Old Ans. The Y data takes negative values. You could try fitting $Y+11.0$ or more generally $Y+Y_{min}$ with a gamma distribution which may also need an offset. Without more information, it is guesswork. Either show tabular data or forget getting a good answer, not clear enough.
After comments, edits and more information.
OK, you can try a less than critically damped sine wave.
Here, for example is $$A \exp (-\lambda t) \sin (\omega t+\theta )$$ where $\{A\to -98.0396,\lambda \to 0.0126426,\omega \to 0.0259994,\theta \to 3.08326\}$
This is the result of assuming that glucose is a less than critically damped negative feedback network moderated by insulin. Caution, this is a quick and dirty answer and attention should be paid to residual and error types, whether or not the data is proportional and myriad other factors.
For example, you might be better served by getting rid of the {0,0} first point, and let the algorithm decide at what time, $t$, the first root of the equation is zero, as this may not be when you think it occurs. Why? If you think about it for a moment, the zero concentration first root is implied by each and every data entry already, so that entering {0,0} is redundant, arbitrary and misleading. Moreover, the feedback loop may not occur in linear time, but in time to some power. Doing those things we obtain
$$A \exp \left(-\lambda x^B\right) \sin \left(\omega x^B+\theta \right)\;,$$
where $\{A\to -104.526,B\to 0.936079,\lambda \to 0.0178241,\omega \to -0.0362561,\theta \to 0.14761\}$
Where the start time is 4.07133 min, and the correlation between model and data is $r=0.99783$. If you want to find any particular predicted concentration, just input that time in the formula.
Keep in mind that this is just an ordinary least squares model, that it is quick and dirty, that I have not examined the original data and its error structure, and that I prefer biological models where $r^2\geq0.999+$ unless there is some outstanding reason (e.g., demonstrated to be noise) why it cannot be that precise. Many of my colleagues are happy when they get $r\geq0.8$, however, that is all too often due to a lack of proper modelling, and not because of noise. In this data, the precision of the concentration data is sometimes only one or two decimal places, and a subtraction (which doubles the error) has been performed. Without better data, it would be hard to exceed the $r^2=0.995665$ found. That is, in this case, the last 1/2% explained fraction would be harder to obtain. Sorry, I use Mathematica, not Python. If you want the Mathematica code, let me know.
