Find the Maximum/Minimum Value -2x^2+9x

$- 2 x^{2} + 9 x$

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

Find the value of $x = - \frac{b}{2 a}$ .

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

$x = - \frac{9}{2 (- 2)}$

Remove parentheses.

$x = - \frac{9}{2 (- 2)}$

Simplify $- \frac{9}{2 (- 2)}$ .

Multiply $2$ by $- 2$ .

$x = - \frac{9}{- 4}$

Move the negative in front of the fraction.

$x = - - \frac{9}{4}$

Multiply $- - \frac{9}{4}$ .

Multiply $- 1$ by $- 1$ .

$x = 1 (\frac{9}{4})$

Multiply $\frac{9}{4}$ by $1$ .

$x = \frac{9}{4}$

Evaluate $f (\frac{9}{4})$ .

Replace the variable $x$ with $\frac{9}{4}$ in the expression.

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

Simplify the result.

Simplify each term.

Apply the product rule to $\frac{9}{4}$ .

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

Raise $9$ to the power of $2$ .

$f (\frac{9}{4}) = - 2 \frac{81}{4^{2}} + 9 (\frac{9}{4})$

Raise $4$ to the power of $2$ .

$f (\frac{9}{4}) = - 2 (\frac{81}{16}) + 9 (\frac{9}{4})$

Cancel the common factor of $2$ .

Factor $2$ out of $- 2$ .

$f (\frac{9}{4}) = 2 (- 1) (\frac{81}{16}) + 9 (\frac{9}{4})$

Factor $2$ out of $16$ .

$f (\frac{9}{4}) = 2 \cdot (- 1 \frac{81}{2 \cdot 8}) + 9 (\frac{9}{4})$

Cancel the common factor.

$f (\frac{9}{4}) = 2 \cdot (- 1 \frac{81}{2 \cdot 8}) + 9 (\frac{9}{4})$

Rewrite the expression.

$f (\frac{9}{4}) = - 1 (\frac{81}{8}) + 9 (\frac{9}{4})$

Rewrite $- 1 (\frac{81}{8})$ as $- (\frac{81}{8})$ .

$f (\frac{9}{4}) = - (\frac{81}{8}) + 9 (\frac{9}{4})$

Multiply $9 (\frac{9}{4})$ .

Combine $9$ and $\frac{9}{4}$ .

$f (\frac{9}{4}) = - \frac{81}{8} + \frac{9 \cdot 9}{4}$

Multiply $9$ by $9$ .

$f (\frac{9}{4}) = - \frac{81}{8} + \frac{81}{4}$

To write $\frac{81}{4}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f (\frac{9}{4}) = - \frac{81}{8} + \frac{81}{4} \cdot \frac{2}{2}$

Write each expression with a common denominator of $8$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{81}{4}$ by $\frac{2}{2}$ .

$f (\frac{9}{4}) = - \frac{81}{8} + \frac{81 \cdot 2}{4 \cdot 2}$

Multiply $4$ by $2$ .

$f (\frac{9}{4}) = - \frac{81}{8} + \frac{81 \cdot 2}{8}$

Combine the numerators over the common denominator.

$f (\frac{9}{4}) = \frac{- 81 + 81 \cdot 2}{8}$

Simplify the numerator.

Multiply $81$ by $2$ .

$f (\frac{9}{4}) = \frac{- 81 + 162}{8}$

Add $- 81$ and $162$ .

$f (\frac{9}{4}) = \frac{81}{8}$

The final answer is $\frac{81}{8}$ .

$\frac{81}{8}$

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

$(\frac{9}{4}, \frac{81}{8})$

Find the Maximum/Minimum Value -2x^2+9x

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