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.

The first one, i believe, would be correct. all the others have one thing in common, and that's that they have repeating numbers. in the second one, x and y have repeating 6's and 3's, and in the third, x and y have repeating 6's as well. in the last one, there are repeating 3's. the first one has no repeating numbers.