Find the Maximum/Minimum Value M(x)=18x-x^2

$M (x) = 18 x - x^{2}$

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 $b$ .

$x = - \frac{18}{2 (- 1)}$

Remove parentheses.

$x = - \frac{18}{2 (- 1)}$

Simplify $- \frac{18}{2 (- 1)}$ .

Cancel the common factor of $18$ and $2$ .

Factor $2$ out of $18$ .

$x = - \frac{2 \cdot 9}{2 \cdot - 1}$

Move the negative one from the denominator of $\frac{9}{- 1}$ .

$x = - (- 1 \cdot 9)$

Multiply.

Multiply $- 1$ by $9$ .

$x = - - 9$

Multiply $- 1$ by $- 9$ .

$x = 9$

Evaluate $f (9)$ .

Replace the variable $x$ with $9$ in the expression.

$M (9) = 18 (9) - {(9)}^{2}$

Simplify the result.

Simplify each term.

Multiply $18$ by $9$ .

$M (9) = 162 - {(9)}^{2}$

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

$M (9) = 162 - 1 \cdot 81$

Multiply $- 1$ by $81$ .

$M (9) = 162 - 81$

Subtract $81$ from $162$ .

$M (9) = 81$

The final answer is $81$ .

$81$

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

$(9, 81)$

Find the Maximum/Minimum Value M(x)=18x-x^2

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