Find the Maximum/Minimum Value -16t^2+96t+112

$- 16 t^{2} + 96 t + 112$

The maximum of a quadratic function occurs at $x = - \frac{b}{2 a}$ . If $a$ is negative, the maximum value of the function is $f (- \frac{b}{2 a})$ .

$f_{max}$ $t = a t^{2} + b t + c$ occurs at $t = - \frac{b}{2 a}$

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

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

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

Remove parentheses.

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

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

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

Factor $2$ out of $96$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Factor $16$ out of $48$ .

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

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

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

Multiply.

Multiply $- 1$ by $3$ .

$t = - - 3$

Multiply $- 1$ by $- 3$ .

$t = 3$

Evaluate $f (3)$ .

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

$f (3) = - 16 {(3)}^{2} + 96 (3) + 112$

Simplify the result.

Simplify each term.

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

$f (3) = - 16 \cdot 9 + 96 (3) + 112$

Multiply $- 16$ by $9$ .

$f (3) = - 144 + 96 (3) + 112$

Multiply $96$ by $3$ .

$f (3) = - 144 + 288 + 112$

Simplify by adding numbers.

Add $- 144$ and $288$ .

$f (3) = 144 + 112$

Add $144$ and $112$ .

$f (3) = 256$

The final answer is $256$ .

$256$

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

$(3, 256)$

Find the Maximum/Minimum Value -16t^2+96t+112

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