finding points on path with given x-coordinate (marking f(x) points for given x's)

Question

A y=f(x) function is given as path; for a set of x-coordinates, I want to place node (x,f(x)) onto the path, and connect it with (x,0) with vertical line.

I have not got very far myself:

\tikz{
  \def\fSpec{ (-2,-2.2) .. controls (-1,2.5) and (.5,-1.7) .. (2,1.6) }
  \def\xxx{{-1.9,-1.3,-.8,-.2,.4,.9,1.5,2.0}}
  \draw[thin] \fSpec;
  \foreach\x in \xxx{
      %% how to find f(\x) here?
  }
}

I've know about the intersections library, but it seems like overkill; I would like to learn something more elegant.

I cannot give parametrically along the path length, because there will be multiple functions with different shapes and the points need to have the same x-coordinates for all of them.

Answer 1

Andrew Stacey said it in the comments: The only alternative to the intersections library is to provide the curve as a function. Here's one that I fitted by eye:

Your curve is shown in light gray. If you don't need to recreate the curve very closely, you could of course make do with a simpler function.

\documentclass{article}
\usepackage{tikz}

\begin{document}
\tikz{
  \def\fSpec{ (-2,-2.2) .. controls (-1,2.5) and (.5,-1.7) .. (2,1.6) }
  \def\xxx{-1.9,-1.3,-.8,-.2,.4,.9,1.5,2.0}
  \draw [gray!50] \fSpec;
  \foreach \x in \xxx{
      \draw (\x,0) -- ({\x},{0.25+1.3*(\x+0.15)^3/5-(\x+0.15)^2/5-\x/10});
  }
  \draw [red] plot [domain=-2:2] ({\x},{0.25+1.3*(\x+0.15)^3/5-(\x+0.15)^2/5-(\x)/10});
}

\end{document}