Find the Maximum/Minimum Value g(x)=-x^2-4x-5

$g (x) = - x^{2} - 4 x - 5$

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 $b$ .

$x = - \frac{- 4}{2 (- 1)}$

Remove parentheses.

$x = - \frac{- 4}{2 (- 1)}$

Simplify $- \frac{- 4}{2 (- 1)}$ .

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

Factor $2$ out of $- 4$ .

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

Move the negative one from the denominator of $\frac{- 2}{- 1}$ .

$x = - (- 1 \cdot - 2)$

Rewrite $- 1 \cdot - 2$ as $- - 2$ .

$x = - - - 2$

Multiply $- - - 2$ .

Multiply $- 1$ by $- 2$ .

$x = - 1 \cdot 2$

Multiply $- 1$ by $2$ .

$x = - 2$

Step 3

Evaluate $f (- 2)$ .

Replace the variable $x$ with $- 2$ in the expression.

$g (- 2) = - {(- 2)}^{2} - 4 \cdot - 2 - 5$

Simplify the result.

Simplify each term.

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

$g (- 2) = - 1 \cdot 4 - 4 \cdot - 2 - 5$

Multiply $- 1$ by $4$ .

$g (- 2) = - 4 - 4 \cdot - 2 - 5$

Multiply $- 4$ by $- 2$ .

$g (- 2) = - 4 + 8 - 5$

Simplify by adding and subtracting.

Add $- 4$ and $8$ .

$g (- 2) = 4 - 5$

Subtract $5$ from $4$ .

$g (- 2) = - 1$

The final answer is $- 1$ .

$- 1$

Step 4

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

$(- 2, - 1)$

Step 5

Find the Maximum/Minimum Value g(x)=-x^2-4x-5

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