Vertex Discovery: Analyzing the Quadratic y = (x-6)² + 1

To find the vertex of the parabola given by the equation $y = (x-6)^2 + 1$, we observe that this equation is already in the vertex form:

$y = (x-h)^2 + k$

where $(h, k)$ is the vertex of the parabola.

From the equation $y = (x-6)^2 + 1$, we can clearly identify:

• $h = 6$, because the expression inside the square is $(x-6)$.
• $k = 1$, the constant added to the squared term.

Therefore, the vertex of the parabola is $(6, 1)$.

Comparing this with the available choices, we see that choice 4, $(6, 1)$, is the correct answer.

In conclusion, the vertex of the parabola is $\boxed{(6, 1)}$.