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I think the following are bugs:

ContourPlot[y (x - 1)^2 == 0, {x, 0, 2}, {y, -2, 10}]

enter image description here

ContourPlot[Exp[-y]*(x - 1)^2 == 0, {x, 0, 2}, {y, -2, 10}]

enter image description here

Please note that I'm not refering to the lack of resolution. I'm talking about the fact that there should be a vertical line $x=1$.

It seems the problem is due to the $(x-1)^2$ term, since without the square, it works fine.

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  • $\begingroup$ why you set them functions == 0 i think you don't need to do this $\endgroup$
    – Alrubaie
    Apr 19 '19 at 19:56
  • $\begingroup$ Do not use the bugs tag for unconfirmed potential bugs. Add PlotRange to your example. Observe. $\endgroup$
    – ciao
    Apr 19 '19 at 20:00
  • $\begingroup$ @ciao I've just added PlotRange and varied xmin, xmax, ymin and ymax, but I don't observe any modification... $\endgroup$
    – AJHC
    Apr 19 '19 at 20:03

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