1. **State the problem:** Find the turning point (vertex) of the curve given by the quadratic function $$y = x^2 - 6x + 8$$ by completing the square. 2. **Rewrite the quadratic expression:** Start with the quadratic and linear terms. We will complete the square on $$x^2 - 6x$$. 3. **Complete the square:** Take half of the coefficient of $$x$$ (which is $$-6$$), divided by 2, and square it: $$\left(\frac{-6}{2}\right)^2 = (-3)^2 = 9$$ Add and subtract 9 inside the expression to keep it equivalent: $$y = (x^2 - 6x + 9) - 9 + 8$$ 4. **Simplify inside parentheses:** The expression inside parentheses is a perfect square trinomial: $$y = (x - 3)^2 - 1$$ 5. **Identify the turning point:** The vertex form $$y = (x - h)^2 + k$$ shows the turning point at $$(h, k)$$. Here, $$h = 3$$ and $$k = -1$$. **Final answer:** The coordinates of the turning point are $$(3, -1)$$.