# How can I plot an implicit function? [duplicate]

Say I have a function

How can I plot the following function

• Look up the documentation for ContourPlot. Feb 10, 2017 at 19:24
• I imagine you want the full heart. You can use the real-valued cube root instead of the principal valued one: ContourPlot[x^2 + (y - CubeRoot[x]^2)^2 == 1, {x, -1, 1}, {y, -1, 2}] Feb 10, 2017 at 21:18

Using the suggestion of @march:

ContourPlot[x^2 + (y - x^(2/3))^2 == 1, {x, -1, 1}, {y, -1, 2}]


Alternatively, if you are more comfortable with the use of Plot:

Plot[Flatten@Solve[x^2 + (y - x^(2/3))^2 == 1, y][[;; , ;; , 2]], {x, -1, 1}]


• Flatten is not necessary in Plot as they formulas aren't to deep, but if you want to see where it transitions from one solution to the other, you can replace it with Evaluate. Feb 10, 2017 at 19:46
• I think the OP meant something like ContourPlot[{x^2 + (y - x^(2/3))^2 == 1, (-x)^2 + (y - (-x)^(2/3))^2 == 1}, {x, -1, 1}, {y, -1, 2}, ContourStyle -> Red]? (+1)
– kglr
Feb 10, 2017 at 19:47
• Thanks for helping me ... excellent answer @Marchi Feb 10, 2017 at 20:10
• Thanks for helping me ... excellent answer @rcollyer Feb 10, 2017 at 20:10
• Thanks for helping me ... excellent answer @kglr Feb 10, 2017 at 20:11