Find the Maximum/Minimum Value -16t^2+320

$- 16 t^{2} + 320$

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 $b$ .

$t = - \frac{0}{2 (- 16)}$

Remove parentheses.

$t = - \frac{0}{2 (- 16)}$

Simplify $- \frac{0}{2 (- 16)}$ .

Cancel the common factor of $0$ and $2$ .

Factor $2$ out of $0$ .

$t = - \frac{2 (0)}{2 (- 16)}$

Cancel the common factors.

Cancel the common factor.

$t = - \frac{2 \cdot 0}{2 \cdot - 16}$

Rewrite the expression.

$t = - \frac{0}{- 16}$

Cancel the common factor of $0$ and $- 16$ .

Factor $16$ out of $0$ .

$t = - \frac{16 (0)}{- 16}$

Move the negative one from the denominator of $\frac{0}{- 1}$ .

$t = - (- 1 \cdot 0)$

Simplify the expression.

Rewrite $- 1 \cdot 0$ as $- 0$ .

$t = - - 0$

Multiply $- 1$ by $0$ .

$t = - 0$

Multiply $- 1$ by $0$ .

$t = 0$

Evaluate $f (0)$ .

Replace the variable $t$ with $0$ in the expression.

$f (0) = - 16 {(0)}^{2} + 320$

Simplify the result.

Simplify each term.

Raising $0$ to any positive power yields $0$ .

$f (0) = - 16 \cdot 0 + 320$

Multiply $- 16$ by $0$ .

$f (0) = 0 + 320$

Add $0$ and $320$ .

$f (0) = 320$

The final answer is $320$ .

$320$

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

$(0, 320)$

Find the Maximum/Minimum Value -16t^2+320

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