Hi, Is it just me or it appears the "temperature" and "probability" should be reversed? -------- Anyhow it should help you to assign your model to a variable (as Joshua did with his own suggestion)
yourmodel <-lm(x[,2] ~ x[,1] + I(x[,1]^2)) # again, taking literally the way you formulated it... And you could then access the coefficients individually: c <- coef(yourmodel)[1] # for the intercept b <- coef(yourmodel)[2] a <- coef(yourmodel)[3] To build yourself a vector of predicted y-values based on a vector of x-values ("dat" as per Joshua's post) and pass them to the "lines" function to be plotted over your existing plot. This is less convenient than using the "predict" but sometimes it helps to do the arithmetic at a lower level. HTH -- View this message in context: http://r.789695.n4.nabble.com/Plotting-a-quadratic-line-on-top-of-an-xy-scatterplot-tp3443018p3443693.html Sent from the R help mailing list archive at Nabble.com. ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.