Find the Maximum/Minimum Value -x^2-2x-5

$- x^{2} - 2 x - 5$

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 $b$ .

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

Remove parentheses.

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

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

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

Factor $2$ out of $- 2$ .

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

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

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

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

$x = - - - 1$

Multiply $- - - 1$ .

Multiply $- 1$ by $- 1$ .

$x = - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$x = - 1$

Step 3

Evaluate $f (- 1)$ .

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

$f (- 1) = - {(- 1)}^{2} - 2 \cdot - 1 - 5$

Simplify the result.

Simplify each term.

Multiply $- 1$ by ${(- 1)}^{2}$ by adding the exponents.

Multiply $- 1$ by ${(- 1)}^{2}$ .

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

$f (- 1) = (- 1) {(- 1)}^{2} - 2 \cdot - 1 - 5$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f (- 1) = {(- 1)}^{1 + 2} - 2 \cdot - 1 - 5$

Add $1$ and $2$ .

$f (- 1) = {(- 1)}^{3} - 2 \cdot - 1 - 5$

Raise $- 1$ to the power of $3$ .

$f (- 1) = - 1 - 2 \cdot - 1 - 5$

Multiply $- 2$ by $- 1$ .

$f (- 1) = - 1 + 2 - 5$

Simplify by adding and subtracting.

Add $- 1$ and $2$ .

$f (- 1) = 1 - 5$

Subtract $5$ from $1$ .

$f (- 1) = - 4$

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 -x^2-2x-5

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