Find the Maximum/Minimum Value f(x)=-x^2-8x-3

$f (x) = - x^{2} - 8 x - 3$

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 $b$ .

$x = - \frac{- 8}{2 (- 1)}$

Remove parentheses.

$x = - \frac{- 8}{2 (- 1)}$

Simplify $- \frac{- 8}{2 (- 1)}$ .

Cancel the common factor of $- 8$ and $2$ .

Factor $2$ out of $- 8$ .

$x = - \frac{2 \cdot - 4}{2 \cdot - 1}$

Move the negative one from the denominator of $\frac{- 4}{- 1}$ .

$x = - (- 1 \cdot - 4)$

Rewrite $- 1 \cdot - 4$ as $- - 4$ .

$x = - - - 4$

Multiply $- - - 4$ .

Multiply $- 1$ by $- 4$ .

$x = - 1 \cdot 4$

Multiply $- 1$ by $4$ .

$x = - 4$

Step 3

Evaluate $f (- 4)$ .

Replace the variable $x$ with $- 4$ in the expression.

$f (- 4) = - {(- 4)}^{2} - 8 \cdot - 4 - 3$

Simplify the result.

Simplify each term.

Raise $- 4$ to the power of $2$ .

$f (- 4) = - 1 \cdot 16 - 8 \cdot - 4 - 3$

Multiply $- 1$ by $16$ .

$f (- 4) = - 16 - 8 \cdot - 4 - 3$

Multiply $- 8$ by $- 4$ .

$f (- 4) = - 16 + 32 - 3$

Simplify by adding and subtracting.

Add $- 16$ and $32$ .

$f (- 4) = 16 - 3$

Subtract $3$ from $16$ .

$f (- 4) = 13$

The final answer is $13$ .

$13$

Step 4

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

$(- 4, 13)$

Step 5

Find the Maximum/Minimum Value f(x)=-x^2-8x-3

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