Find the Maximum/Minimum Value h(t)=-16t^2+136t+9

$h (t) = - 16 t^{2} + 136 t + 9$

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 $b$ .

$t = - \frac{136}{2 (- 16)}$

Remove parentheses.

$t = - \frac{136}{2 (- 16)}$

Simplify $- \frac{136}{2 (- 16)}$ .

Cancel the common factor of $136$ and $2$ .

Factor $2$ out of $136$ .

$t = - \frac{2 \cdot 68}{2 \cdot - 16}$

Cancel the common factors.

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

$t = - \frac{2 \cdot 68}{2 (- 16)}$

Cancel the common factor.

$t = - \frac{2 \cdot 68}{2 \cdot - 16}$

Rewrite the expression.

$t = - \frac{68}{- 16}$

Cancel the common factor of $68$ and $- 16$ .

Factor $4$ out of $68$ .

$t = - \frac{4 (17)}{- 16}$

Cancel the common factors.

Factor $4$ out of $- 16$ .

$t = - \frac{4 \cdot 17}{4 \cdot - 4}$

Cancel the common factor.

$t = - \frac{4 \cdot 17}{4 \cdot - 4}$

Rewrite the expression.

$t = - \frac{17}{- 4}$

Move the negative in front of the fraction.

$t = - - \frac{17}{4}$

Multiply $- - \frac{17}{4}$ .

Multiply $- 1$ by $- 1$ .

$t = 1 (\frac{17}{4})$

Multiply $\frac{17}{4}$ by $1$ .

$t = \frac{17}{4}$

Step 3

Evaluate $f (\frac{17}{4})$ .

Replace the variable $t$ with $\frac{17}{4}$ in the expression.

$h (\frac{17}{4}) = - 16 {(\frac{17}{4})}^{2} + 136 (\frac{17}{4}) + 9$

Simplify the result.

Simplify each term.

Apply the product rule to $\frac{17}{4}$ .

$h (\frac{17}{4}) = - 16 \frac{17^{2}}{4^{2}} + 136 (\frac{17}{4}) + 9$

Raise $17$ to the power of $2$ .

$h (\frac{17}{4}) = - 16 \frac{289}{4^{2}} + 136 (\frac{17}{4}) + 9$

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

$h (\frac{17}{4}) = - 16 (\frac{289}{16}) + 136 (\frac{17}{4}) + 9$

Cancel the common factor of $16$ .

Factor $16$ out of $- 16$ .

$h (\frac{17}{4}) = 16 (- 1) (\frac{289}{16}) + 136 (\frac{17}{4}) + 9$

Cancel the common factor.

$h (\frac{17}{4}) = 16 \cdot (- 1 (\frac{289}{16})) + 136 (\frac{17}{4}) + 9$

Rewrite the expression.

$h (\frac{17}{4}) = - 1 \cdot 289 + 136 (\frac{17}{4}) + 9$

Multiply $- 1$ by $289$ .

$h (\frac{17}{4}) = - 289 + 136 (\frac{17}{4}) + 9$

Cancel the common factor of $4$ .

Factor $4$ out of $136$ .

$h (\frac{17}{4}) = - 289 + 4 (34) (\frac{17}{4}) + 9$

Cancel the common factor.

$h (\frac{17}{4}) = - 289 + 4 \cdot (34 (\frac{17}{4})) + 9$

Rewrite the expression.

$h (\frac{17}{4}) = - 289 + 34 \cdot 17 + 9$

Multiply $34$ by $17$ .

$h (\frac{17}{4}) = - 289 + 578 + 9$

Simplify by adding numbers.

Add $- 289$ and $578$ .

$h (\frac{17}{4}) = 289 + 9$

Add $289$ and $9$ .

$h (\frac{17}{4}) = 298$

The final answer is $298$ .

$298$

Step 4

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

$(\frac{17}{4}, 298)$

Step 5

Find the Maximum/Minimum Value h(t)=-16t^2+136t+9

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