The equation given is $y = (x - 3)^2 + 1$. This is the equation of a parabola in vertex form, $y = a(x - h)^2 + k$, where the vertex is at the point $(h, k)$.

In this case, the vertex is at $(3, 1)$. The parabola opens upwards because the coefficient of the squared term is positive (i.e., 1). Now, let's find the points to graph the parabola:

1. Vertex: $(3, 1)$

2. Two points to the left of the vertex:

   • For $x = 2$: $y = (2 - 3)^2 + 1 = 1^2 + 1 = 2$, so $(2, 2)$
   • For $x = 1$: $y = (1 - 3)^2 + 1 = 2^2 + 1 = 5$, so $(1, 5)$

3. Two points to the right of the vertex:

   • For $x = 4$: $y = (4 - 3)^2 + 1 = 1^2 + 1 = 2$, so $(4, 2)$
   • For $x = 5$: $y = (5 - 3)^2 + 1 = 2^2 + 1 = 5$, so $(5, 5)$

So, the five points to plot on the graph are:

• $(3, 1)$ (vertex)
• $(2, 2)$
• $(1, 5)$
• $(4, 2)$
• $(5, 5)$

Would you like further clarification or a graph of this parabola?

Here are five related questions to deepen understanding:

1. What is the general form of a quadratic equation?
2. How can we determine whether a parabola opens upwards or downwards?
3. How does changing the value of $h$ in the vertex form affect the graph?
4. What is the axis of symmetry for this parabola?
5. How does the value of $k$ affect the position of the parabola?

Tip: The vertex form $y = a(x-h)^2 + k$ makes it easy to identify the vertex and understand transformations like shifting.