Find the Maximum/Minimum Value h(t)=-16t^2+64t+4

$h (t) = - 16 t^{2} + 64 t + 4$

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 $b$ .

$t = - \frac{64}{2 (- 16)}$

Remove parentheses.

$t = - \frac{64}{2 (- 16)}$

Simplify $- \frac{64}{2 (- 16)}$ .

Cancel the common factor of $64$ and $2$ .

Factor $2$ out of $64$ .

$t = - \frac{2 \cdot 32}{2 \cdot - 16}$

Cancel the common factors.

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

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

Cancel the common factor.

$t = - \frac{2 \cdot 32}{2 \cdot - 16}$

Rewrite the expression.

$t = - \frac{32}{- 16}$

Cancel the common factor of $32$ and $- 16$ .

Factor $16$ out of $32$ .

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

Move the negative one from the denominator of $\frac{2}{- 1}$ .

$t = - (- 1 \cdot 2)$

Multiply.

Multiply $- 1$ by $2$ .

$t = - - 2$

Multiply $- 1$ by $- 2$ .

$t = 2$

Evaluate $f (2)$ .

Replace the variable $t$ with $2$ in the expression.

$h (2) = - 16 {(2)}^{2} + 64 (2) + 4$

Simplify the result.

Simplify each term.

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

$h (2) = - 16 \cdot 4 + 64 (2) + 4$

Multiply $- 16$ by $4$ .

$h (2) = - 64 + 64 (2) + 4$

Multiply $64$ by $2$ .

$h (2) = - 64 + 128 + 4$

Simplify by adding numbers.

Add $- 64$ and $128$ .

$h (2) = 64 + 4$

Add $64$ and $4$ .

$h (2) = 68$

The final answer is $68$ .

$68$

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

$(2, 68)$

Find the Maximum/Minimum Value h(t)=-16t^2+64t+4

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