MATLAB: Fitting data into a quadratic curve

curve fittingcurvesdata fitquadratic

Hello,
I have some data that can be fitted into a quadratic curve of y=ax^2 + bx + c. My data passes through the origin, and has a horizontal slope near the origin too. So my question is how do i force b and c to be zero in the qudratic equation ? while doing that i want to estimate the error too, so how do i keep track of the error if i have more than one quadratic equation.
I tried using fittype and polyval, and i am familiar with them, but that doesn't do my task.
I attached my data for better understanding.
Thanks for any help.

Best Answer

To force ‘b’ and ‘c’ to be zero: 
B = volume_flow_rate1(:).^2 \ influence(:)                              % Estimate Regression Parameter
vfrv = linspace(min(volume_flow_rate1), max(volume_flow_rate1));        % Vector For Regression Evaluation

inflv = vfrv(:).^2 * B;                                                 % Evaluate Regression

figure();
plot(volume_flow_rate1,influence, '--r*' );
hold on
plot(vfrv, inflv, '-.k')                                                % Plot Regression

hold off
xlabel ('Q (uL/s)');
ylabel ('P(Pa)');
legend ('Via');
text(25, 1200, sprintf('y = %.6f\\cdotx^2', B))                         % Regression Equation

The regression parameter is: 
B =
        0.0266527362537937
To force only ‘c’ to be zero: 
B = [volume_flow_rate1(:).^2, volume_flow_rate1(:)] \ influence(:);     % Estimate Regression Parameters
vfrv = linspace(min(volume_flow_rate1), max(volume_flow_rate1));        % Vector For Regression Evaluation
inflv = [vfrv(:).^2, vfrv(:)] * B;                                      % Evaluate Regression
figure();
plot(volume_flow_rate1,influence, '--r*' );
hold on
plot(vfrv, inflv, '-.k')                                                % Plot Regression
hold off
xlabel ('Q (uL/s)');
ylabel ('P(Pa)');
legend ('Via');
text(25, 1200, sprintf('y = %.6f\\cdotx^2  %+.6f\\cdotx', B))           % Regression Equation
The regression parameter estimates are: 
B =
        0.0304986880886496
        -0.722814145120915
Related Question