Graph this quadratic funtion f (x) = x^2+1

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.