Making a Perfect Cusp Tikz












8














This Cusp I made came out awkward and I can not fix it.



documentclass{article}
usepackage{tikz}
begin{document}
begin{tikzpicture}[scale=1,cap=round]
tikzset{axes/.style={}}
% The graphic
begin{scope}[style=axes]
draw[->] (-.5,0) -- (3,0) node[below] {$x$};
draw[->] (0,-.5)-- (0,3) node[left] {$y$};
draw (0.25,0.4) to [out=10,in=80] (1.5,2.5);
draw (1.5,2.5) to [out=-80, in=175] (2.75,.4);
%%%
end{scope}
end{tikzpicture}
end{document}


This outputs:



enter image description here



I am trying to achieve:



enter image description here



Note it is not exactly symmetric either but the head is cleaner.










share|improve this question
























  • What precisely do you want to achieve? (The top is tilted since you have in=80 and out=-80 instead of 90.)
    – marmot
    Nov 30 at 14:30










  • @Marmot, I posted above what I am trying to achieve, I didn't think I'd have to ask a question on this but I also tried 90 and could not get the look I wanted.
    – MathScholar
    Nov 30 at 15:09












  • How about draw[thick] (0.25,0.4) to [out=10,in=-90] (1.5,2.5) to [out=-90, in=175] (2.75,.4);?
    – marmot
    Nov 30 at 15:13
















8














This Cusp I made came out awkward and I can not fix it.



documentclass{article}
usepackage{tikz}
begin{document}
begin{tikzpicture}[scale=1,cap=round]
tikzset{axes/.style={}}
% The graphic
begin{scope}[style=axes]
draw[->] (-.5,0) -- (3,0) node[below] {$x$};
draw[->] (0,-.5)-- (0,3) node[left] {$y$};
draw (0.25,0.4) to [out=10,in=80] (1.5,2.5);
draw (1.5,2.5) to [out=-80, in=175] (2.75,.4);
%%%
end{scope}
end{tikzpicture}
end{document}


This outputs:



enter image description here



I am trying to achieve:



enter image description here



Note it is not exactly symmetric either but the head is cleaner.










share|improve this question
























  • What precisely do you want to achieve? (The top is tilted since you have in=80 and out=-80 instead of 90.)
    – marmot
    Nov 30 at 14:30










  • @Marmot, I posted above what I am trying to achieve, I didn't think I'd have to ask a question on this but I also tried 90 and could not get the look I wanted.
    – MathScholar
    Nov 30 at 15:09












  • How about draw[thick] (0.25,0.4) to [out=10,in=-90] (1.5,2.5) to [out=-90, in=175] (2.75,.4);?
    – marmot
    Nov 30 at 15:13














8












8








8







This Cusp I made came out awkward and I can not fix it.



documentclass{article}
usepackage{tikz}
begin{document}
begin{tikzpicture}[scale=1,cap=round]
tikzset{axes/.style={}}
% The graphic
begin{scope}[style=axes]
draw[->] (-.5,0) -- (3,0) node[below] {$x$};
draw[->] (0,-.5)-- (0,3) node[left] {$y$};
draw (0.25,0.4) to [out=10,in=80] (1.5,2.5);
draw (1.5,2.5) to [out=-80, in=175] (2.75,.4);
%%%
end{scope}
end{tikzpicture}
end{document}


This outputs:



enter image description here



I am trying to achieve:



enter image description here



Note it is not exactly symmetric either but the head is cleaner.










share|improve this question















This Cusp I made came out awkward and I can not fix it.



documentclass{article}
usepackage{tikz}
begin{document}
begin{tikzpicture}[scale=1,cap=round]
tikzset{axes/.style={}}
% The graphic
begin{scope}[style=axes]
draw[->] (-.5,0) -- (3,0) node[below] {$x$};
draw[->] (0,-.5)-- (0,3) node[left] {$y$};
draw (0.25,0.4) to [out=10,in=80] (1.5,2.5);
draw (1.5,2.5) to [out=-80, in=175] (2.75,.4);
%%%
end{scope}
end{tikzpicture}
end{document}


This outputs:



enter image description here



I am trying to achieve:



enter image description here



Note it is not exactly symmetric either but the head is cleaner.







