Question: I am preparing a figure for an article, and I would like very much to be able to fill a curve, but not under it, but to the vertical axis, i.e. to the left. Due to the meaning of the data I present, I cannot exchange the axes.

Here is a minimal example:

```
Plot[ArcCot[x], {x, 0, 1}, PlotRange -> {{0, 1.1}, {0.5, 2}}, AxesLabel -> {x, y}]

```

Can anyone think of a way of filling the curve to the`y` axis? I find no mention of this kind of filling in the standard documentation.

Answer: 
Adding a line at the minimum value and shading as normal:-

```
min = MinValue[ArcCot[x], 0 <= x <= 1, x];
Plot[{ArcCot[x], min}, {x, 0, 1}, PlotRange -> {{0, 1.1}, {0.5, 2}}, 
 AxesLabel -> {x, y}, Filling -> {1 -> {2}}, PlotStyle -> {Automatic, None}]

```

![enter image description here](https://i.stack.imgur.com/2EIdA.png)
I know this thread is long time inactive, but **Michael E2**'s excellent fillVertical function above helped me out a lot. I did need a more "density"-like fill, so I'm sharing my variant of it (using my novice-level Mathematica programming skills) which splits the Line into monotonic segments:

```
fillVertical2[plot_, x0_: 0.] := plot /. Line[p_] :>
  Function[{mp},{{Opacity[0.2],
      Polygon[mp ~Join~ {{N@x0, mp[[-1, 2]]}, {N@x0, mp[[1, 2]]}}]}, Line[mp]}] /@ 
  SequenceCases[p, lp_ /; Less @@ lp[[All, 2]] || Greater @@ lp[[All, 2]]];

```

This gives a plot like

```
fillVertical2[Plot[Sin[x] + 0.3 x, {x, 0, 10}, PlotStyle -> Red]]

```

[![overlayed fillVertical](https://i.stack.imgur.com/4RZvD.png)](https://i.stack.imgur.com/4RZvD.png)