shift trendline up and down

[jonr123]

Hello,

I have a exponential trendline of my data. My data is pressure in function of the temperature in of a gas in a container. Now I want to shift the trendline based on the values on 20°C. I want to shift the trendline +0.2 and -0.2 at 20°C so I can read on 50°C what my pressure can be without measuring at 50°C. Is it possible to draw a trendline with a formula (e^(x+0.2)) than?

thanks in advance!

[AliGW]

1. You failed to thank your helpers in your first thread here, and you didn't mark it as solved. Please do this now: https://www.excelforum.com/excel-pro...dex-sheet.html
2. There are instructions at the top of the page explaining how to attach your sample workbook.

[MrShorty]

If you are not for some reason required to use the chart trendline feature, the easiest approach would be to use the LINEST() or LOGEST() functions to generate the regression coefficients directly in the spreadsheet. Then it will be relatively simple to add the desired offset to the final formula to calculate the desired values.
LOGEST() help file: https://support.microsoft.com/en-us/...b-a272c1d18b4b

[jonr123]

Thank you for your answer, i can calculate now the trendline, but do you know how I can draw this trendline on the graph, wihout showing any datapoint? I basically want to have my measured values, than make a trendline and than shift this trendline 0.2 up and 0.2 down so I can make margins for the QC department who measure pressure on every container.

[MrShorty]

how I can draw this trendline on the graph, wihout showing any datapoint?

What do you mean by this?

The only way I know to do this is to calculate data points for the offset curves and then add those values to the chart as a new data series. So a lot depends on what you mean by "without showing any datapoint."

[jonr123]

I already fixed that (I added new dataset and than maked the data markes invisible) problem is that when I add +0.2 and -0.2 to all the data that doesn't work for an exponential trendline. The higher the x value the more the curves shift to eachother,... I don't think this problem is solvable...

[davsth]

I think you are not saying what you want

if you followed MrShorty's suggestion you would have got there. the function is better than the tool on a graph, but more complex to explain using the logest function

or crudely taken form your graph say in H3 and copy down
=3.1518*EXP(0.0216*D3) to work out your y values then have columns with this + or - 0.2 and this will give uniform gaps

i3=h3+0.2 (or whatever you fancy)
j3=h3-0.2