Find the Maximum/Minimum Value y=-16t^2+128t

$y = - 16 t^{2} + 128 t$

Step 1

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 $t = - \frac{b}{2 a}$

Step 2

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

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

$t = - \frac{128}{2 (- 16)}$

Remove parentheses.

$t = - \frac{128}{2 (- 16)}$

Simplify $- \frac{128}{2 (- 16)}$ .

Cancel the common factor of $128$ and $2$ .

Factor $2$ out of $128$ .

$t = - \frac{2 \cdot 64}{2 \cdot - 16}$

Cancel the common factors.

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

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

Cancel the common factor.

$t = - \frac{2 \cdot 64}{2 \cdot - 16}$

Rewrite the expression.

$t = - \frac{64}{- 16}$

Cancel the common factor of $64$ and $- 16$ .

Factor $16$ out of $64$ .

$t = - \frac{16 (4)}{- 16}$

Move the negative one from the denominator of $\frac{4}{- 1}$ .

$t = - (- 1 \cdot 4)$

Multiply $- (- 1 \cdot 4)$ .

Multiply $- 1$ by $4$ .

$t = - - 4$

Multiply $- 1$ by $- 4$ .

$t = 4$

Step 3

Evaluate $f (4)$ .

Replace the variable $t$ with $4$ in the expression.

$f (4) = - 16 {(4)}^{2} + 128 (4)$

Simplify the result.

Simplify each term.

Raise $4$ to the power of $2$ .

$f (4) = - 16 \cdot 16 + 128 (4)$

Multiply $- 16$ by $16$ .

$f (4) = - 256 + 128 (4)$

Multiply $128$ by $4$ .

$f (4) = - 256 + 512$

Add $- 256$ and $512$ .

$f (4) = 256$

The final answer is $256$ .

$256$

Step 4

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

$(4, 256)$

Step 5

Find the Maximum/Minimum Value y=-16t^2+128t

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