Math geeks - help needed

If you had a third point it is easy to model with simple algebra.    As it is with  points on the graph and the first derivative of first point it is a differential equations problem.  What is the particular shape or curve that you wish to model?  Also, do you know the slope of the function at the second point?

y= (5/4)x^3, with only a little understanding of calculus needed.

Start with the generic formula y = ax^3 + bx^2 + cx + d.    There is enough information given  to generate this answer.

We see d = 0 because given the point (0,0)  we have 0 = a(0)^3 + b(0)^2 +c(0) +d

this gives 0 = 0 + 0 + 0 + d,  implying d = 0.

the derivative of our generic function is y' = 3ax^2 + 2bx + c.

The slope of the first derivation is zero so at x = 0 then y' = 0.

0 = 3a(0)^2 + 2b(0) + c  implies that also c = 0.

So our generic function is now defined as y = ax^3 + bx^2.    

Since we want as smooth a curve as possible we make an assumption that b = 0.   This, because with two terms in our equation we get an oscillation that degrades "smooth."   With this assumption we can use the point (10,2)  and plug into y = ax^3 to solve for a.

This satisfies all given conditions.

Bythe way, my fee for tutoring is $50/hr.      Buy me a beer and we will call it even....  

P.S.   I, by no stretch of anyone's imagination, can be called a  "geek."    

P.S.S     Lately I'm partial to anything Blue Moon.   Any of the seasonal brews will do nicely....  

You are correct in that we can pick an equation of any degree and fit according.   Not knowing any of the points in between the two given means we have no clue as to how accurate any equation of any degree will be.   Therefore any degree we wish to use will be just as good as another.