Find the Maximum/Minimum Value f(x)=-2x^2-12x-20

$f (x) = - 2 x^{2} - 12 x - 20$

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

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

$x = - \frac{- 12}{2 (- 2)}$

Remove parentheses.

$x = - \frac{- 12}{2 (- 2)}$

Simplify $- \frac{- 12}{2 (- 2)}$ .

Cancel the common factor of $- 12$ and $2$ .

Factor $2$ out of $- 12$ .

$x = - \frac{2 \cdot - 6}{2 \cdot - 2}$

Cancel the common factors.

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

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

Cancel the common factor.

$x = - \frac{2 \cdot - 6}{2 \cdot - 2}$

Rewrite the expression.

$x = - \frac{- 6}{- 2}$

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

Factor $- 2$ out of $- 6$ .

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

Cancel the common factors.

Factor $- 2$ out of $- 2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $3$ by $1$ .

$x = - 1 \cdot 3$

Multiply $- 1$ by $3$ .

$x = - 3$

Step 3

Evaluate $f (- 3)$ .

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

$f (- 3) = - 2 {(- 3)}^{2} - 12 \cdot - 3 - 20$

Simplify the result.

Simplify each term.

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

$f (- 3) = - 2 \cdot 9 - 12 \cdot - 3 - 20$

Multiply $- 2$ by $9$ .

$f (- 3) = - 18 - 12 \cdot - 3 - 20$

Multiply $- 12$ by $- 3$ .

$f (- 3) = - 18 + 36 - 20$

Simplify by adding and subtracting.

Add $- 18$ and $36$ .

$f (- 3) = 18 - 20$

Subtract $20$ from $18$ .

$f (- 3) = - 2$

The final answer is $- 2$ .

$- 2$

Step 4

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

$(- 3, - 2)$

Step 5

Find the Maximum/Minimum Value f(x)=-2x^2-12x-20

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