Hello there buddy

I have attached a copy of the graph of the data to explain a little more what I am talking about. Basically, I think that it is unlikely that you would be asked to perform numerical curve fitting until you get to college as the process usually requires the computer number crunching or fairly technical statistical processes. However, that's not to say that you couldn't answer the question this way.

You could, for example, do your best to actively interpolate your data to the y-axis and thus find the y-intercept and then guesstimate the a and b coefficients in the equation where c is the value of the y-intercept that you have found and a is positive. This is because your data looks a lot like a quadratic curve. It shouldn't take you too long to do this as your data looks fairly symmetrical around age 50.

If you assume that the graph is symmetrical about age 50 (as it appears to be) you could then use the equation where to find b in terms of a (in this case . This would then leave you with the equation and, since you have already estimated c, you would only need to vary a to obtain different values for your graph. I found that setting c at about 1200 and a at about 0.41 gave a pretty good result, though of course you will need to vary this. If you follow this method then you should then compare the fitted graph with the graph from your data and interperate from there

However, you will need to be careful to state that the fitted model doesn't neccersarily reflect the data from real life and that you are only extrapolating from information that you have

I hope that helped,

Paul

PS: If you didn't want to use the method above then you could simply do something similar to what I have done on the graph, although the method of fitting numerically will likely get you far more marks