How to find the y from given x on fit line?

I have a fit line of a graph and I need to find the spesific y value for a x value that I determine. Is it possible?

Answer 1

[Image Analyst]

You can use polyfit and polyval. Here is a full demo. Don't be afraid, the actual code is only 2 lines. The rest is just code to create the variables and do some fancy plotting.

x = sort(10 * rand(1, 10));
y = 0.5 * x + 0.3 * rand(1, length(x));
% Plot the original points.
plot(x, y, 'b.', 'MarkerSize', 30);
grid on;
xlabel('x', 'FontSize',20)
ylabel('y', 'FontSize',20)

% Fit a line through the points.
coefficients = polyfit(x, y, 1);     % YOU NEED THIS LINE!

% Get a fit.  Find the fit values everywhere, even between the original points, just for fun.
xFit = linspace(min(x), max(x), 1000);
yFit = polyval(coefficients, xFit);
hold on;
plot(xFit, yFit, 'r-', 'LineWidth', 2)

% Find the y location where x equals exactly 6.
xDesired = 6;
xline(xDesired, 'Color', 'g', 'LineWidth',2)

yDesired = polyval(coefficients, xDesired)  %  YOU NEED THIS LINE!

yDesired = 3.1371

% Draw a line from the y axis to the fit line and report what the y value
% is in the title of the graph.
line([0, xDesired], [yDesired, yDesired], 'Color', 'g', 'LineWidth', 2)
caption = sprintf('y = %f when x = %.1f', yDesired, xDesired);
title(caption, 'FontSize',20)

Answer 2

[KSSV]

Yes very much it is possible. Read about interp1. Also have a look on polyfit and polyval.

Answer 3

[Jeffrey Clark]

@kivanc Koca, if you have a fit using spline, pchip, makima, interp1, or the spline utility function mkpp you can use Evaluate piecewise polynomial - MATLAB ppval (mathworks.com).

If you have a polynomial fit using Create and Evaluate Polynomials you can use Polynomial evaluation - MATLAB polyval (mathworks.com).

Answer 4

[John D'Errico]

It completely depends on how you perform the fit. For example, with the curve fitting toolbox, just do this:

X = rand(10,1);
Y = cos(X);

mdl = fit(X,Y,'poly2') % quadratic fit to a cosine is probably adequate over a reasonably small interval

mdl = 
     Linear model Poly2:
     mdl(x) = p1*x^2 + p2*x + p3
     Coefficients (with 95% confidence bounds):
       p1 =     -0.4559  (-0.4637, -0.4481)
       p2 =    -0.01716  (-0.02471, -0.009607)
       p3 =       1.002  (1, 1.003)

Now we can just use the resulting fitted model to predict any point as I do here:

mdl(.3)

ans = 0.9554

Other fitting tools, for example, polyfit, also have evaluation tools provided with them, so you can use polyval to evaluate a polynomial model from polyfit. Read the help for the tool you used to perform the fit.