Precalculus Examples

$h (x) = 10 x - 0.01 x^{2}$

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 $f (- \frac{b}{2 a})$ .

$f_{max}$ $x = a x^{2} + b x + c$ occurs at $x = - \frac{b}{2 a}$

Step 2

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

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Step 2.1

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

$x = - \frac{10}{2 (- 0.01)}$

Step 2.2

Remove parentheses.

$x = - \frac{10}{2 (- 0.01)}$

Step 2.3

Simplify $- \frac{10}{2 (- 0.01)}$ .

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Step 2.3.1

Cancel the common factor of $10$ and $2$ .

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Step 2.3.1.1

Factor $2$ out of $10$ .

$x = - \frac{2 \cdot 5}{2 \cdot - 0.01}$

Step 2.3.1.2

Cancel the common factors.

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Step 2.3.1.2.1

Factor $2$ out of $2 \cdot - 0.01$ .

$x = - \frac{2 \cdot 5}{2 (- 0.01)}$

Step 2.3.1.2.2

Cancel the common factor.

$x = - \frac{2 \cdot 5}{2 \cdot - 0.01}$

Step 2.3.1.2.3

Rewrite the expression.

$x = - \frac{5}{- 0.01}$

Step 2.3.2

Simplify the expression.

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Step 2.3.2.1

Divide $5$ by $- 0.01$ .

$x = - - 500$

Step 2.3.2.2

Multiply $- 1$ by $- 500$ .

$x = 500$

Step 3

Evaluate $f (500)$ .

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Step 3.1

Replace the variable $x$ with $500$ in the expression.

$h (500) = 10 (500) - 0.01 {(500)}^{2}$

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

Multiply $10$ by $500$ .

$h (500) = 5000 - 0.01 {(500)}^{2}$

Step 3.2.1.2

Raise $500$ to the power of $2$ .

$h (500) = 5000 - 0.01 \cdot 250000$

Step 3.2.1.3

Multiply $- 0.01$ by $250000$ .

$h (500) = 5000 - 2500$

Step 3.2.2

Subtract $2500$ from $5000$ .

$h (500) = 2500$

Step 3.2.3

The final answer is $2500$ .

$2500$

Step 4

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

$(500, 2500)$

Step 5