Find the Maximum/Minimum Value h(t)=-16t^2+144t+20

$h (t) = - 16 t^{2} + 144 t + 20$

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 $b$ .

$t = - \frac{144}{2 (- 16)}$

Remove parentheses.

$t = - \frac{144}{2 (- 16)}$

Simplify $- \frac{144}{2 (- 16)}$ .

Cancel the common factor of $144$ and $2$ .

Factor $2$ out of $144$ .

$t = - \frac{2 \cdot 72}{2 \cdot - 16}$

Cancel the common factors.

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

$t = - \frac{2 \cdot 72}{2 (- 16)}$

Cancel the common factor.

$t = - \frac{2 \cdot 72}{2 \cdot - 16}$

Rewrite the expression.

$t = - \frac{72}{- 16}$

Cancel the common factor of $72$ and $- 16$ .

Factor $8$ out of $72$ .

$t = - \frac{8 (9)}{- 16}$

Cancel the common factors.

Factor $8$ out of $- 16$ .

$t = - \frac{8 \cdot 9}{8 \cdot - 2}$

Cancel the common factor.

$t = - \frac{8 \cdot 9}{8 \cdot - 2}$

Rewrite the expression.

$t = - \frac{9}{- 2}$

Move the negative in front of the fraction.

$t = - - \frac{9}{2}$

Multiply $- - \frac{9}{2}$ .

Multiply $- 1$ by $- 1$ .

$t = 1 (\frac{9}{2})$

Multiply $\frac{9}{2}$ by $1$ .

$t = \frac{9}{2}$

Step 3

Evaluate $f (\frac{9}{2})$ .

Replace the variable $t$ with $\frac{9}{2}$ in the expression.

$h (\frac{9}{2}) = - 16 {(\frac{9}{2})}^{2} + 144 (\frac{9}{2}) + 20$

Simplify the result.

Simplify each term.

Apply the product rule to $\frac{9}{2}$ .

$h (\frac{9}{2}) = - 16 \frac{9^{2}}{2^{2}} + 144 (\frac{9}{2}) + 20$

Raise $9$ to the power of $2$ .

$h (\frac{9}{2}) = - 16 \frac{81}{2^{2}} + 144 (\frac{9}{2}) + 20$

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

$h (\frac{9}{2}) = - 16 (\frac{81}{4}) + 144 (\frac{9}{2}) + 20$

Cancel the common factor of $4$ .

Factor $4$ out of $- 16$ .

$h (\frac{9}{2}) = 4 (- 4) (\frac{81}{4}) + 144 (\frac{9}{2}) + 20$

Cancel the common factor.

$h (\frac{9}{2}) = 4 \cdot (- 4 (\frac{81}{4})) + 144 (\frac{9}{2}) + 20$

Rewrite the expression.

$h (\frac{9}{2}) = - 4 \cdot 81 + 144 (\frac{9}{2}) + 20$

Multiply $- 4$ by $81$ .

$h (\frac{9}{2}) = - 324 + 144 (\frac{9}{2}) + 20$

Cancel the common factor of $2$ .

Factor $2$ out of $144$ .

$h (\frac{9}{2}) = - 324 + 2 (72) (\frac{9}{2}) + 20$

Cancel the common factor.

$h (\frac{9}{2}) = - 324 + 2 \cdot (72 (\frac{9}{2})) + 20$

Rewrite the expression.

$h (\frac{9}{2}) = - 324 + 72 \cdot 9 + 20$

Multiply $72$ by $9$ .

$h (\frac{9}{2}) = - 324 + 648 + 20$

Simplify by adding numbers.

Add $- 324$ and $648$ .

$h (\frac{9}{2}) = 324 + 20$

Add $324$ and $20$ .

$h (\frac{9}{2}) = 344$

The final answer is $344$ .

$344$

Step 4

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

$(\frac{9}{2}, 344)$

Step 5

Find the Maximum/Minimum Value h(t)=-16t^2+144t+20

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