1. The problem is to find the coordinates of the vertex of the parabola given by the equation $$y = x^2 + 14x + 50$$. 2. The vertex of a parabola in the form $$y = ax^2 + bx + c$$ can be found using the formula for the x-coordinate of the vertex: $$x = -\frac{b}{2a}$$. 3. Here, $$a = 1$$ and $$b = 14$$. Substitute these values into the formula: $$x = -\frac{14}{2 \times 1} = -\frac{14}{2} = -7$$. 4. To find the y-coordinate of the vertex, substitute $$x = -7$$ back into the original equation: $$y = (-7)^2 + 14(-7) + 50 = 49 - 98 + 50 = 1$$. 5. Therefore, the vertex of the parabola is at the point $$(-7, 1)$$. This point represents the minimum of the parabola since the coefficient of $$x^2$$ is positive, indicating it opens upwards.