Calculate the Positive Area of y=(x-4)²+1: Shifted Parabola Problem

To solve this problem, we will analyze the given quadratic function:

The function is $y = (x-4)^2 + 1$.

Step 1: Identify the vertex of the parabola.
The function $y = (x-4)^2 + 1$ is in the form $y = (x-h)^2 + k$, where $h = 4$ and $k = 1$. This gives the vertex at the point $(4, 1)$.

Step 2: Analyze the shape and direction of the parabola.
This parabola opens upwards because the coefficient of $(x-4)^2$ is positive. Hence, the vertex is the minimum point of the parabola.

Step 3: Determine when the function is positive.
Since the minimum value of the function at the vertex is $1$, and all quadratic functions in the form of $(x-h)^2 + k$ are non-negative, this means the function $y = (x-4)^2 + 1$ is always positive for any real number $x$.

Step 4: Comparison with the choices:
From the explanation, the function is always positive for all $x$. Thus, the correct choice is: For all X.

The positive area of the function covers all real numbers $x$.

Therefore, the function is positive for all $x$.