Precalculus Examples

$y = - \frac{1}{16} x^{2} + 6 x + 5$

Step 1

Combine $x^{2}$ and $\frac{1}{16}$ .

$f (x) = - \frac{x^{2}}{16} + 6 x + 5$

Step 2

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 3

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

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Step 3.1

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

$x = - \frac{6}{2 (- 0.0625)}$

Step 3.2

Remove parentheses.

$x = - \frac{6}{2 (- 0.0625)}$

Step 3.3

Simplify $- \frac{6}{2 (- 0.0625)}$ .

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Step 3.3.1

Cancel the common factor of $6$ and $2$ .

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Step 3.3.1.1

Factor $2$ out of $6$ .

$x = - \frac{2 \cdot 3}{2 \cdot - 0.0625}$

Step 3.3.1.2

Cancel the common factors.

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Step 3.3.1.2.1

Factor $2$ out of $2 \cdot - 0.0625$ .

$x = - \frac{2 \cdot 3}{2 (- 0.0625)}$

Step 3.3.1.2.2

Cancel the common factor.

$x = - \frac{2 \cdot 3}{2 \cdot - 0.0625}$

Step 3.3.1.2.3

Rewrite the expression.

$x = - \frac{3}{- 0.0625}$

Step 3.3.2

Simplify the expression.

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Step 3.3.2.1

Divide $3$ by $- 0.0625$ .

$x = - - 48$

Step 3.3.2.2

Multiply $- 1$ by $- 48$ .

$x = 48$

Step 4

Evaluate $f (48)$ .

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Step 4.1

Replace the variable $x$ with $48$ in the expression.

$f (48) = - \frac{{(48)}^{2}}{16} + 6 (48) + 5$

Step 4.2

Simplify the result.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Raise $48$ to the power of $2$ .

$f (48) = - \frac{2304}{16} + 6 (48) + 5$

Step 4.2.1.2

Divide $2304$ by $16$ .

$f (48) = - 1 \cdot 144 + 6 (48) + 5$

Step 4.2.1.3

Multiply $- 1$ by $144$ .

$f (48) = - 144 + 6 (48) + 5$

Step 4.2.1.4

Multiply $6$ by $48$ .

$f (48) = - 144 + 288 + 5$

Step 4.2.2

Simplify by adding numbers.

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Step 4.2.2.1

Add $- 144$ and $288$ .

$f (48) = 144 + 5$

Step 4.2.2.2

Add $144$ and $5$ .

$f (48) = 149$

Step 4.2.3

The final answer is $149$ .

$149$

Step 5

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

$(48, 149)$

Step 6