Find the Maximum/Minimum Value -16t^2+32t+560

$- 16 t^{2} + 32 t + 560$

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 $b$ .

$t = - \frac{32}{2 (- 16)}$

Remove parentheses.

$t = - \frac{32}{2 (- 16)}$

Simplify $- \frac{32}{2 (- 16)}$ .

Cancel the common factor of $32$ and $2$ .

Factor $2$ out of $32$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Factor $16$ out of $16$ .

$t = - \frac{16 (1)}{- 16}$

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

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

Multiply.

Multiply $- 1$ by $1$ .

$t = - - 1$

Multiply $- 1$ by $- 1$ .

$t = 1$

Evaluate $f (1)$ .

Replace the variable $t$ with $1$ in the expression.

$f (1) = - 16 {(1)}^{2} + 32 (1) + 560$

Simplify the result.

Simplify each term.

One to any power is one.

$f (1) = - 16 \cdot 1 + 32 (1) + 560$

Multiply $- 16$ by $1$ .

$f (1) = - 16 + 32 (1) + 560$

Multiply $32$ by $1$ .

$f (1) = - 16 + 32 + 560$

Simplify by adding numbers.

Add $- 16$ and $32$ .

$f (1) = 16 + 560$

Add $16$ and $560$ .

$f (1) = 576$

The final answer is $576$ .

$576$

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

$(1, 576)$

Find the Maximum/Minimum Value -16t^2+32t+560

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