Question 1005593
This response is not a complete answer nor full solution...


There are two data points of  (2, 120411) and  (7, 1083699).   The model you are using is {{{y=p*e^(rt)}}} which is exponential.  What you want is something LINEAR from this exponential equation, and some of your data, specifically the second coordinates, need to be modified.  Use base for the Natural Logarithm, and find natural log of both sides of the equation.  That will give you, upon appropriate simplification, the linear form of equation to use.


Your two data points would be   (2, ln(120411) )  and  (7, ln(1083699) ).


Much of the process is what you learned in "Algebra 1".


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ADDITIONAL PROCESS DESCRIPTION


{{{ln(y)=ln(p)+rt*ln(e)}}}
{{{ln(y)=ln(p)+rt}}}
{{{highlight_green(ln(y)=rt+ln(p))}}}-----Linear Form as Slope-Intercept form, slope is r, vertical axis intercept is {{{ln(p)}}}.
The points to use for help either in graphing or finding r, and p, are ( 2, 11.69867 ) and
(7,  13.85989).


You can use those two points, in that form, to find slope, and this slope will be the value for r.


Use the found slope and either of the treated points to calculate the vertical axis intercept, which will be the value of  ln(p); and then use your calculator or log table to get the value of p.


Once those are found, you have a more specific model for {{{y=p*e^(rt)}}}.