Interpolation in negative axis?

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Raju
Raju 2018-1-16
评论： Star Strider 2018-1-16
I have fitted my data with a gaussion function. the curve is basically lying in first coordinate and not starting from zero. I want to calculate the value of x for y=0 (x would then definitely come in 2nd coordinate). How to find this value of x. In figure, the first fitted curve (not starting from zero)
fun1 = @(p,xdata) p(1).*(1-exp(-xdata./p(2)))+5000;
I want to a value of x for y=0 (when curve touches x axis). How to find it. Any lead will be appricated.

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Star Strider
Star Strider 2018-1-16
Once you have your fitted ‘p’ values, you can find the x value at y=0 with the fzero function.
Try this:
fun1 = @(p,xdata) p(1).*(1-exp(-xdata./p(2)))+5000;
x_intercept = fzero(@(xdata) fun1(p,xdata), -1);
  2 个评论
Raju
Raju 2018-1-16
Thank you so much for your answer. It's working.
Star Strider
Star Strider 2018-1-16
As always, my pleasure.
If my Answer helped you solve your problem, please Accept it!

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