In this MWE


\usepackage{amsmath, amssymb}



\item La fonction $f$ est déjà sous forme canonique. On reconnaît $\alpha = -1$ et $\beta = 3$. De plus, le coefficient dominant de $f$ est $-2>0$. Donc, la courbe $\mathscr{C}_f$ de $f$ est une parabole tournée vers le bas et de sommet $(-1,3)$.

On calcule de plus : $f(0)=1$. Cela nous permet de dessiner : 

\definecolor{sexdts}{rgb}{0.1803921568627451,0.49019607843137253,0.19607843137254902}\definecolor{cqcqcq}{rgb}{0.7529411764705882,0.7529411764705882,0.7529411764705882}\begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1cm,y=1cm]\draw [color=cqcqcq,, xstep=1cm,ystep=1cm] (-3.63555555555557,-2.45) grid (2.3022222222221753,4.474444444444443);\draw[->,color=black] (-3.63555555555557,0) -- (2.3022222222221753,0);\foreach \x in {-3,-2,-1,1,2}\draw[shift={(\x,0)},color=black] (0pt,2pt) -- (0pt,-2pt) node[below] {\footnotesize $\x$};\draw[->,color=black] (0,-2.45) -- (0,4.474444444444443);\foreach \y in {-2,-1,1,2,3,4}\draw[shift={(0,\y)},color=black] (2pt,0pt) -- (-2pt,0pt) node[left] {\footnotesize $\y$};\draw[color=black] (0pt,-10pt) node[right] {\footnotesize $0$};\clip(-3.63555555555557,-2.45) rectangle (2.3022222222221753,4.474444444444443);\draw[line width=2pt,color=sexdts,smooth,samples=100,domain=-3.63555555555557:2.3022222222221753] plot(\x,{0-2*((\x)+1)^(2)+3});\end{tikzpicture}



enter image description here

the first column is vertically justified. I would like to have a normal vertical spacing in this first column.

Is it possible to break a column but without vertically justifying the broken column?


Just before \columnbreak insert \vfill\mbox{} I don't know if it always works, but works with your MWE.

  • it should work generally in such a context – Frank Mittelbach Oct 5 '17 at 6:55

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