# Mathematica Basic Plotting Question

Is it possible to plot y = x^2 and x = y^2 on the some graph? For some reason, I can't get it to plot the x = y^2 properly. Here is what I get:

Show[{ Plot[x == y^2, {y, -1, 1}], Plot[y = x^2, {x, -1, 1}]}]


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Yes, you just have to change how you think about plotting, a little. Specifically, you are looking for ParametricPlot.

ParametricPlot[{{t, t^2}, {t^2, t}}, {t, -1, 1}]


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This is good to see. Thank you! – Oliver Spryn Nov 22 '12 at 5:26
Just wondering, is it possible to fill in the enclosed region? – Oliver Spryn Nov 22 '12 at 5:28
@spryno724, that requires an entirely different approach... RegionPlot[x^2 < y && y^2 < x, {x, 0, 1}, {y, 0, 1}] – J. M. Nov 22 '12 at 5:29
Ah... ok. Just thought I would check. – Oliver Spryn Nov 22 '12 at 5:30

For completeness we can mention finding all branches of inverse function. It won't always work, but it is conceptually instructive for simple cases.

f[x_] = x^2; g[x_] = InverseFunction[f][x]


-Sqrt[x]

Plot[{f[x], g[x], -g[x]}, {x, -1, 1}, AspectRatio -> 1, Filling -> {1 -> {3}}]


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ParametricPlot needs Filling as an option! :) – rcollyer Nov 22 '12 at 6:02
@rcollyer: orienting the filling would be a bit problematic, tho... – J. M. Nov 22 '12 at 6:03
@rcollyer MeshShading as a remedy ;) – Vitaliy Kaurov Nov 22 '12 at 6:04
@J.M. maybe. But, filling between lines would be less difficult. – rcollyer Nov 22 '12 at 6:10
@VitaliyKaurov why yes it is, but I'll have to play with that tomorrow. – rcollyer Nov 22 '12 at 6:10

rcollyer's method is the best way of going about it. Here's an alternative:

g = Plot[x^2, {x, -1, 1}];
Show[g, g /. v_ /; VectorQ[v, NumericQ] && Length[v] == 2 :> Reverse[v],
PlotRange -> All]


Here's a more conventional route, tho:

ContourPlot[{y == x^2, x == y^2}, {x, -1, 1}, {y, -1, 1}, Axes -> True, Frame -> False]


Show[ContourPlot[{y == x^2, x == y^2}, {x, -1, 1}, {y, -1, 1}],