Find the Maximum/Minimum Value 9-4x-x^2

$9 - 4 x - x^{2}$

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.

$f (- 2) = 9 - 4 \cdot - 2 - {(- 2)}^{2}$

Simplify the result.

Simplify each term.

Multiply $- 4$ by $- 2$ .

$f (- 2) = 9 + 8 - {(- 2)}^{2}$

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

$f (- 2) = 9 + 8 - 1 \cdot 4$

Multiply $- 1$ by $4$ .

$f (- 2) = 9 + 8 - 4$

Simplify by adding and subtracting.

Add $9$ and $8$ .

$f (- 2) = 17 - 4$

Subtract $4$ from $17$ .

$f (- 2) = 13$

The final answer is $13$ .

$13$

Step 4

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

$(- 2, 13)$

Step 5

Find the Maximum/Minimum Value 9-4x-x^2

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