Get tangents (+ linear equation of tangents) of turning points in cfit graph

Hello,

I plotted a cfit graph with the matlab curveFitterTool. The original graph was fitted via polynomial fit of the 9th grade.

Now I am trying to plot and get the equation of the tangents of turning points of the derivated graph of the cfit graph.

I already got the 2nd derivation graph via "differentiate" function. There are turning points, if the second derivation = 0 and the third derivation =/ 0. But theres no clear polynomial equation for the fitted graph, so there are problems to filter the turning points.

Is there any way to get the turning point tangents and their linear equation of my derivated cfit-graph without having constand polynomial variable values?

%% Polynomial Fit Grad 9
[xData, yData] = prepareCurveData( x, y );

% Set up fittype and options.
ft = fittype( 'poly9' );

% Fit model to data.
[fitresult, gof] = fit( xData, yData, ft, 'Normalize', 'on' );

% Plot fit with data.
figure( 'Name', 'untitled fit 1' );
h = plot( fitresult, xData, yData );
legend( h, 'y vs. x', 'untitled fit 1', 'Location', 'NorthEast', 'Interpreter', 'none' );
% Label axes
xlabel( 'x', 'Interpreter', 'none' );
ylabel( 'y', 'Interpreter', 'none' );
grid on


%% Derivation
[fStrich,fStrichStrich] = differentiate(fitresult,xData);

Answer 1

[Matt J]

p=flip(coeffvalues(ft));

dp=polyder(p);
ddp=polyder(dp);
dddp=polyder(ddp); %3rd derivative

r=roots(ddp);

turingpoints=r(abs(polyval(dddp,r))>=tolerance)

[Torsten]

turningpoints=r(abs(polyval(dddp,r))>=tolerance & abs(imag(r )) < eps)

[Jerry]

Unfortunaly, this isn't working.

I have already tried getting the graph equation with polyder, but because of the curve fittin, the coefficients p1-p10 are not constant.

the graph p isn't equal to the fitted graph.

[Matt J]

No, p is a vector containing the polynomial coefficients. It is not a vector of curve samples. It is not clear why you would be trying to plot p since it is nothing more than an alternative representation of the fit you've already performed, but if you wanted to plot it, you would do,

plot(xData, polyval(p,xData))

[Jerry]

It isn't an alternative representation of the fit I already performed, it isn't even a similar graph:

[Matt J]

No way to know what's happening without you posting xData and yData.

[Matt J]

Jerry's comment moved here:

these are the x and y value points

Torsten's comment moved here:

At row 638, there is a sudden jump in the x-values from 1.0 to 1e14 in your data.

Please check and format the columns (Scientific format with comma instead of decimal point)

[Matt J]

I brushed out the defective-looking x,y data and reattached it here. I apply the method that I originally posted (plus Torsten's modification) and get the results below. There seem to be 3 turning points

load data
s=1e14; %scale factor
x=x/s;
y=y/s;

p=polyfit(x,y,9);

plot(s*x(1:10:end),s*y(1:10:end),'.',s*x,s*polyval(p,x),'-'); 
legend('Original Data','Fit','location','southeast')

dp=polyder(p);
ddp=polyder(dp);
dddp=polyder(ddp); %3rd derivative

r=roots(ddp);
turningpoints=r(abs(polyval(dddp,r))>=1e-6 & abs(imag(r )) < 1e-6)*s

turningpoints = 3×1

1.0e+14 * 2.4232 1.3979 1.0582

[Jerry]

This is working perfectly, thank you very much!