The given quadratic function is $f(x) = (x - 4)^2 + 1$.

1. Domain:

The domain of a quadratic function is all real numbers because there are no restrictions on the values that $x$ can take. Therefore, the domain is: $\text{Domain}: (-\infty, \infty)$

2. Range:

The range is determined by the values that $f(x)$ can take.

The function is in vertex form $f(x) = (x - 4)^2 + 1$, where the vertex is at $(4, 1)$. Since this is a quadratic function that opens upwards (because the coefficient of $(x - 4)^2$ is positive), the minimum value of the function occurs at $x = 4$, where $f(4) = 1$.

Thus, the function's output (range) will always be greater than or equal to 1. Therefore, the range is: $\text{Range}: [1, \infty)$

Final Answer:

• Domain: $(-\infty, \infty)$
• Range: $[1, \infty)$

Would you like further details or have any questions?

Here are 5 related questions to explore further:

1. How do you find the vertex of any quadratic function?
2. How would the domain and range change if the quadratic opened downward?
3. How do transformations affect the graph of a quadratic function?
4. What is the axis of symmetry for the given quadratic function?
5. How does completing the square help in finding the vertex form?

Tip: Always check the sign of the leading coefficient to determine if a parabola opens upwards or downwards.