Algebra Examples

$g (x) = x^{2} - 6 x - 12$

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

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Step 2.1

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

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

Step 2.2

Remove parentheses.

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

Step 2.3

Simplify $- \frac{- 6}{2 (1)}$ .

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Step 2.3.1

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

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Step 2.3.1.1

Factor $2$ out of $- 6$ .

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

Step 2.3.1.2

Cancel the common factors.

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Step 2.3.1.2.1

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

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

Step 2.3.1.2.2

Cancel the common factor.

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

Step 2.3.1.2.3

Rewrite the expression.

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

Step 2.3.1.2.4

Divide $- 3$ by $1$ .

$x = - - 3$

Step 2.3.2

Multiply $- 1$ by $- 3$ .

$x = 3$

Step 3

Evaluate $f (3)$ .

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Step 3.1

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

$g (3) = {(3)}^{2} - 6 \cdot 3 - 12$

Step 3.2

Simplify the result.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

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

$g (3) = 9 - 6 \cdot 3 - 12$

Step 3.2.1.2

Multiply $- 6$ by $3$ .

$g (3) = 9 - 18 - 12$

Step 3.2.2

Simplify by subtracting numbers.

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Step 3.2.2.1

Subtract $18$ from $9$ .

$g (3) = - 9 - 12$

Step 3.2.2.2

Subtract $12$ from $- 9$ .

$g (3) = - 21$

Step 3.2.3

The final answer is $- 21$ .

$- 21$

Step 4

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

$(3, - 21)$

Step 5