Find the Maximum/Minimum Value y=-6x^2-12x-1

$y = - 6 x^{2} - 12 x - 1$

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 $b$ .

$x = - \frac{- 12}{2 (- 6)}$

Remove parentheses.

$x = - \frac{- 12}{2 (- 6)}$

Simplify $- \frac{- 12}{2 (- 6)}$ .

Cancel the common factor of $- 12$ and $2$ .

Factor $2$ out of $- 12$ .

$x = - \frac{2 \cdot - 6}{2 \cdot - 6}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 6$ .

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

Cancel the common factor.

$x = - \frac{2 \cdot - 6}{2 \cdot - 6}$

Rewrite the expression.

$x = - \frac{- 6}{- 6}$

Cancel the common factor of $- 6$ .

Cancel the common factor.

$x = - \frac{- 6}{- 6}$

Rewrite the expression.

$x = - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$x = - 1$

Step 3

Evaluate $f (- 1)$ .

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

$f (- 1) = - 6 {(- 1)}^{2} - 12 \cdot - 1 - 1$

Simplify the result.

Simplify each term.

Raise $- 1$ to the power of $2$ .

$f (- 1) = - 6 \cdot 1 - 12 \cdot - 1 - 1$

Multiply $- 6$ by $1$ .

$f (- 1) = - 6 - 12 \cdot - 1 - 1$

Multiply $- 12$ by $- 1$ .

$f (- 1) = - 6 + 12 - 1$

Simplify by adding and subtracting.

Add $- 6$ and $12$ .

$f (- 1) = 6 - 1$

Subtract $1$ from $6$ .

$f (- 1) = 5$

The final answer is $5$ .

$5$

Step 4

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

$(- 1, 5)$

Step 5

Find the Maximum/Minimum Value y=-6x^2-12x-1

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