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Initially I thought that the function AspectRatio-> calibrated the ratio between the units axes, e.g. putting AspectRatio->1/2 I thought that the ratio x:y between units of the axes would be 1:2, but I surprisingly discovered that this is false: indeed, in the following plot

f[x_] := ((3/20)^(471/100)*20/29*(x - 80))^(50/
     91) + ((3/20)^(471/100)*20/29*(x - 60))^(50/91) - 1/25
Subscript[x, 0] = Extract[{x} /. FindRoot[f[x] == 0, {x, 80}], 1];
Subscript[P, 0] = Point[{Subscript[x, 0], 0}];
Subscript[G, f] = 
  Subscript[G, f] = 
   Plot[f[x], {x, 80, 100}, PlotStyle -> Red, AspectRatio -> 1, 
    AxesLabel -> {"x", "y"}, 
    PlotLegends -> {LineLegend[{Red}, {"f(x)"}], 
      Placed[Row[{"\!\(\*SubscriptBox[\(P\), \(0\)]\) = (", 
         Round[Subscript[x, 0], .001], ",", 0, ")"}], Below]}];
Show[Subscript[G, f], Graphics[{PointSize[0.015], Subscript[P, 0]}]]

enter image description here

the ratio between the units of the axes is not 1:1.The more relevant problem is that I do not know the ratio between the units of the axes in the image and this is a problem because I have to put it into the graphic. So how calibrate the ratio between the units of the axes? Could anyone help me, please?

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  • $\begingroup$ With v12.3.1 on a Mac your code produces a ratio of 1:1 comparing the total axes lengths. $\endgroup$
    – Bob Hanlon
    Nov 25 '21 at 18:59
  • $\begingroup$ So you are saying that $$\frac{\text{unit }x}{\text{unit }y}=1$$ right? $\endgroup$ Nov 25 '21 at 19:00
  • $\begingroup$ The more relevant problem is that I do not know the ratio between axes in the image and this is a problem because I have to put it into the graphic. $\endgroup$ Nov 25 '21 at 19:05
  • $\begingroup$ No, I'm saying the (measured length of the total x axis) == (measured length of the total y axis). If you want the length of 1 unit to be the same on each axis, use AspectRatio -> Identity; however, in this case the scales are radically different and this would cause the y axis to be collapsed to a line. $\endgroup$
    – Bob Hanlon
    Nov 25 '21 at 19:11
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    $\begingroup$ (PlotRange /. Cases[Subscript[G, f], _Rule, Infinity]) extracts the PlotRange, i.e., {{80, 100}, {-0.00881036, 0.036832}}; reversing and subtracting gives the extent of each range, i.e, {20, 0.0456423}; and the ratio is 438.19. If that or its reciprocal is not what you want, I don't understand what you want. $\endgroup$
    – Bob Hanlon
    Nov 25 '21 at 19:55

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