Find the Maximum/Minimum Value f(x)=1/6x^2-2x-9

$f (x) = \frac{1}{6} x^{2} - 2 x - 9$

Step 1

Combine $\frac{1}{6}$ and $x^{2}$ .

$f (x) = \frac{x^{2}}{6} - 2 x - 9$

Step 2

The minimum of a quadratic function occurs at $x = - \frac{b}{2 a}$ . If $a$ is positive, the minimum value of the function is $f (- \frac{b}{2 a})$ .

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

Step 3

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

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

$x = - \frac{- 2}{2 (0.1\overline{6})}$

Remove parentheses.

$x = - \frac{- 2}{2 (0.1\overline{6})}$

Simplify $- \frac{- 2}{2 (0.1\overline{6})}$ .

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

Factor $2$ out of $- 2$ .

$x = - \frac{2 \cdot - 1}{2 \cdot 0.1\overline{6}}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 0.1\overline{6}$ .

$x = - \frac{2 \cdot - 1}{2 (0.1\overline{6})}$

Cancel the common factor.

$x = - \frac{2 \cdot - 1}{2 \cdot 0.1\overline{6}}$

Rewrite the expression.

$x = - \frac{- 1}{0.1\overline{6}}$

Simplify the expression.

Divide $- 1$ by $0.1\overline{6}$ .

$x = - - 6$

Multiply $- 1$ by $- 6$ .

$x = 6$

Step 4

Evaluate $f (6)$ .

Replace the variable $x$ with $6$ in the expression.

$f (6) = \frac{{(6)}^{2}}{6} - 2 \cdot 6 - 9$

Simplify the result.

Simplify each term.

Raise $6$ to the power of $2$ .

$f (6) = \frac{36}{6} - 2 \cdot 6 - 9$

Divide $36$ by $6$ .

$f (6) = 6 - 2 \cdot 6 - 9$

Multiply $- 2$ by $6$ .

$f (6) = 6 - 12 - 9$

Simplify by subtracting numbers.

Subtract $12$ from $6$ .

$f (6) = - 6 - 9$

Subtract $9$ from $- 6$ .

$f (6) = - 15$

The final answer is $- 15$ .

$- 15$

Step 5

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

$(6, - 15)$

Step 6

Find the Maximum/Minimum Value f(x)=1/6x^2-2x-9

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