1. **State the problem:** Find the minimum value of the quadratic function $$y = x^2 + 6x + 10$$. 2. **Recall the formula:** A quadratic function $$y = ax^2 + bx + c$$ has its vertex at $$x = -\frac{b}{2a}$$. The vertex represents the minimum point if $$a > 0$$. 3. **Identify coefficients:** Here, $$a = 1$$, $$b = 6$$, and $$c = 10$$. 4. **Calculate the x-coordinate of the vertex:** $$x = -\frac{6}{2 \times 1} = -\frac{6}{2} = -3$$. 5. **Find the minimum value by substituting $$x = -3$$ into the function:** $$y = (-3)^2 + 6(-3) + 10 = 9 - 18 + 10$$ $$y = 1$$. 6. **Conclusion:** The minimum value of $$y = x^2 + 6x + 10$$ is $$1$$, which occurs at $$x = -3$$.