1. **Stating the problem:** Find the vertex of the quadratic function $$f(x) = 4(x-3)^2 - 6$$. 2. **Formula and explanation:** The vertex form of a quadratic function is $$f(x) = a(x-h)^2 + k$$, where $(h, k)$ is the vertex. 3. **Identify vertex coordinates:** Comparing, we see $$a=4$$, $$h=3$$, and $$k=-6$$. 4. **Conclusion:** Therefore, the vertex of the function is at the point $$\boxed{(3, -6)}$$. This means the parabola opens upwards (since $$a=4>0$$) and its lowest point is at $$x=3$$ with value $$-6$$.