Precalculus Examples

$f (x) = - x^{2} - 6 x$

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 $b$ .

$x = - \frac{- 6}{2 (- 1)}$

Step 2.2

Remove parentheses.

$x = - \frac{- 6}{2 (- 1)}$

Step 2.3

Simplify $- \frac{- 6}{2 (- 1)}$ .

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Step 2.3.1

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

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Step 2.3.1.1

Factor $2$ out of $- 6$ .

$x = - \frac{2 \cdot - 3}{2 \cdot - 1}$

Step 2.3.1.2

Move the negative one from the denominator of $\frac{- 3}{- 1}$ .

$x = - (- 1 \cdot - 3)$

Step 2.3.2

Rewrite $- 1 \cdot - 3$ as $- - 3$ .

$x = - - - 3$

Step 2.3.3

Multiply $- - - 3$ .

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Step 2.3.3.1

Multiply $- 1$ by $- 3$ .

$x = - 1 \cdot 3$

Step 2.3.3.2

Multiply $- 1$ by $3$ .

$x = - 3$

Step 3

Evaluate $f (- 3)$ .

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Step 3.1

Replace the variable $x$ with $- 3$ in the expression.

$f (- 3) = - {(- 3)}^{2} - 6 \cdot - 3$

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

Raise $- 3$ to the power of $2$ .

$f (- 3) = - 1 \cdot 9 - 6 \cdot - 3$

Step 3.2.1.2

Multiply $- 1$ by $9$ .

$f (- 3) = - 9 - 6 \cdot - 3$

Step 3.2.1.3

Multiply $- 6$ by $- 3$ .

$f (- 3) = - 9 + 18$

Step 3.2.2

Add $- 9$ and $18$ .

$f (- 3) = 9$

Step 3.2.3

The final answer is $9$ .

$9$

Step 4

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

$(- 3, 9)$

Step 5