I have the following plot:
\begin{tikzpicture}
\begin{axis}[grmin=-5:5 -5:5]
\addplot[mark=none] {x^2/4};
\addplot[mark=none] {-1};
\addplot[mark=*] coordinates {(0,1)};
\end{axis}
\end{tikzpicture}
Which looks like this:
My question is: how can I label the point (0,1) as "(0,1)", such that the label appears either to the right or left of the point?
You can use the axis coordinate system to refer to actual plot coordinates and it would be great if you can include full compilable code examples as below.
\documentclass{standalone}
\usepackage{pgfplots}
\pgfplotsset{compat=1.7}
\begin{document}
\begin{tikzpicture}
\begin{axis}[axis y line=middle,axis x line=bottom]%[grmin=-5:5 -5:5]
\addplot[mark=none] {x^2/4};
\addplot[mark=none] {-1};
\node[label={180:{(0,1)}},circle,fill,inner sep=2pt] at (axis cs:0,1) {};
\end{axis}
\end{tikzpicture}
\end{document}
You may be looking for TikZ's pin option, used like so:
\begin{tikzpicture}
\begin{axis}[grmin=-5:5 -5:5]
\addplot[mark=none] {x^2/4};
\addplot[mark=none] {-1};
\addplot[mark=*] coordinates {(0,1)} node[pin=150:{$(0,1)$}]{} ;
\end{axis}
\end{tikzpicture}
With PSTricks. I show you how to put a point at any location: on the curve or not on the curve.
\documentclass[pstricks,border=12pt]{standalone}
\usepackage{pst-plot}
\usepackage{pst-eucl}
\def\f(#1){x^2+0.5}
\begin{document}
\begin{pspicture}(-2.5,-0.5)(3,6)
\psaxes[linecolor=gray,tickcolor=gray]{->}(0,0)(-2.5,-0.5)(2.5,5.5)[$x$,0][$y$,90]
\psplot[algebraic]{-2.2}{2.2}{\f(x)}
\pstGeonode[PosAngle={45,0}](0,1){F}(*1.1 {\f(x)}){A}
\end{pspicture}
\end{document}
or
\documentclass[pstricks,border=12pt]{standalone}
\usepackage{pst-plot}
\usepackage{pst-eucl}
\usepackage{mathpazo}
\def\f(#1){x^2+0.5}
\begin{document}
\begin{pspicture}(-2.5,-0.5)(3,6)
\psaxes[linecolor=gray,tickcolor=gray]{->}(0,0)(-2.5,-0.5)(2.5,5.5)[$x$,0][$y$,90]
\psplot[algebraic]{-2.2}{2.2}{\f(x)}
\pstGeonode[PosAngle=-30,PointNameSep=15pt,PointName={{\scriptstyle(0,0.5)},{\scriptstyle(1,f(1))}}]
(0,0.5){F}(*1 {\f(x)}){A}
\end{pspicture}
\end{document}
\documentclass[pstricks,border=12pt]{standalone}
\usepackage{pst-plot}
\usepackage{pst-eucl}
\usepackage{mathpazo}
\def\f(#1){x^2+0.5}
\def\Picture#1{%
\begin{pspicture}(-2.5,-0.5)(3,6)
\psaxes[linecolor=gray,tickcolor=gray]{->}(0,0)(-2.5,-0.5)(2.5,5.5)[$x$,0][$y$,90]
\psplot[algebraic]{-2.2}{2.2}{\f(x)}
\pstGeonode[PosAngle={-15,135},PointNameSep=20pt,PointName={{\scriptscriptstyle(0.5,f(0.5))},{\scriptscriptstyle(#1,f(#1))}}]
(*0.5 {\f(x)}){F}(*#1 {\f(x)}){A}
\pstLineAB[nodesep=-1,linecolor=blue]{F}{A}
\end{pspicture}}
\begin{document}
\multido{\n=2.0+-0.1}{15}{\Picture{\n}}
\multido{\n=0.7+0.1}{13}{\Picture{\n}}
\end{document}
The point (*1 {\f(x)}){A} uses the following weird "RPN-algebraic" syntax
(*<a constant numerical value in RPN expression> {<symbolic algebraic expression in x>})
No "algebraic-algebraic" syntax such as
({<a constant numerical algebraic expression>}| {<symbolic algebraic expression in x>})
is available.
(0.5,f(0.5)) because it needs another trick!
– kiss my armpit
Mar 24 '13 at 17:31
I wanted to add a single point based on a predefined function but couldn't find an appropriate answer among this question and related questions.
Therefore, for future reference, if you have your own function defined as e.g.:
\begin[tikzpicture][declare function={y(\x) = \x^2;}] ...
And you want to add a single point based on this function, you should use:
\addplot [only marks,samples at={5}] {y(x)} node[pin=150:{$(0,1)$}]{};
That's the best way to add a single point to an existing function, regardless of your axis setup or of other functions that have been added to the plot (you need pgfplots for this).
Hope this might help someone out in the future.
label={180:{$(0,1)$}}-- putting the coordinate in math mode. – Xenon Mar 24 '13 at 12:50