Find the Maximum/Minimum Value x^2+4x-12

$x^{2} + 4 x - 12$

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 $x = - \frac{b}{2 a}$

Find the value of $x = - \frac{b}{2 a}$ .

Substitute in the values of $a$ and $b$ .

$x = - \frac{4}{2 (1)}$

Remove parentheses.

$x = - \frac{4}{2 (1)}$

Simplify $- \frac{4}{2 (1)}$ .

Cancel the common factor of $4$ and $2$ .

Factor $2$ out of $4$ .

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

Cancel the common factors.

Factor $2$ out of $2 \cdot 1$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $2$ by $1$ .

$x = - 1 \cdot 2$

Multiply $- 1$ by $2$ .

$x = - 2$

Evaluate $f (- 2)$ .

Replace the variable $x$ with $- 2$ in the expression.

$f (- 2) = {(- 2)}^{2} + 4 (- 2) - 12$

Simplify the result.

Simplify each term.

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

$f (- 2) = 4 + 4 (- 2) - 12$

Multiply $4$ by $- 2$ .

$f (- 2) = 4 - 8 - 12$

Simplify by subtracting numbers.

Subtract $8$ from $4$ .

$f (- 2) = - 4 - 12$

Subtract $12$ from $- 4$ .

$f (- 2) = - 16$

The final answer is $- 16$ .

$- 16$

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

$(- 2, - 16)$

Find the Maximum/Minimum Value x^2+4x-12

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