Find the Maximum/Minimum Value f(x)=-2x^2-4x-6

$f (x) = - 2 x^{2} - 4 x - 6$

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 $b$ .

$x = - \frac{- 4}{2 (- 2)}$

Remove parentheses.

$x = - \frac{- 4}{2 (- 2)}$

Simplify $- \frac{- 4}{2 (- 2)}$ .

Cancel the common factor of $- 4$ and $2$ .

Factor $2$ out of $- 4$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $- 2$ .

Cancel the common factor.

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

Rewrite the expression.

$x = - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$x = - 1$

Step 3

Evaluate $f (- 1)$ .

Replace the variable $x$ with $- 1$ in the expression.

$f (- 1) = - 2 {(- 1)}^{2} - 4 \cdot - 1 - 6$

Simplify the result.

Simplify each term.

Raise $- 1$ to the power of $2$ .

$f (- 1) = - 2 \cdot 1 - 4 \cdot - 1 - 6$

Multiply $- 2$ by $1$ .

$f (- 1) = - 2 - 4 \cdot - 1 - 6$

Multiply $- 4$ by $- 1$ .

$f (- 1) = - 2 + 4 - 6$

Simplify by adding and subtracting.

Add $- 2$ and $4$ .

$f (- 1) = 2 - 6$

Subtract $6$ from $2$ .

$f (- 1) = - 4$

The final answer is $- 4$ .

$- 4$

Step 4

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

$(- 1, - 4)$

Step 5

Find the Maximum/Minimum Value f(x)=-2x^2-4x-6

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