To find the minimum value of the quadratic function $g(x) = x^2 - 6x - 12$, we can use the method of completing the square or applying the vertex formula. Here's the solution:

Step 1: Write the quadratic in vertex form

The general quadratic equation is $ax^2 + bx + c$. The vertex form is given by:

$g(x) = a(x - h)^2 + k$

where $h = -\frac{b}{2a}$ and $k$ is the minimum value.

In this case:

• $a = 1$, $b = -6$, and $c = -12$.

Step 2: Find $h$ (x-coordinate of the vertex)

$h = -\frac{b}{2a} = -\frac{-6}{2(1)} = 3$

Step 3: Find $k$ (minimum value of the function)

Substitute $h = 3$ into the original equation to find the corresponding $g(h)$: $g(3) = (3)^2 - 6(3) - 12$ $g(3) = 9 - 18 - 12 = -21$

Final Answer:

The minimum value of $g(x)$ is $-21$.

Do you want more details or clarification?

Here are some related questions to explore:

1. How is completing the square used to find the vertex?
2. Why is the vertex the minimum point for a parabola that opens upwards?
3. How can we verify this result using calculus?
4. What happens if $a < 0$ in the quadratic equation?
5. Can this method apply to non-integer coefficients?

Tip: Remember, the vertex formula simplifies solving quadratic equations without graphing!