# Plot an inverse function

I have a phrase that in it x is a function of y but I need to plot y as a function of x. I tried InverseFunction but because my phrase is not single-valued in some points Mathematica can not give me a plot. After that, I tried ParametricPlot but this code has not any answer. In fact I wanna inverse x and y axes. what should I do?

0.008 y +
125. (2. -
0.0000271476 (659.714/(1 - 0.169673 y) + (
2 (2.30388*10^-6 + 0.169673 y)^2)/(1 - 0.169673 y)^2) - (
0.00271476 (2.30388*10^-6 + 0.169673 y))/(1 - 0.169673 y))^2 y=x

Plot[{0.008 y +
125. (2. -
0.0000271476 (659.714/(1 - 0.169673 y) + (
2 (2.30388*10^-6 + 0.169673 y)^2)/(1 - 0.169673 y)^2) - (
0.00271476 (2.30388*10^-6 + 0.169673 y))/(1 - 0.169673 y))^2 y}, {y, 5.6, 6}]

• I change the coefficients in my code: Plot[2.3864*10^-15 y + 9.5456*10^-15 (25. - 2.78813*10^-9 (659.714/(1 - 5.57627*10^-9 y) + ( 2 (9.21551*10^-6 + 1.39407*10^-6 y)^2)/(1 - 5.57627*10^-9 y)^2) - ( 1.39407*10^-6 (9.21551*10^-6 + 1.39407*10^-6 y))/( 1 - 5.57627*10^-9 y))^2 y, {y, 1.7*10^8, 1.9*10^8}] when I plotted these function no minus region appear but when I plotted inverse function with ParametricPlot some region that are minus in sign will appear, why is that? Dec 9, 2020 at 22:46

f[y_] := 0.008 y +
125. (2. -
0.0000271476 (659.714/(1 -
0.169673 y) + (2 (2.30388*10^-6 + 0.169673 y)^2)/(1 -
0.169673 y)^2) - (0.00271476 (2.30388*10^-6 +
0.169673 y))/(1 - 0.169673 y))^2 y;
fun = Plot[f[y], {y, 0, 6}, PlotStyle -> Blue, AspectRatio -> 1];
invfun = ParametricPlot[{f[y], y}, {y, 0, 6}, PlotStyle -> Red,
AspectRatio -> 1];
GraphicsRow[{fun, invfun}] • ContourPlot is another way. Nov 30, 2020 at 11:22
• I change the coefficients in my code: Plot[2.3864*10^-15 y + 9.5456*10^-15 (25. - 2.78813*10^-9 (659.714/(1 - 5.57627*10^-9 y) + ( 2 (9.21551*10^-6 + 1.39407*10^-6 y)^2)/(1 - 5.57627*10^-9 y)^2) - ( 1.39407*10^-6 (9.21551*10^-6 + 1.39407*10^-6 y))/( 1 - 5.57627*10^-9 y))^2 y, {y, 1.7*10^8, 1.9*10^8}] when I plotted these function no minus region appear but when I plotted inverse function with ParametricPlot some region that are minus in sign will appear, why is that? Dec 12, 2020 at 10:34
• @nafisgh Since MMA change the origin automatic. you can set AxesOrigin -> {0, 0} etc. Dec 12, 2020 at 10:43