Precalculus Examples

$H (t) = - 16 t^{2} + 1000 t + 15$

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

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Step 2.1

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

$t = - \frac{1000}{2 (- 16)}$

Step 2.2

Remove parentheses.

$t = - \frac{1000}{2 (- 16)}$

Step 2.3

Simplify $- \frac{1000}{2 (- 16)}$ .

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Step 2.3.1

Cancel the common factor of $1000$ and $2$ .

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Step 2.3.1.1

Factor $2$ out of $1000$ .

$t = - \frac{2 \cdot 500}{2 \cdot - 16}$

Step 2.3.1.2

Cancel the common factors.

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Step 2.3.1.2.1

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

$t = - \frac{2 \cdot 500}{2 (- 16)}$

Step 2.3.1.2.2

Cancel the common factor.

$t = - \frac{2 \cdot 500}{2 \cdot - 16}$

Step 2.3.1.2.3

Rewrite the expression.

$t = - \frac{500}{- 16}$

Step 2.3.2

Cancel the common factor of $500$ and $- 16$ .

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Step 2.3.2.1

Factor $4$ out of $500$ .

$t = - \frac{4 (125)}{- 16}$

Step 2.3.2.2

Cancel the common factors.

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Step 2.3.2.2.1

Factor $4$ out of $- 16$ .

$t = - \frac{4 \cdot 125}{4 \cdot - 4}$

Step 2.3.2.2.2

Cancel the common factor.

$t = - \frac{4 \cdot 125}{4 \cdot - 4}$

Step 2.3.2.2.3

Rewrite the expression.

$t = - \frac{125}{- 4}$

Step 2.3.3

Move the negative in front of the fraction.

$t = - - \frac{125}{4}$

Step 2.3.4

Multiply $- - \frac{125}{4}$ .

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Step 2.3.4.1

Multiply $- 1$ by $- 1$ .

$t = 1 (\frac{125}{4})$

Step 2.3.4.2

Multiply $\frac{125}{4}$ by $1$ .

$t = \frac{125}{4}$

Step 3

Evaluate $f (\frac{125}{4})$ .

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Step 3.1

Replace the variable $t$ with $\frac{125}{4}$ in the expression.

$H (\frac{125}{4}) = - 16 {(\frac{125}{4})}^{2} + 1000 (\frac{125}{4}) + 15$

Step 3.2

Simplify the result.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Apply the product rule to $\frac{125}{4}$ .

$H (\frac{125}{4}) = - 16 \frac{125^{2}}{4^{2}} + 1000 (\frac{125}{4}) + 15$

Step 3.2.1.2

Raise $125$ to the power of $2$ .

$H (\frac{125}{4}) = - 16 \frac{15625}{4^{2}} + 1000 (\frac{125}{4}) + 15$

Step 3.2.1.3

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

$H (\frac{125}{4}) = - 16 (\frac{15625}{16}) + 1000 (\frac{125}{4}) + 15$

Step 3.2.1.4

Cancel the common factor of $16$ .

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Step 3.2.1.4.1

Factor $16$ out of $- 16$ .

$H (\frac{125}{4}) = 16 (- 1) (\frac{15625}{16}) + 1000 (\frac{125}{4}) + 15$

Step 3.2.1.4.2

Cancel the common factor.

$H (\frac{125}{4}) = 16 \cdot (- 1 (\frac{15625}{16})) + 1000 (\frac{125}{4}) + 15$

Step 3.2.1.4.3

Rewrite the expression.

$H (\frac{125}{4}) = - 1 \cdot 15625 + 1000 (\frac{125}{4}) + 15$

Step 3.2.1.5

Multiply $- 1$ by $15625$ .

$H (\frac{125}{4}) = - 15625 + 1000 (\frac{125}{4}) + 15$

Step 3.2.1.6

Cancel the common factor of $4$ .

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Step 3.2.1.6.1

Factor $4$ out of $1000$ .

$H (\frac{125}{4}) = - 15625 + 4 (250) (\frac{125}{4}) + 15$

Step 3.2.1.6.2

Cancel the common factor.

$H (\frac{125}{4}) = - 15625 + 4 \cdot (250 (\frac{125}{4})) + 15$

Step 3.2.1.6.3

Rewrite the expression.

$H (\frac{125}{4}) = - 15625 + 250 \cdot 125 + 15$

Step 3.2.1.7

Multiply $250$ by $125$ .

$H (\frac{125}{4}) = - 15625 + 31250 + 15$

Step 3.2.2

Simplify by adding numbers.

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Step 3.2.2.1

Add $- 15625$ and $31250$ .

$H (\frac{125}{4}) = 15625 + 15$

Step 3.2.2.2

Add $15625$ and $15$ .

$H (\frac{125}{4}) = 15640$

Step 3.2.3

The final answer is $15640$ .

$15640$

Step 4

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

$(\frac{125}{4}, 15640)$

Step 5