Although I already worked around this problem by finding the x-intercept with the linear regression equation I calculated w/gnuplot, I'm still curious to know how pgfplots
can do it without me manually computing it.
\documentclass[12pt]{article}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{lmodern,tikz,pgfplots,pgfplotstable}
\begin{document}
\begin{center}
\begin{tikzpicture}
\begin{axis}[axis on top=false, axis x line=middle, axis y line=middle,xlabel=$\mathrm{\frac{1}{[S]}}$,ylabel=$\mathrm{\frac{1}{\textit{V}_0}}$,
xmin=-10000,xmax=10000,ymin=-0.01,ymax=0.035]
\addplot table [y={create col/linear regression={}}]
{
X Y
10000 0.030
5000 0.02
2000 0.014
1000 0.012
500 0.0110
200 0.0104
100 0.0102
50 0.010
20 0.01
10 0.01
5 0.01
-4999.85 0
-9995.35 -0.01
};
\xdef\slope{\pgfplotstableregressiona}
\xdef\slope{\pgfplotstableregressionb}
\end{axis}
\end{tikzpicture}
\end{center}
\end{document}
Best Answer
The linear regression line is only valid for the domain of the given data, and hence by default it is only drawn from the minimum and maximum x in the given data.
If you want to extrapolate for points outside the domain of the data, you can use
\pgfplotstableregressiona
and\pgfplotstableregressionb
to draw the best fit line and specify the desired domain.I commented out the last two lines that I assume you manually added, and corrected the second
\xdef
as that is the y-intercept and not the slope.