Let's examine the given function first: f(x) = x^2 + 1 is the same as f(x) = 1(x-0)^2 + 1. The vertex of the graph of this function is at (0, 1). Let x=0 to find the y-intercept: f(0)=0^2+1 = 1; y-int. is at (0,1) (which happens to be the vertex also)
Comparing f(x) = x^2 + 1 to y = x^2, we see that the only difference is that f(x) has a vertical offset of 1. So: Graph y=x^2. Then translate the whole graph UP by 1 unit. That's it. Note (again) that the vertex will be at (0,1), and (0,1) is also the y-intercept.
