What are the new values? How do they compare to the old values?

Also, the formulas you shared are all missing the square term and the=20
constant term:

=3D1.8485*(G1)+641.06*(G1)
=3D1.8485*(G2)+641.06*(G2)
=3D1.8485*(G3)+641.06*(G3)
etc.
=3D1.8485*(G288)+641.06*(G288)

You should have a2*x^2 + a1*x + a0
--=20
Regards,

Tushar Mehta
www.tushar-mehta.com
Excel, PowerPoint, and VBA add-ins, tutorials
Custom MS Office productivity solutions

In article ,=20
says...
Hi Tushar,
=20
Referencing the first example, the log trendline: the "extract" is a way =
to=20
make a duplicate of the Excel-created trendline, using the Excel-generate=
d=20
equation. You now have two trendlines - one on top of the other; however=
, on=20
the extract series, all plotted values can be determined - not possible w=
ith=20
normal function of Excel.
=20
That's what I want to do in the second example, the 2nd order polynomial=
=20
trendline, is create a second trendline, where I can access the values of=
=20
each point on the plotted curve. Do you know how to do this?
=20
As to changing number format, this has no impact on the issue that I can=
=20
rationalize. Can you help me further?
=20
Thanks, Phil
=20
Tshar Mehta" wrote:
=20
On the chart, double click the trendline equation, then select the=20
Number tab. Change the format to something that shows lots of decimal=
=20
places (scientific with 14 decimal places, for example).
=20
--=20
Regards,
=20
Tushar Mehta
www.tushar-mehta.com
Excel, PowerPoint, and VBA add-ins, tutorials
Custom MS Office productivity solutions
=20
In article ,=20
says...
I am calculating/plotting a log trendline extract on the equation
y =3D 533371Ln(x) - 3E+06. Ln(x) is the period (extract) number, sta=
rting at=20
1 and ending at 288. There are 288 plotted points. The value for 533=
=C3=A2=E2=3D3F=AC=3D3F and=20
3E=C3=A2=E2=3D3F=AC=3D3F are derived from the trendline equation each=
month, when a new value (and=20
account balance) is added to the series. The plotted extracted value=
s (y)=20
give me an exact duplicate of the Excel log trendline curve, and, a v=
alue=20
(forcast) for period 288, where the trendline crosses the right side =
of the=20
chart. The series looks like this, where cell L2 holds the 533 value=
, M2=20
holds the 3E value, and G* is the extract number. Works fine.
=20
=3D(L2)*LN(G1)-M2
=3D(L2)*LN(G2)-M2 =20
=3D(L2)*LN(G3)-M2
etc.
=3D(L2)*LN(G288)-M2
=20
Now I want to do a second series, using a 2nd order poly trendline wi=
th the=20
equation y =3D 1.8485x2+641.06x (given by Excel). The extract dimens=
ion=20
remains 1 through 288, and the series is:
=20
=3D1.8485*(G1)+641.06*(G1)
=3D1.8485*(G2)+641.06*(G2)
=3D1.8485*(G3)+641.06*(G3)
etc.
=3D1.8485*(G288)+641.06*(G288)
=20
The final value (G288) is coming up 183,228, when it should be in the=
region=20
of 340,000, the approximate value where the Excel-plotted trendline c=
rosses=20
the right side of the chart.
=20
Where is the logic flaw?
=20
=20
=20
=20