# How to configure callouts in pgfplots?

In this answer I found a callout in a tikz graph.

 \node [above, callout relative pointer={(-2,1)},rounded corners,note=green!50, opacity=.5, overlay] at (F) {\scriptsize $y=(x+1)^2$};


However I can't replicate this in a pgfplots since I get an error:

! Package pgfkeys Error: I do not know the key '/tikz/note', to which you passe d 'green!50', and I am going to ignore it. Perhaps you misspelled it.


I've looked in the pgf/tiks manual and in the library source code but I found no documentation, related to the note key, or any other keys related to call outs (like callout relative pointer). They are there, just no explanation is presented.

Is there documentation available somewhere for tiks libraries? Or does some one wants to leave here some guidelines for callouts?

So, in order:

1. Is there documentation available somewhere for tiks libraries?

2. What am I doing wrong in my code? (where is note defined?)

3. Does some one wants to leave here some guidelines for callouts?

• Look at the note/.style at the tikzpicture environment start – percusse Oct 2 '15 at 16:28

1. The PGF manual (or texdoc pgf in a terminal) has explanations and numerous examples about libraries.

2. You need to load the shapes.callouts library as well as have defined the note style (the definition can be found in the linked post); also, since I don't know whether you have a previously defined (F) point, I changed the location to (axis cs:0,0.5):

3. For the specific case of callouts, please refer to Section 67.7 Callout Shapes (page 729-733) of the PGF manual (for version 3.0.1); there you will find the available options, with explanations and examples

A working version of the code:

\documentclass{article}
\usepackage{pgfplots}
\usetikzlibrary{shapes.callouts}
\begin{document}

\begin{tikzpicture}[note/.style={rectangle callout, fill=#1}]
\begin{axis}
\node [above, callout relative pointer={(axis cs:-3,0.25)},rounded corners,note=green!50, opacity=.5, overlay] at (axis cs:0,0.5) {\scriptsize $y=(x+1)^2$};