Find the Maximum/Minimum Value f(x)=6x^2+12x-7

$f (x) = 6 x^{2} + 12 x - 7$

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 $b$ .

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

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.

$f (- 1) = 6 {(- 1)}^{2} + 12 (- 1) - 7$

Simplify the result.

Simplify each term.

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

$f (- 1) = 6 \cdot 1 + 12 (- 1) - 7$

Multiply $6$ by $1$ .

$f (- 1) = 6 + 12 (- 1) - 7$

Multiply $12$ by $- 1$ .

$f (- 1) = 6 - 12 - 7$

Simplify by subtracting numbers.

Subtract $12$ from $6$ .

$f (- 1) = - 6 - 7$

Subtract $7$ from $- 6$ .

$f (- 1) = - 13$

The final answer is $- 13$ .

$- 13$

Step 4

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

$(- 1, - 13)$

Step 5

Find the Maximum/Minimum Value f(x)=6x^2+12x-7

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