tikz-pgf






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 30 at 15:08

























asked Nov 30 at 14:19









MathScholar

5848




5848












  • What precisely do you want to achieve? (The top is tilted since you have in=80 and out=-80 instead of 90.)
    – marmot
    Nov 30 at 14:30










  • @Marmot, I posted above what I am trying to achieve, I didn't think I'd have to ask a question on this but I also tried 90 and could not get the look I wanted.
    – MathScholar
    Nov 30 at 15:09












  • How about draw[thick] (0.25,0.4) to [out=10,in=-90] (1.5,2.5) to [out=-90, in=175] (2.75,.4);?
    – marmot
    Nov 30 at 15:13


















  • What precisely do you want to achieve? (The top is tilted since you have in=80 and out=-80 instead of 90.)
    – marmot
    Nov 30 at 14:30










  • @Marmot, I posted above what I am trying to achieve, I didn't think I'd have to ask a question on this but I also tried 90 and could not get the look I wanted.
    – MathScholar
    Nov 30 at 15:09












  • How about draw[thick] (0.25,0.4) to [out=10,in=-90] (1.5,2.5) to [out=-90, in=175] (2.75,.4);?
    – marmot
    Nov 30 at 15:13
















What precisely do you want to achieve? (The top is tilted since you have in=80 and out=-80 instead of 90.)
– marmot
Nov 30 at 14:30




What precisely do you want to achieve? (The top is tilted since you have in=80 and out=-80 instead of 90.)
– marmot
Nov 30 at 14:30












@Marmot, I posted above what I am trying to achieve, I didn't think I'd have to ask a question on this but I also tried 90 and could not get the look I wanted.
– MathScholar
Nov 30 at 15:09






@Marmot, I posted above what I am trying to achieve, I didn't think I'd have to ask a question on this but I also tried 90 and could not get the look I wanted.
– MathScholar
Nov 30 at 15:09














How about draw[thick] (0.25,0.4) to [out=10,in=-90] (1.5,2.5) to [out=-90, in=175] (2.75,.4);?
– marmot
Nov 30 at 15:13




How about draw[thick] (0.25,0.4) to [out=10,in=-90] (1.5,2.5) to [out=-90, in=175] (2.75,.4);?
– marmot
Nov 30 at 15:13










3 Answers
3






active

oldest

votes


















7














I guess the sign of 80 in draw (0.25,0.4) to [out=10,in=80] (1.5,2.5); is the culprit.



documentclass{article}
usepackage{tikz}
begin{document}
begin{tikzpicture}[scale=1,cap=round]
tikzset{axes/.style={}}
% The graphic
begin{scope}[style=axes]
draw[->] (-.5,0) -- (3,0) node[below] {$x$};
draw[->] (0,-.5)-- (0,3) node[left] {$y$};
draw[thick] (0.25,0.4) to [out=10,in=-90] (1.5,2.5)
to [out=-90, in=175] (2.75,.4);
draw[thin] (1.5,2.5) -- ++ (0.2,0.6) node[above]{$f'(a)$ does not exist};
draw (1.5,0.2) -- (1.5,-0.2) node[below]{$a$};
%%%
end{scope}
end{tikzpicture}
end{document}


enter image description here






share|improve this answer



















  • 1




    marmot, why does it have to be 90? (Side note $f'(a)$ does not exist).
    – gusbrs
    Nov 30 at 15:17












  • @gusbrs Well, that's how I interpreted "Note it is not exactly symmetric either but the head is cleaner.". I thought the OP wants to have it symmetric. (And thanks for the note!)
    – marmot
    Nov 30 at 15:20












  • @Marmot, that works but I thought I tried that. Eyes playing tricks on me. Not sure exactly what we changed Thanks Marmot!
    – MathScholar
    Nov 30 at 15:21










  • @MathScholar I guess the sign of in in draw (0.25,0.4) to [out=10,in=80] (1.5,2.5); is crucial. I have a - there.
    – marmot
    Nov 30 at 15:22










  • Well, conceptually, we could still have side derivatives (is this how this is said in english?...) reaching "a" at 80 degrees and the derivative in a would not exist. The graphic would be symmetric like this. But the question is why doesn't tikz accept it?
    – gusbrs
    Nov 30 at 15:23



















