1. The problem is to find the vertex of the quadratic function given by the equation $$y = x^2 - 8x + 14$$. 2. Recall that the vertex form of a quadratic function is $$y = a(x-h)^2 + k$$, where $$(h,k)$$ is the vertex. 3. To find the vertex, we complete the square for the quadratic expression: $$y = x^2 - 8x + 14$$ 4. Take the coefficient of $x$, which is $-8$, divide it by 2, and square it: $$\left(\frac{-8}{2}\right)^2 = (-4)^2 = 16$$ 5. Add and subtract 16 inside the equation to complete the square: $$y = (x^2 - 8x + 16) + 14 - 16$$ 6. Rewrite the perfect square trinomial as a squared binomial: $$y = (x - 4)^2 - 2$$ 7. From the vertex form, the vertex is at $$(h,k) = (4, -2)$$. Final answer: The vertex of the parabola is at $$\boxed{(4, -2)}$$.