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The asymptote occurs at a turning point when its first derivative (a quadratic) equals zero. After differentiating we have -0.000015x^2 + 0.0006x - 0.0026 =0. This quadratic has a turning point in the sampled interval [10,60] at 43.94 obtained by inputting the regression coefficients into the [http://www.mathsisfun.com/quadratic-equation-solver.html: quadatic root solver here.] The asymptote occurs at a turning point when its first derivative (a quadratic) equals zero. For the curve in red this quadratic, obtained from differentiating the above cubic, is fo form -0.000015x^2 + 0.0006x - 0.0026 =0. This quadratic has a turning point in the sampled interval [10,60] at 43.94 obtained by inputting the regression coefficients into the [http://www.mathsisfun.com/quadratic-equation-solver.html: quadatic root solver here.]

Summary measures for learning curves

Suppose we wish to look at change in processing success (y) with ease of task (x).

We fit a cubic polynomial and obtain using multiple regression a best fitting cubic of form -0.000005x3 + 0.0003x2 + 0.0026x + 0.7508. This is the curve in red in this [attachment:plot.jpeg plot.]

The asymptote occurs at a turning point when its first derivative (a quadratic) equals zero. For the curve in red this quadratic, obtained from differentiating the above cubic, is fo form -0.000015x^2 + 0.0006x - 0.0026 =0. This quadratic has a turning point in the sampled interval [10,60] at 43.94 obtained by inputting the regression coefficients into the [http://www.mathsisfun.com/quadratic-equation-solver.html: quadatic root solver here.]

The predicted slope to asymptote is obtained by subtracting the predicted responses, obtaine dusing the regression coefficients in the cubic, at the start (x=10) from that at the asymptote(x=43.94) of the learning line and dividing by the change in x (43.94-10),

So for the red curve the change from baseline to asymptote = (1.02 – 0.80)/(43.94-10)=0.0064.

None: FAQ/polyeqn (last edited 2015-04-15 14:41:49 by PeterWatson)