4














Here is a solution using the pgfplots package, which extends tikz to include a wide variety of plotting options:



documentclass{article}
usepackage{pgfplots}
begin{document}
begin{tikzpicture}begin{axis}[xmin=0,xmax=1,ymin=0,xtick={0.5},xticklabels=$a$,axis lines=left,ymajorticks=false,xlabel=$x$,ylabel=$y$,clip=false]
addplot [domain=0.1:0.5] {0.1+x^2} node [pos=0.2,above left] {$f$};
addplot [domain=0.5:0.9] {0.1+(x-1)^2} node[pin={90: $f'(a)$ does not exist.} ] at (axis cs:0.5,0.35) {};
end{axis}
end{tikzpicture}
end{document}


enter image description here






share|improve this answer





















  • @Ubiquitos Thank you for your answer. I needed a corner with a more vertical right and left tangent, but someone else will find this very useful to them. Thanks for sharing
    – MathScholar
    Nov 30 at 15:41





















2














A PSTricks solution just for comparison purposes.



documentclass[pstricks]{standalone}
usepackage{pst-plot,pst-plot}
deff#1{(x-#1)^2+1}
begin{document}
begin{pspicture}[algebraic,ticks=none,labels=none](-.4,-.5)(4.5,3.5)
psaxes{->}(0,0)(-.3,-.4)(4,3)[$x$,0][$y$,90]
psplot{1}{2}{f{1}}
psplot{2}{3}{f{3}}
psxTick[labelsep=1pt](2){a}
rput(2.2,3){rnode[b]{a}{$f'(a)$ does not exist}}
pcline[nodesep=2pt]{->}(a)(*2 {f{1}})
end{pspicture}
end{document}


enter image description here






share|improve this answer





















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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    7














    I guess the sign of 80 in draw (0.25,0.4) to [out=10,in=80] (1.5,2.5); is the culprit.



    documentclass{article}
    usepackage{tikz}
    begin{document}
    begin{tikzpicture}[scale=1,cap=round]
    tikzset{axes/.style={}}
    % The graphic
    begin{scope}[style=axes]
    draw[->] (-.5,0) -- (3,0) node[below] {$x$};
    draw[->] (0,-.5)-- (0,3) node[left] {$y$};
    draw[thick] (0.25,0.4) to [out=10,in=-90] (1.5,2.5)
    to [out=-90, in=175] (2.75,.4);
    draw[thin] (1.5,2.5) -- ++ (0.2,0.6) node[above]{$f'(a)$ does not exist};
    draw (1.5,0.2) -- (1.5,-0.2) node[below]{$a$};
    %%%
    end{scope}
    end{tikzpicture}
    end{document}


    enter image description here






    share|improve this answer



















    • 1




      marmot, why does it have to be 90? (Side note $f'(a)$ does not exist).
      – gusbrs
      Nov 30 at 15:17












    • @gusbrs Well, that's how I interpreted "Note it is not exactly symmetric either but the head is cleaner.". I thought the OP wants to have it symmetric. (And thanks for the note!)
      – marmot
      Nov 30 at 15:20












    • @Marmot, that works but I thought I tried that. Eyes playing tricks on me. Not sure exactly what we changed Thanks Marmot!
      – MathScholar
      Nov 30 at 15:21










    • @MathScholar I guess the sign of in in draw (0.25,0.4) to [out=10,in=80] (1.5,2.5); is crucial. I have a - there.
      – marmot
      Nov 30 at 15:22










    • Well, conceptually, we could still have side derivatives (is this how this is said in english?...) reaching "a" at 80 degrees and the derivative in a would not exist. The graphic would be symmetric like this. But the question is why doesn't tikz accept it?
      – gusbrs
      Nov 30 at 15:23
















    7














    I guess the sign of 80 in draw (0.25,0.4) to [out=10,in=80] (1.5,2.5); is the culprit.



    documentclass{article}
    usepackage{tikz}
    begin{document}
    begin{tikzpicture}[scale=1,cap=round]
    tikzset{axes/.style={}}
    % The graphic
    begin{scope}[style=axes]
    draw[->] (-.5,0) -- (3,0) node[below] {$x$};
    draw[->] (0,-.5)-- (0,3) node[left] {$y$};
    draw[thick] (0.25,0.4) to [out=10,in=-90] (1.5,2.5)
    to [out=-90, in=175] (2.75,.4);
    draw[thin] (1.5,2.5) -- ++ (0.2,0.6) node[above]{$f'(a)$ does not exist};
    draw (1.5,0.2) -- (1.5,-0.2) node[below]{$a$};
    %%%
    end{scope}
    end{tikzpicture}
    end{document}


    enter image description here






    share|improve this answer



















    • 1




      marmot, why does it have to be 90? (Side note $f'(a)$ does not exist).
      – gusbrs
      Nov 30 at 15:17












