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I want to plot this: area below parabola

I know, that I can find intersections using TikZ (tried \shade but don't know how to plot parabola only in a segment), but I don't know how to plot this R figure, and S+T, and U+V. How can I plot this? New detail: How [domain] works?

\tipc{[x=1cm,y=1cm]

  \def\xmin{0}
  \def\xmax{10}
  \def\ymin{0}
  \def\ymax{10}
  \draw[style=help lines, ystep=1, xstep=1] (\xmin,\ymin) grid
  (\xmax,\ymax);

  % axes
  \draw (-.25,-.25) node[auto] {0};
  \draw[->] (\xmin,\ymin) -- (\xmax,\ymin) node[right] {$Q$};
  \draw[->] (\xmin,\ymin) -- (\xmin,\ymax) node[above] {$P$};
  \draw[red] (2,2) parabola (8,8) node[right,black] {$S$};
  \draw[blue] (8,2) parabola (2,8) node[left,black] {$D$};
  \draw[dashed]  (5,0) node[below] {$q_A$} -- (5,3.5);
  \draw[dashed] (3.72,-0.5) -- (3.72,2.5) -- (6.27,2.5) -- (6.27,-0.5);
  \draw[<->] (3.72,-0.5) node [below] {$q_s$}  -- ++(2.55,0) node [midway,below] {$Im$}
  node [below] {$q_d$};
  \begin{scope}
        \draw[color=red!30,domain=1.72:3]
            (5,3.5) parabola  (2,8)  |- (3.72,3.5);
    \end{scope}
}

enter image description here

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2  
I guess a nice tutorial for your purpose is TikZ for economists. –  Claudio Fiandrino Jan 24 '13 at 10:17
    
Just from looking at it, i would say this would be much easier to do with pgfplots (if you have mathematical expressions for the curves). You can handle the problem of getting the intersection with the sugestions from tex.stackexchange.com/questions/21408/intersections-in-pgfplots –  joão gândara Jan 24 '13 at 10:50
    
I'm not sure if I've understand your question correctly... Do you want to create exactly the plot you've shown above? –  Pouya Jan 24 '13 at 11:52
    
It would be great. –  Ptech Jan 24 '13 at 11:52
    
@Ptech can you provide exact coordinates or and approximation of the plot would be ok? –  Pouya Jan 24 '13 at 11:56

1 Answer 1

up vote 7 down vote accepted

Here is my solution. Please note that I have not used the parabola function of tikz because I failed to define the domain (not the end-points) and instead plotted two quadratic functions:

\documentclass{minimal}
\usepackage{tikz}

\begin{document}
    \begin{tikzpicture}

        \def\xmin{0}
        \def\xmax{10}
        \def\ymin{0}
        \def\ymax{10}
        \draw[style=help lines, ystep=1, xstep=1] (\xmin,\ymin) grid
        (\xmax,\ymax);

        \draw (-.25,-.25) node[auto] {0};
        \draw[->] (\xmin,\ymin) -- (\xmax,\ymin) node[right] {$Q$};
        \draw[->] (\xmin,\ymin) -- (\xmin,\ymax) node[above] {$P$};

        \def\intersectX{4.76}
        \def\intersectY{4.26}
        \def\QPX{4}
        \def\QPY{5}
        \draw[color=red] plot [domain=0:8] (\x,{((\x)^2)/10 +2)});
        \draw[color=blue] plot [domain=0:8] (\x,{((\x-14)^2)/20)});

        \fill[fill=pink,opacity=0.7] (0,\QPY) -- plot [domain=0:\QPY] (\x,{((\x-14)^2)/20)}) -- (\QPX,\QPY) -- cycle;
        \fill[fill=cyan,opacity=0.7] (0,\QPY) -- plot [domain=0:\QPX] (\x,{((\x)^2)/10 +2)}) -- (\QPX,\QPY) -- cycle;

        \draw [domain=\QPX:\intersectX] 
               plot(\x,{((\x-14)^2)/20)}) -- (\QPX,\QPY) -- (\QPX,\QPY) -- cycle; 

        \draw [fill=green,opacity=0.7,domain=\QPX:\intersectX] 
               plot(\x,{((\x)^2)/10 +2)}) -- (\QPX,\QPY) -- cycle;

        \draw[dashed]  (\intersectX,0) node[below] {$Q_1$} -- (\intersectX,\intersectY);
        \draw[dashed]  (0,\intersectY) node[below] {$P_1$} -- (\intersectX,\intersectY);
        \draw[dashed]  (0,\intersectY) node[below] {$P_1$} -- (\intersectX,\intersectY);
        \draw[dashed]  (\QPX,0) node[below] {$Q_2$} -- (\QPX,\QPY);
        \draw[dashed]  (0,\QPY) node[below] {$P_2$} -- (\QPX,\QPY);

    \end{tikzpicture}
\end{document}

Resulting in this:

enter image description here

I used some help from this answer as well.

share|improve this answer
    
This answer is great, I think, that i should get rid of parabola in tikz, and use plot instead (I had this question because parabola doesn't understand [domain]) –  Ptech Jan 24 '13 at 13:54

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