Find the Maximum/Minimum Value -x^2+91x-769

$- x^{2} + 91 x - 769$

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 $b$ .

$x = - \frac{91}{2 (- 1)}$

Remove parentheses.

$x = - \frac{91}{2 (- 1)}$

Simplify $- \frac{91}{2 (- 1)}$ .

Multiply $2$ by $- 1$ .

$x = - \frac{91}{- 2}$

Move the negative in front of the fraction.

$x = - - \frac{91}{2}$

Multiply $- - \frac{91}{2}$ .

Multiply $- 1$ by $- 1$ .

$x = 1 (\frac{91}{2})$

Multiply $\frac{91}{2}$ by $1$ .

$x = \frac{91}{2}$

Step 3

Evaluate $f (\frac{91}{2})$ .

Replace the variable $x$ with $\frac{91}{2}$ in the expression.

$f (\frac{91}{2}) = - {(\frac{91}{2})}^{2} + 91 (\frac{91}{2}) - 769$

Simplify the result.

Simplify each term.

Apply the product rule to $\frac{91}{2}$ .

$f (\frac{91}{2}) = - \frac{91^{2}}{2^{2}} + 91 (\frac{91}{2}) - 769$

Raise $91$ to the power of $2$ .

$f (\frac{91}{2}) = - \frac{8281}{2^{2}} + 91 (\frac{91}{2}) - 769$

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

$f (\frac{91}{2}) = - \frac{8281}{4} + 91 (\frac{91}{2}) - 769$

Multiply $91 (\frac{91}{2})$ .

Combine $91$ and $\frac{91}{2}$ .

$f (\frac{91}{2}) = - \frac{8281}{4} + \frac{91 \cdot 91}{2} - 769$

Multiply $91$ by $91$ .

$f (\frac{91}{2}) = - \frac{8281}{4} + \frac{8281}{2} - 769$

Find the common denominator.

Multiply $\frac{8281}{2}$ by $\frac{2}{2}$ .

$f (\frac{91}{2}) = - \frac{8281}{4} + \frac{8281}{2} \cdot \frac{2}{2} - 769$

Multiply $\frac{8281}{2}$ by $\frac{2}{2}$ .

$f (\frac{91}{2}) = - \frac{8281}{4} + \frac{8281 \cdot 2}{2 \cdot 2} - 769$

Write $- 769$ as a fraction with denominator $1$ .

$f (\frac{91}{2}) = - \frac{8281}{4} + \frac{8281 \cdot 2}{2 \cdot 2} + \frac{- 769}{1}$

Multiply $\frac{- 769}{1}$ by $\frac{4}{4}$ .

$f (\frac{91}{2}) = - \frac{8281}{4} + \frac{8281 \cdot 2}{2 \cdot 2} + \frac{- 769}{1} \cdot \frac{4}{4}$

Multiply $\frac{- 769}{1}$ by $\frac{4}{4}$ .

$f (\frac{91}{2}) = - \frac{8281}{4} + \frac{8281 \cdot 2}{2 \cdot 2} + \frac{- 769 \cdot 4}{4}$

Multiply $2$ by $2$ .

$f (\frac{91}{2}) = - \frac{8281}{4} + \frac{8281 \cdot 2}{4} + \frac{- 769 \cdot 4}{4}$

Combine the numerators over the common denominator.

$f (\frac{91}{2}) = \frac{- 8281 + 8281 \cdot 2 - 769 \cdot 4}{4}$

Simplify each term.

Multiply $8281$ by $2$ .

$f (\frac{91}{2}) = \frac{- 8281 + 16562 - 769 \cdot 4}{4}$

Multiply $- 769$ by $4$ .

$f (\frac{91}{2}) = \frac{- 8281 + 16562 - 3076}{4}$

Simplify by adding and subtracting.

Add $- 8281$ and $16562$ .

$f (\frac{91}{2}) = \frac{8281 - 3076}{4}$

Subtract $3076$ from $8281$ .

$f (\frac{91}{2}) = \frac{5205}{4}$

The final answer is $\frac{5205}{4}$ .

$\frac{5205}{4}$

Step 4

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

$(\frac{91}{2}, \frac{5205}{4})$

Step 5

Find the Maximum/Minimum Value -x^2+91x-769

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