Find the Maximum/Minimum Value f(x)=-2x^2-20x-50

$f (x) = - 2 x^{2} - 20 x - 50$

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 $b$ .

$x = - \frac{- 20}{2 (- 2)}$

Remove parentheses.

$x = - \frac{- 20}{2 (- 2)}$

Simplify $- \frac{- 20}{2 (- 2)}$ .

Cancel the common factor of $- 20$ and $2$ .

Factor $2$ out of $- 20$ .

$x = - \frac{2 \cdot - 10}{2 \cdot - 2}$

Cancel the common factors.

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

$x = - \frac{2 \cdot - 10}{2 (- 2)}$

Cancel the common factor.

$x = - \frac{2 \cdot - 10}{2 \cdot - 2}$

Rewrite the expression.

$x = - \frac{- 10}{- 2}$

Cancel the common factor of $- 10$ and $- 2$ .

Factor $- 2$ out of $- 10$ .

$x = - \frac{- 2 (5)}{- 2}$

Cancel the common factors.

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

$x = - \frac{- 2 \cdot 5}{- 2 \cdot 1}$

Cancel the common factor.

$x = - \frac{- 2 \cdot 5}{- 2 \cdot 1}$

Rewrite the expression.

$x = - \frac{5}{1}$

Divide $5$ by $1$ .

$x = - 1 \cdot 5$

Multiply $- 1$ by $5$ .

$x = - 5$

Step 3

Evaluate $f (- 5)$ .

Replace the variable $x$ with $- 5$ in the expression.

$f (- 5) = - 2 {(- 5)}^{2} - 20 \cdot - 5 - 50$

Simplify the result.

Simplify each term.

Raise $- 5$ to the power of $2$ .

$f (- 5) = - 2 \cdot 25 - 20 \cdot - 5 - 50$

Multiply $- 2$ by $25$ .

$f (- 5) = - 50 - 20 \cdot - 5 - 50$

Multiply $- 20$ by $- 5$ .

$f (- 5) = - 50 + 100 - 50$

Simplify by adding and subtracting.

Add $- 50$ and $100$ .

$f (- 5) = 50 - 50$

Subtract $50$ from $50$ .

$f (- 5) = 0$

The final answer is $0$ .

$0$

Step 4

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

$(- 5, 0)$

Step 5

Find the Maximum/Minimum Value f(x)=-2x^2-20x-50

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