Find the Maximum/Minimum Value 16x-x^2

$16 x - x^{2}$

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 $b$ .

$x = - \frac{16}{2 (- 1)}$

Remove parentheses.

$x = - \frac{16}{2 (- 1)}$

Simplify $- \frac{16}{2 (- 1)}$ .

Cancel the common factor of $16$ and $2$ .

Factor $2$ out of $16$ .

$x = - \frac{2 \cdot 8}{2 \cdot - 1}$

Move the negative one from the denominator of $\frac{8}{- 1}$ .

$x = - (- 1 \cdot 8)$

Multiply $- (- 1 \cdot 8)$ .

Multiply $- 1$ by $8$ .

$x = - - 8$

Multiply $- 1$ by $- 8$ .

$x = 8$

Step 3

Evaluate $f (8)$ .

Replace the variable $x$ with $8$ in the expression.

$f (8) = 16 (8) - {(8)}^{2}$

Simplify the result.

Simplify each term.

Multiply $16$ by $8$ .

$f (8) = 128 - {(8)}^{2}$

Raise $8$ to the power of $2$ .

$f (8) = 128 - 1 \cdot 64$

Multiply $- 1$ by $64$ .

$f (8) = 128 - 64$

Subtract $64$ from $128$ .

$f (8) = 64$

The final answer is $64$ .

$64$

Step 4

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

$(8, 64)$

Step 5

Find the Maximum/Minimum Value 16x-x^2

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