how to label x-intercept in plot

Questions
,
1

Hi All,

I would like to label x-intercept as k_c. Any suggestion.

Thanks in advance..

Manipulate[
 Plot[B k^4 (-M + E^-Subscript[h, 0] M - E^-Subscript[h, 0] G M + 
      15 M \[Epsilon] - 15 E^-Subscript[h, 0] M \[Epsilon]) + 
   k^2 (B E^-Subscript[h, 0] M - B E^-Subscript[h, 0] G M - A Mp + 
      B E^-Subscript[h, 0] M \[Epsilon]), {k, 0, 7}, Frame -> True, 
  FrameLabel -> {"k", \[Omega]}, RotateLabel -> False, 
  FrameTicks -> None], {A, -1, 100}, {Mp, -5, 10}, {Subscript[h, 0], 
  1, 4}, {G, 1, 4}, {\[Epsilon], 0, 1/15}, {B, 8, 8}, {M, 1, 1}]

every value is left value but just change A=70

2

Hello,
Are you sure your x-intercept has real values?

y = B k^4 (-M + E^-Subscript[h, 0] M - E^-Subscript[h, 0] G M + 
    15 M \[Epsilon] - 15 E^-Subscript[h, 0] M \[Epsilon]) + 
 k^2 (B E^-Subscript[h, 0] M - B E^-Subscript[h, 0] G M - A Mp + 
    B E^-Subscript[h, 0] M \[Epsilon])

Solve[y==0,k]
3

Hi,

It this what you need?

Manipulate[

  VAL = Evaluate[ k /. NSolve[
     B*k^4*(-M + M/E^Subscript[h, 0] - (G*M)/E^Subscript[h, 0] + 
          15*M*\[Epsilon] - (15*M*\[Epsilon])/E^Subscript[h, 0]) + 
       k^2*((B*M)/E^Subscript[h, 0] - (B*G*M)/E^Subscript[h, 0] - 
          A*Mp + (B*M*\[Epsilon])/E^Subscript[h, 0]) == 0, k, Reals]];
 POINTS = Graphics[{PointSize[Large], Pink, 
    Point[Table[{VAL[[i]], 0}, {i, 1, Length@VAL}]]}];
 GRAPH = Plot[
   B*k^4*(-M + M/E^Subscript[h, 0] - (G*M)/E^Subscript[h, 0] + 
       15*M*\[Epsilon] - (15*M*\[Epsilon])/E^Subscript[h, 0]) + 
    k^2*((B*M)/E^Subscript[h, 0] - (B*G*M)/E^Subscript[h, 0] - 
       A*Mp + (B*M*\[Epsilon])/E^Subscript[h, 0]), {k, 0, 7},
    Frame -> True, FrameLabel -> {"k", \[Omega]}, 
   RotateLabel -> False];
 Show[GRAPH, POINTS]
 , {A, -1, 100}, {Mp, -5, 10}, {Subscript[h, 0], 1, 4}, {G, 1,4}, {\[Epsilon], 0, 1/15}, 
   {B, 8, 8}, {M, 1, 1}]

I.M.

4

Thank you for replying me.

It is kinda right but also I would like to see “k_c” on my graph at the x-intercept

5

Ok, you are quite demanding :slight_smile:

Manipulate[

 VAL = Evaluate[
   k /. NSolve[
     B*k^4*(-M + M/E^Subscript[h, 0] - (G*M)/E^Subscript[h, 0] + 
          15*M*\[Epsilon] - (15*M*\[Epsilon])/E^Subscript[h, 0]) + 
       k^2*((B*M)/E^Subscript[h, 0] - (B*G*M)/E^Subscript[h, 0] - 
          A*Mp + (B*M*\[Epsilon])/E^Subscript[h, 0]) == 0, k, Reals]];
 POINTS =   Graphics[{PointSize[Large], Pink, 
    Point[Table[{VAL[[i]], 0}, {i, 1, Length@VAL}]]}];
 GRAPH = Plot[
   B*k^4*(-M + M/E^Subscript[h, 0] - (G*M)/E^Subscript[h, 0] + 
       15*M*\[Epsilon] - (15*M*\[Epsilon])/E^Subscript[h, 0]) + 
    k^2*((B*M)/E^Subscript[h, 0] - (B*G*M)/E^Subscript[h, 0] - 
       A*Mp + (B*M*\[Epsilon])/E^Subscript[h, 0]), {k, 0, 7}, 
   Frame -> True, FrameLabel -> {"k", \[Omega]}, RotateLabel -> False,
    GridLines -> {{Max@VAL}, {}}];
 LETTER = ListPlot[{{1.1 Max@VAL, 0}}, PlotMarkers -> {"\!\(\*
StyleBox[\"\[HappySmiley]\",\nFontSize->18]\)"}];
 Show[GRAPH, POINTS, LETTER], {A, -1, 100}, {Mp, -5,10}, {Subscript[h, 0], 1, 4}, {G, 1, 4}, {\[Epsilon], 0, 1/15}, {B, 8, 8}, {M, 1, 1}]

Also you can add LINE = Graphics[Graphics[Line[{{0, 0}, {Max@VAL, 0}}]]]; instead x=0 axes

I.M.

