Find the Maximum/Minimum Value f(x)=-16x^2+96x

$f (x) = - 16 x^{2} + 96 x$

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 $b$ .

$x = - \frac{96}{2 (- 16)}$

Remove parentheses.

$x = - \frac{96}{2 (- 16)}$

Simplify $- \frac{96}{2 (- 16)}$ .

Cancel the common factor of $96$ and $2$ .

Factor $2$ out of $96$ .

$x = - \frac{2 \cdot 48}{2 \cdot - 16}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 16$ .

$x = - \frac{2 \cdot 48}{2 (- 16)}$

Cancel the common factor.

$x = - \frac{2 \cdot 48}{2 \cdot - 16}$

Rewrite the expression.

$x = - \frac{48}{- 16}$

Cancel the common factor of $48$ and $- 16$ .

Factor $16$ out of $48$ .

$x = - \frac{16 (3)}{- 16}$

Move the negative one from the denominator of $\frac{3}{- 1}$ .

$x = - (- 1 \cdot 3)$

Multiply.

Multiply $- 1$ by $3$ .

$x = - - 3$

Multiply $- 1$ by $- 3$ .

$x = 3$

Evaluate $f (3)$ .

Replace the variable $x$ with $3$ in the expression.

$f (3) = - 16 {(3)}^{2} + 96 (3)$

Simplify the result.

Simplify each term.

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

$f (3) = - 16 \cdot 9 + 96 (3)$

Multiply $- 16$ by $9$ .

$f (3) = - 144 + 96 (3)$

Multiply $96$ by $3$ .

$f (3) = - 144 + 288$

Add $- 144$ and $288$ .

$f (3) = 144$

The final answer is $144$ .

$144$

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

$(3, 144)$

Find the Maximum/Minimum Value f(x)=-16x^2+96x

Related Questions

Support LCMGCF.com by disabling your adblocker.