Find the Maximum/Minimum Value y=-x^2-6x-5

$y = - x^{2} - 6 x - 5$

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 $b$ .

$x = - \frac{- 6}{2 (- 1)}$

Remove parentheses.

$x = - \frac{- 6}{2 (- 1)}$

Simplify $- \frac{- 6}{2 (- 1)}$ .

Cancel the common factor of $- 6$ and $2$ .

Factor $2$ out of $- 6$ .

$x = - \frac{2 \cdot - 3}{2 \cdot - 1}$

Move the negative one from the denominator of $\frac{- 3}{- 1}$ .

$x = - (- 1 \cdot - 3)$

Rewrite $- 1 \cdot - 3$ as $- - 3$ .

$x = - - - 3$

Multiply $- - - 3$ .

Multiply $- 1$ by $- 3$ .

$x = - 1 \cdot 3$

Multiply $- 1$ by $3$ .

$x = - 3$

Step 3

Evaluate $f (- 3)$ .

Replace the variable $x$ with $- 3$ in the expression.

$f (- 3) = - {(- 3)}^{2} - 6 \cdot - 3 - 5$

Simplify the result.

Simplify each term.

Raise $- 3$ to the power of $2$ .

$f (- 3) = - 1 \cdot 9 - 6 \cdot - 3 - 5$

Multiply $- 1$ by $9$ .

$f (- 3) = - 9 - 6 \cdot - 3 - 5$

Multiply $- 6$ by $- 3$ .

$f (- 3) = - 9 + 18 - 5$

Simplify by adding and subtracting.

Add $- 9$ and $18$ .

$f (- 3) = 9 - 5$

Subtract $5$ from $9$ .

$f (- 3) = 4$

The final answer is $4$ .

$4$

Step 4

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

$(- 3, 4)$

Step 5

Find the Maximum/Minimum Value y=-x^2-6x-5

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