplify the expression.

Multiply $3$ by $2$ .

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

Add $- 1$ and $2$ .

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

Add $1$ and $6$ .

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

The final answer is $\frac{7}{2}$ .

$\frac{7}{2}$

Step 4

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

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

Step 5

Find the Maximum/Minimum Value f(x)=-2x^2-2x+3

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