Let's just put it in vertex form so we know what it is
y=x^2+2x+1
y-1=x^2+2x
y-1+1=x^2+2x+1
y=(x+1)^2
So the vertex is (-1, 0), and since the coefficient is 1 is positive this vertex is the absolute minimum for the parabola.
So plot points for x={-3,-2,0,1} to get
(-3, 4),(-2, 1),(-1, 0),(0, 1),(1, 4)