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I'd like to show you what the "cut out" design looks like.

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Is it possible to plot a graph like this Plot[{8 x^2, 0.9 x^2 + 600}, {x, -10, 10}, AspectRatio -> 1] but with a single implicit formula? Or how do I plot a graph of two functions bounded only by interseptions?

8 x^2, 0.9 x^2 + 600

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  • $\begingroup$ Try different powers of x, like e.g.:` t = Table[Plot[(5 - y) x^2 + y x^6, {x, -1, 1}], {y, 0, 5, .5}]; Show[t]` $\endgroup$ Dec 23, 2020 at 8:39

2 Answers 2

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(Not really a Mathematica question, but o well...)

An easy way is to use a piecewise combination of sine curves:

With[{a = 20, b = 5}, 
     ParametricPlot[{10 Cos[t], ((b - a) UnitStep[t] - b) Sin[t]^2}, {t, -π, π}]]

modified bicorn

where the two parameters control where the curve intersects the $y$-axis.

Just to explain where I got this from: I was inspired by the parametric equations of Sylvester's bicorn, and started modifying the equation for that accordingly. (Translate/rescale the resulting curve as seen fit.)

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We can use BezierCurve to get the curve. And use Manipulate to get the control points.

a = {{-2, 0}, {-2, -8}, {2, -8}, {2, 0}};
Manipulate[
 Show[Graphics[{Text[pts // TraditionalForm, {1, 1}], EdgeForm[White],
     CMYKColor[4/100, 7/100, 19/100, 0, .5], 
    FilledCurve[{BSplineCurve[{{2, 0}, Sequence @@ pts, {-2, 0}}, 
       SplineDegree -> 3], Line[a]}]}], 
  ParametricPlot[
   BSplineFunction[{{2, 0}, Sequence @@ pts, {-2, 0}}][t], {t, 0, 1}],
   PlotRange -> {{-2.5, 
     2.5}, {2, -10}}], {{pts, {{0.29, -1.44}, {-1.34, -2.32}}}, 
  Locator}]

enter image description here

pts = {{0.29, -1.44}, {-1.34, -2.32}};
f = BSplineFunction[{{2, 0}, Sequence @@ pts, {-2, 0}}];
ParametricPlot[f[t], {t, 0, 1}]

enter image description here

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