Find the Maximum/Minimum Value 2x^2+12x-1

$2 x^{2} + 12 x - 1$

Step 1

The minimum of a quadratic function occurs at $x = - \frac{b}{2 a}$ . If $a$ is positive, the minimum value of the function is $f (- \frac{b}{2 a})$ .

$f_{min}$ $x = a x^{2} + b x + c$ 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 $b$ .

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

Remove parentheses.

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

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

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

Factor $2$ out of $12$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Factor $2$ out of $6$ .

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

Cancel the common factors.

Factor $2$ out of $2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $3$ by $1$ .

$x = - 1 \cdot 3$

Multiply $- 1$ by $3$ .

$x = - 3$

Step 3

Evaluate $f (- 3)$ .

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

$f (- 3) = 2 {(- 3)}^{2} + 12 (- 3) - 1$

Simplify the result.

Simplify each term.

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

$f (- 3) = 2 \cdot 9 + 12 (- 3) - 1$

Multiply $2$ by $9$ .

$f (- 3) = 18 + 12 (- 3) - 1$

Multiply $12$ by $- 3$ .

$f (- 3) = 18 - 36 - 1$

Simplify by subtracting numbers.

Subtract $36$ from $18$ .

$f (- 3) = - 18 - 1$

Subtract $1$ from $- 18$ .

$f (- 3) = - 19$

The final answer is $- 19$ .

$- 19$

Step 4

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

$(- 3, - 19)$

Step 5

Find the Maximum/Minimum Value 2x^2+12x-1

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