Find the Ascending Area of y=(x-6)² + 4: Quadratic Function Analysis

To solve this problem, we first note that the function is expressed in vertex form as $y = (x-6)^2 + 4$. The vertex form of a quadratic function is $y = a(x - p)^2 + k$, where the vertex is located at $(p, k)$. For our function, this vertex is at $(6, 4)$.

Since the coefficient of $(x-6)^2$ is positive, the parabola opens upwards. This means that the lowest point on the parabola is the vertex, making $x = 6$ the axis of symmetry and the point at which the function changes from decreasing to increasing.

Thus, the function is increasing on the interval where $x > 6$. When $x$ is greater than 6, the value of $y$ increases as $x$ increases.

By examining the multiple-choice answers, we find that the choice that matches $6 < x$ is the correct one. Therefore, the ascending area of the function is for $x > 6$.

Therefore, the solution to the problem is $6 < x$.