My first thought would be to assume a linear function but then the second derivative would be 0, automatically. The next simplest thing to do is to assume a quadratic function. I would be inclined to write the function as so when x= 0, this is just , when x= 2, this is just , and when x= 4, this is just [tex]a(2)^2+ b(2)+ c= 4a+ 2b+ c= 43.
That is, you have 4a- 2b+ c= 7, c= 13, and 4a+ 2b+ c= 43. It should be easy to solve those for a, b, and c and so write the quadratic function.
Of course, if , then and . Evaluate those at x= 0 and x= 1.
Now, that does not use the values at x= 6, or 8 at all. You could, if you wished, write a degree polynomial passing through those 5 points.
Again, the question, as you posed it, is impossible. There exist an infinite number of twice-differentiable functions passing through those 5 points. Unless there is more to this problem than you have told us, what I am suggesting is the simplest way to get an answer.