6

This is what I want.. Thank you… Do you think we can remove left dot and also write k_c below the right dot. if it is hard to do that dont worry, this is quite enough :slight_smile:

Manipulate[
 VAL = Evaluate[
   k /. NSolve[
     B*k^4*(-M + M/E^Subscript[h, 0] - (G*M)/E^Subscript[h, 0] + 
          15*M*\[Epsilon] - (15*M*\[Epsilon])/E^Subscript[h, 0]) + 
       k^2*((B*M)/E^Subscript[h, 0] - (B*G*M)/E^Subscript[h, 0] - 
          A*Mp + (B*M*\[Epsilon])/E^Subscript[h, 0]) == 0, k, Reals]];
 POINTS = 
  Graphics[{PointSize[Large], 
    Point[Table[{VAL[[i]], 0}, {i, 1, Length@VAL}]]}];
 GRAPH = Plot[
   B*k^4*(-M + M/E^Subscript[h, 0] - (G*M)/E^Subscript[h, 0] + 
       15*M*\[Epsilon] - (15*M*\[Epsilon])/E^Subscript[h, 0]) + 
    k^2*((B*M)/E^Subscript[h, 0] - (B*G*M)/E^Subscript[h, 0] - 
       A*Mp + (B*M*\[Epsilon])/E^Subscript[h, 0]), {k, 0, 7}, 
   Frame -> True, FrameLabel -> {"k", \[Omega]}, RotateLabel -> False,
    FrameTicks -> None];
 LETTER = ListPlot[{{1.1 Max@VAL, 0}}, PlotMarkers -> {"\!\(\*
     StyleBox[\"\!\(\*SubscriptBox[\(k\), \(c\)]\)\",\n\
FontSize->15]\)"}];
 Show[GRAPH, POINTS, LETTER], {A, -1, 100}, {Mp, -5, 
  10}, {Subscript[h, 0], 1, 4}, {G, 1, 4}, {\[Epsilon], 0, 1/15}, {B, 
  8, 8}, {M, 1, 1}]
7

>> Do you think we can remove left
Yes, it is possible. To delete the left dote you shoud take only max in VAL for POINTS .
>> k_c below the right dot
Define the “below” to be - max of y scale and modify the second arg in the LETTER

I suggest you to do it on your own, see Graphics[], ListPlot[], Point[] and other related manuals.

I.M.

8

I figured it out to remove left point. I just modified nsolve like bla bla bla==0&&k>0,k] instead of all reals..

I could not handle to change position of k_c

Thanks again for helping me..

9

when I try to define

f[k_]:=B k^4 (-M+E^-Subscript[h, 0] M-E^-Subscript[h, 0] G M+15 M \[Epsilon]-15 E^-Subscript[h, 0] M \[Epsilon])+k^2 (B E^-Subscript[h, 0] M-B E^-Subscript[h, 0] G M-A Mp+B E^-Subscript[h, 0] M \[Epsilon])

Manipulate does not work, I do not know why. That’s why I pasted long function in code..

Manipulate[
 Plot[B k^4 (-M + E^-Subscript[h, 0] M - E^-Subscript[h, 0] G M + 
      15 M \[Epsilon] - 15 E^-Subscript[h, 0] M \[Epsilon]) + 
   k^2 (B E^-Subscript[h, 0] M - B E^-Subscript[h, 0] G M - A Mp + 
      B E^-Subscript[h, 0] M \[Epsilon]), {k, 0, 7}, Frame -> True, 
  FrameLabel -> {"k", \[Omega]}, RotateLabel -> False, 
  FrameTicks -> None, 
  Epilog -> {PointSize[
     Large], (Point[{#, (B k^4 (-M + E^-Subscript[h, 0] M - 
               E^-Subscript[h, 0] G M + 15 M \[Epsilon] - 
               15 E^-Subscript[h, 0] M \[Epsilon]) + 
            k^2 (B E^-Subscript[h, 0] M - B E^-Subscript[h, 0] G M - 
               A Mp + B E^-Subscript[h, 0] M \[Epsilon])) /. 
          k -> #}] &) /@ (k /. 
       NSolve[(B k^4 (-M + E^-Subscript[h, 0] M - 
               E^-Subscript[h, 0] G M + 15 M \[Epsilon] - 
               15 E^-Subscript[h, 0] M \[Epsilon]) + 
            k^2 (B E^-Subscript[h, 0] M - B E^-Subscript[h, 0] G M - 
               A Mp + B E^-Subscript[h, 0] M \[Epsilon])) == 0 && 
         k > 0, k])}], {A, -1, 100}, {Mp, -5, 10}, {Subscript[h, 0], 
  1, 4}, {G, 1, 4}, {\[Epsilon], 0, 1/15}, {B, 8, 8}, {M, 1, 1}]
10

That is due to Plot[], because**:**
Plot has attribute HoldAll and evaluates f only after assigning specific numerical values to x.

This is how you deal with it:

f[x_] := A x^2;
Manipulate[
 LIST = {A -> a};
 Plot[f[x] /. LIST, {x, -2, 2}] ,
 {a, 0, 1}]

I.M.

11

I have learned shorter way to handle it.

f[x_] := a x^2;
Manipulate[Plot[#, {x, -2, 2}], {a, 0, 1}]&@f[x]