To find the range of the function $f(x) = (x - 4)^2 + 1$, let's analyze it step by step:

1. Form of the function:

The function is a quadratic function, written as $f(x) = (x - 4)^2 + 1$. This is a transformation of the basic quadratic function $y = x^2$, which is a parabola opening upwards.

2. Vertex:

The function $f(x) = (x - 4)^2 + 1$ is in vertex form, $f(x) = (x - h)^2 + k$, where $(h, k)$ is the vertex of the parabola.

• In this case, $h = 4$ and $k = 1$, so the vertex is at $(4, 1)$.

3. Behavior of the function:

Since the parabola opens upwards (because the coefficient of $(x - 4)^2$ is positive), the minimum value of $f(x)$ occurs at the vertex. At the vertex, the function takes the value $f(4) = (4 - 4)^2 + 1 = 0 + 1 = 1$.

4. Range:

The minimum value of the function is $1$ (at $x = 4$), and as $x$ moves away from 4 in either direction, the value of $f(x)$ increases. Thus, the range of the function is all values greater than or equal to 1.

Final Answer:

The range of the function is $[1, \infty)$.

Would you like more details or have any questions?

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Here are 5 related questions to further explore the concept:

1. What is the domain of the function $f(x) = (x - 4)^2 + 1$?
2. How would the range change if the function were $f(x) = -(x - 4)^2 + 1$?
3. Can you find the x-intercepts of the function $f(x) = (x - 4)^2 + 1$?
4. What is the axis of symmetry for the function $f(x) = (x - 4)^2 + 1$?
5. How would shifting the vertex affect the range of this quadratic function?

Tip: In vertex form $f(x) = a(x-h)^2 + k$, the value of $k$ always determines the minimum or maximum value depending on the direction the parabola opens.