Find the Maximum/Minimum Value x^2+12x+10

$x^{2} + 12 x + 10$

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 $b$ .

$x = - \frac{12}{2 (1)}$

Remove parentheses.

$x = - \frac{12}{2 (1)}$

Simplify $- \frac{12}{2 (1)}$ .

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

Factor $2$ out of $12$ .

$x = - \frac{2 \cdot 6}{2 \cdot 1}$

Cancel the common factors.

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

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

Cancel the common factor.

$x = - \frac{2 \cdot 6}{2 \cdot 1}$

Rewrite the expression.

$x = - \frac{6}{1}$

Divide $6$ by $1$ .

$x = - 1 \cdot 6$

Multiply $- 1$ by $6$ .

$x = - 6$

Step 3

Evaluate $f (- 6)$ .

Replace the variable $x$ with $- 6$ in the expression.

$f (- 6) = {(- 6)}^{2} + 12 (- 6) + 10$

Simplify the result.

Simplify each term.

Raise $- 6$ to the power of $2$ .

$f (- 6) = 36 + 12 (- 6) + 10$

Multiply $12$ by $- 6$ .

$f (- 6) = 36 - 72 + 10$

Simplify by adding and subtracting.

Subtract $72$ from $36$ .

$f (- 6) = - 36 + 10$

Add $- 36$ and $10$ .

$f (- 6) = - 26$

The final answer is $- 26$ .

$- 26$

Step 4

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

$(- 6, - 26)$

Step 5

Find the Maximum/Minimum Value x^2+12x+10

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