# Find this polynomial using this graph.

• Aug 9th 2011, 04:22 PM
pieman91
Find this polynomial using this graph.
G'day,
I have created a graph on excel and have tried to extract the polynomial (I think that is what it is) but excel tells me the wrong polynomial.

Please open the doc file and extract the polynomial from the graph's data. The first graph is the graph I need the polynomial for. The second is the polynomial that excel gives me which is incorrect.

The first lot of data is for the graph that I want the polynomial for. The second lot of data is the data from the polynomial that excel gave which is incorrect.

This is for private purposes and not for schooling of any type.

I would greatly appreciate your help.
• Aug 10th 2011, 03:39 AM
Bacterius
Re: Find this polynomial using this graph.
Extract the polynomial? I think you mean approximate your data using a polynomial. You cannot model your data perfectly using a polynomial, because your data is rough and not smooth while a polynomial is a perfect mathematical curve, so you are just telling excel to try and find a polynomial that fits your data as closely as possible. Here the coefficient of determination for the polynomial \$\displaystyle 0.0088 x^2 - 0.11x + 16.565\$ is \$\displaystyle 0.9707\$ which means it is a very good (almost perfect, in fact) approximation of your data.

At least that's what I understand from your question. Feel free to clarify if I missed your point.
• Aug 11th 2011, 12:01 AM
pieman91
Re: Find this polynomial using this graph.
Quote:

Originally Posted by Bacterius
Extract the polynomial? I think you mean approximate your data using a polynomial. You cannot model your data perfectly using a polynomial, because your data is rough and not smooth while a polynomial is a perfect mathematical curve, so you are just telling excel to try and find a polynomial that fits your data as closely as possible. Here the coefficient of determination for the polynomial \$\displaystyle 0.0088 x^2 - 0.11x + 16.565\$ is \$\displaystyle 0.9707\$ which means it is a very good (almost perfect, in fact) approximation of your data.

At least that's what I understand from your question. Feel free to clarify if I missed your point.