Plotting on the y-axis in Mathematica?

One possibility is to use a ParametricPlot like this: ParametricPlot {-y*Exp-y^2, y}, {y, -0.3, 4}, PlotRange -> {{-2, 2}, All}, AxesLabel -> {"x", "y"}, AspectRatio -> 1/4 .

You can flip the axes after plotting with Reverse: g = PlotSinx, {x, 0, 9}; Showg /. X_Line :> Reversex, 3, PlotRange -> Automatic With a minor change this works for plots using Filling as well: g1 = Plot{Sinx, .5 Sin2 x}, {x, 0, 2 \Pi}; g2 = Plot{Sinx, .5 Sin2 x}, {x, 0, 2 \Pi}, Filling -> {1 -> {2}}; Show# /. X_Line | x_GraphicsComplex :> x~Reverse~3, PlotRange -> Automatic & /@ {g1, g2} (It may be more robust to replace the RHS of :> with MapAt#~Reverse~2 &, x, 1) As a Function Here is the form I recommend one use.It includes flipping of the original PlotRange rather than forcing PlotRange -> All: axisFlip = # /.

{ x_Line | x_GraphicsComplex :> MapAt#~Reverse~2 &, x, 1, x : (PlotRange -> _) :> x~Reverse~2 } &; To be used like: axisFlip @ g1 or axisFlip @ {g1, g2} A different effect can be had with Rotate: Showg /. X_Line :> Rotatex, Pi/2, {0,0}, PlotRange -> Automatic.

1 Nice trick .. – belisarius Jun 2 at 2:39 @belisarius thanks! I added a couple of variants. – Mr.Wizard Jun 2 at 12:07 +1 I feel the version that includes filling should be the definitive answer.

The Rotate version is wrong in the same way as Alexey's: wrong sign of y. – Sjoerd C. De Vries Jun 3 at 20:50.

ParametricPlot{5 Siny, y}, {y, -2 \Pi, 2 \Pi}, Frame -> True, AxesLabel -> {"x", "y"} EDIT None of the answers given thus far can work with Plot's Filling option. Plot's output contains a GraphicsComplex in that case (which, incidentally, breaks Mr.Wizard's replacements). To get the filling capability (it doesn't work for a standard plot without filling) you could use the following: PlotSinx, {x, 0, 2 \Pi}, Filling -> Axis /.

Listx_, y_ -> Listy, x Plot{Sinx, .5 Sin2 x}, {x, 0, 2 \Pi}, Filling -> {1 -> {2}} /. Listx_, y_ -> Listy, x.

Oh, you beat me to it ;) – Thies Heidecke Jun 1 at 17:32 1 @Thies It's a jungle out there ;-) – Sjoerd C. De Vries Jun 1 at 17:34 1 @Thies and, apparently, it doesn't help to be first (even by 2 min only). Yours got accepted.

– Sjoerd C. De Vries Jun 1 at 17:51 @Sjoerd I knew that the function reverse engineering skill would come in handy some day ;) – Thies Heidecke Jun 2 at 0:27 1 I already voted for this, so I cannot vote again, but good update! – Mr.Wizard Jun 2 at 21:30.

Just for fun: ContourPlot is another alternative. Using Thies function: ContourPlot-y*Exp-y^2/2 - x == 0, {x, -2, 2}, {y, 0, 4}, Axes -> True, Frame -> None RegionPlot is another RegionPlot-y*Exp-y^2/2 > x, {x, -2.1, 2.1}, {y, -.1, 4.1}, Axes -> True, Frame -> None, PlotStyle -> White, PlotRange -> {{-2, 2}, {0, 4}} And finally, a REALLY convoluted way using ListCurvePathPlot and Solve: OffSolve::ifun, FindMaxValue::fmgz; ListCurvePathPlot Join @@ Table {x, y} /. Solve-y*Exp-y^2/2 == x, y, {x, FindMaxValue-y*Exp-y^2/2, y, 0, .01}, PlotRange -> {{-2, 2}, {0, 4}} OnSolve::ifun, FindMaxValue::fmgz; Off Topic Answer to Sjoerd's None of the answers given thus far can work with Plot's Filling option.

Reply: Not necessary f={.5 Sin2 y,Siny}; RegionPlotMin@fTrue,Frame->None, PlotRange->{{-2,2},{0,2 Pi}}, PlotPoints->500.

I thought about RegionPlot as well, but the syntax for that is different from Plot and more complex. It is more familiar to be able to simply add: Filling -> True – Mr.Wizard Jun 3 at 20:14 @Mr. True! But this is just a little joke we are enjoying with Sjoerd :) – belisarius Jun 3 at 20:18 Yes @mr.Wizard, don't spoil our fun.

This is our battle of wits. – Sjoerd C. De Vries Jun 3 at 20:44 Your almost there :-D Perhaps 1000 points will do?

– Sjoerd C. De Vries Jun 4 at 15:18 @Sjoerd You shouldn't laugh at my efforts. I am now exploring AxesStyle -> Thickness.1 with very good results – belisarius Jun 4 at 15:40.

Depending on how you wanted the axis labels to show, you could just wrap the code for the original Plot in the Rotate function.

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