Find the Maximum/Minimum Value f(x)=-2x^2+6x-3

$f (x) = - 2 x^{2} + 6 x - 3$

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 $b$ .

$x = - \frac{6}{2 (- 2)}$

Remove parentheses.

$x = - \frac{6}{2 (- 2)}$

Simplify $- \frac{6}{2 (- 2)}$ .

Cancel the common factor of $6$ and $2$ .

Factor $2$ out of $6$ .

$x = - \frac{2 \cdot 3}{2 \cdot - 2}$

Cancel the common factors.

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

$x = - \frac{2 \cdot 3}{2 (- 2)}$

Cancel the common factor.

$x = - \frac{2 \cdot 3}{2 \cdot - 2}$

Rewrite the expression.

$x = - \frac{3}{- 2}$

Move the negative in front of the fraction.

$x = - - \frac{3}{2}$

Multiply $- - \frac{3}{2}$ .

Multiply $- 1$ by $- 1$ .

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

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

$x = \frac{3}{2}$

Step 3

Evaluate $f (\frac{3}{2})$ .

Replace the variable $x$ with $\frac{3}{2}$ in the expression.

$f (\frac{3}{2}) = - 2 {(\frac{3}{2})}^{2} + 6 (\frac{3}{2}) - 3$

Simplify the result.

Simplify each term.

Apply the product rule to $\frac{3}{2}$ .

$f (\frac{3}{2}) = - 2 \frac{3^{2}}{2^{2}} + 6 (\frac{3}{2}) - 3$

Raise $3$ to the power of $2$ .

$f (\frac{3}{2}) = - 2 \frac{9}{2^{2}} + 6 (\frac{3}{2}) - 3$

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

$f (\frac{3}{2}) = - 2 (\frac{9}{4}) + 6 (\frac{3}{2}) - 3$

Cancel the common factor of $2$ .

Factor $2$ out of $- 2$ .

$f (\frac{3}{2}) = 2 (- 1) (\frac{9}{4}) + 6 (\frac{3}{2}) - 3$

Factor $2$ out of $4$ .

$f (\frac{3}{2}) = 2 \cdot (- 1 \frac{9}{2 \cdot 2}) + 6 (\frac{3}{2}) - 3$

Cancel the common factor.

$f (\frac{3}{2}) = 2 \cdot (- 1 \frac{9}{2 \cdot 2}) + 6 (\frac{3}{2}) - 3$

Rewrite the expression.

$f (\frac{3}{2}) = - 1 (\frac{9}{2}) + 6 (\frac{3}{2}) - 3$

Rewrite $- 1 (\frac{9}{2})$ as $- (\frac{9}{2})$ .

$f (\frac{3}{2}) = - (\frac{9}{2}) + 6 (\frac{3}{2}) - 3$

Cancel the common factor of $2$ .

Factor $2$ out of $6$ .

$f (\frac{3}{2}) = - \frac{9}{2} + 2 (3) (\frac{3}{2}) - 3$

Cancel the common factor.

$f (\frac{3}{2}) = - \frac{9}{2} + 2 \cdot (3 (\frac{3}{2})) - 3$

Rewrite the expression.

$f (\frac{3}{2}) = - \frac{9}{2} + 3 \cdot 3 - 3$

Multiply $3$ by $3$ .

$f (\frac{3}{2}) = - \frac{9}{2} + 9 - 3$

Find the common denominator.

Write $9$ as a fraction with denominator $1$ .

$f (\frac{3}{2}) = - \frac{9}{2} + \frac{9}{1} - 3$

Multiply $\frac{9}{1}$ by $\frac{2}{2}$ .

$f (\frac{3}{2}) = - \frac{9}{2} + \frac{9}{1} \cdot \frac{2}{2} - 3$

Multiply $\frac{9}{1}$ by $\frac{2}{2}$ .

$f (\frac{3}{2}) = - \frac{9}{2} + \frac{9 \cdot 2}{2} - 3$

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

$f (\frac{3}{2}) = - \frac{9}{2} + \frac{9 \cdot 2}{2} + \frac{- 3}{1}$

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

$f (\frac{3}{2}) = - \frac{9}{2} + \frac{9 \cdot 2}{2} + \frac{- 3}{1} \cdot \frac{2}{2}$

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

$f (\frac{3}{2}) = - \frac{9}{2} + \frac{9 \cdot 2}{2} + \frac{- 3 \cdot 2}{2}$

Combine the numerators over the common denominator.

$f (\frac{3}{2}) = \frac{- 9 + 9 \cdot 2 - 3 \cdot 2}{2}$

Simplify each term.

Multiply $9$ by $2$ .

$f (\frac{3}{2}) = \frac{- 9 + 18 - 3 \cdot 2}{2}$

Multiply $- 3$ by $2$ .

$f (\frac{3}{2}) = \frac{- 9 + 18 - 6}{2}$

Simplify by adding and subtracting.

Add $- 9$ and $18$ .

$f (\frac{3}{2}) = \frac{9 - 6}{2}$

Subtract $6$ from $9$ .

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

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

$\frac{3}{2}$

Step 4

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

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

Step 5

Find the Maximum/Minimum Value f(x)=-2x^2+6x-3

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