To solve the problem of determining the range of the function $f(x) = (x-4)^2 + 1$, we can proceed as follows:

Step-by-Step Solution

1. Analyze the Function:

   • The given function is of the form $f(x) = a(x-h)^2 + k$, which represents a quadratic equation.
   • This is a parabola that opens upwards because the coefficient of $(x-4)^2$ (i.e., $a = 1$) is positive.
   • The vertex of the parabola is at $(h, k) = (4, 1)$.

2. Minimum Value:

   • Since the parabola opens upwards, the minimum value of $f(x)$ occurs at the vertex, where $f(4) = 1$.
   • For any other value of $x$, $(x-4)^2 \geq 0$. Therefore, $f(x) = (x-4)^2 + 1 \geq 1$.

3. Range:

   • The function takes all values $f(x) \geq 1$.
   • Hence, the range of $f(x)$ is $[1, \infty)$.

Correct Answer:

The range is $f(x) \geq 1$, which matches option (4).

Would you like additional details or have any questions?
Here are five related questions to explore:

1. How do you find the domain of functions like this one?
2. What changes in the range if the function is shifted downwards, such as $f(x) = (x-4)^2 - 3$?
3. How do you graph this function step-by-step to confirm the range visually?
4. What happens to the range if $f(x)$ becomes $-(x-4)^2 + 1$?
5. Can we determine the symmetry of the graph and how it relates to the vertex?

Tip: The range of a parabola depends on whether it opens upwards ($a > 0$) or downwards ($a < 0$). Always focus on the vertex for this.