    • @gusbrs Well, that's how I interpreted "Note it is not exactly symmetric either but the head is cleaner.". I thought the OP wants to have it symmetric. (And thanks for the note!)
      – marmot
      Nov 30 at 15:20












    • @Marmot, that works but I thought I tried that. Eyes playing tricks on me. Not sure exactly what we changed Thanks Marmot!
      – MathScholar
      Nov 30 at 15:21










    • @MathScholar I guess the sign of in in draw (0.25,0.4) to [out=10,in=80] (1.5,2.5); is crucial. I have a - there.
      – marmot
      Nov 30 at 15:22










    • Well, conceptually, we could still have side derivatives (is this how this is said in english?...) reaching "a" at 80 degrees and the derivative in a would not exist. The graphic would be symmetric like this. But the question is why doesn't tikz accept it?
      – gusbrs
      Nov 30 at 15:23














    7












    7








    7






    I guess the sign of 80 in draw (0.25,0.4) to [out=10,in=80] (1.5,2.5); is the culprit.



    documentclass{article}
    usepackage{tikz}
    begin{document}
    begin{tikzpicture}[scale=1,cap=round]
    tikzset{axes/.style={}}
    % The graphic
    begin{scope}[style=axes]
    draw[->] (-.5,0) -- (3,0) node[below] {$x$};
    draw[->] (0,-.5)-- (0,3) node[left] {$y$};
    draw[thick] (0.25,0.4) to [out=10,in=-90] (1.5,2.5)
    to [out=-90, in=175] (2.75,.4);
    draw[thin] (1.5,2.5) -- ++ (0.2,0.6) node[above]{$f'(a)$ does not exist};
    draw (1.5,0.2) -- (1.5,-0.2) node[below]{$a$};
    %%%
    end{scope}
    end{tikzpicture}
    end{document}


    enter image description here






    share|improve this answer














    I guess the sign of 80 in draw (0.25,0.4) to [out=10,in=80] (1.5,2.5); is the culprit.



    documentclass{article}
    usepackage{tikz}
    begin{document}
    begin{tikzpicture}[scale=1,cap=round]
    tikzset{axes/.style={}}
    % The graphic
    begin{scope}[style=axes]
    draw[->] (-.5,0) -- (3,0) node[below] {$x$};
    draw[->] (0,-.5)-- (0,3) node[left] {$y$};
    draw[thick] (0.25,0.4) to [out=10,in=-90] (1.5,2.5)
    to [out=-90, in=175] (2.75,.4);
    draw[thin] (1.5,2.5) -- ++ (0.2,0.6) node[above]{$f'(a)$ does not exist};
    draw (1.5,0.2) -- (1.5,-0.2) node[below]{$a$};
    %%%
    end{scope}
    end{tikzpicture}
    end{document}


    enter image description here







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Nov 30 at 15:24

























    answered Nov 30 at 15:16









    marmot

    84.9k495179




    84.9k495179








    • 1




      marmot, why does it have to be 90? (Side note $f'(a)$ does not exist).
      – gusbrs
      Nov 30 at 15:17












    • @gusbrs Well, that's how I interpreted "Note it is not exactly symmetric either but the head is cleaner.". I thought the OP wants to have it symmetric. (And thanks for the note!)
      – marmot
      Nov 30 at 15:20












    • @Marmot, that works but I thought I tried that. Eyes playing tricks on me. Not sure exactly what we changed Thanks Marmot!
      – MathScholar
      Nov 30 at 15:21










    • @MathScholar I guess the sign of in in draw (0.25,0.4) to [out=10,in=80] (1.5,2.5); is crucial. I have a - there.
      – marmot
      Nov 30 at 15:22










    • Well, conceptually, we could still have side derivatives (is this how this is said in english?...) reaching "a" at 80 degrees and the derivative in a would not exist. The graphic would be symmetric like this. But the question is why doesn't tikz accept it?
      – gusbrs
      Nov 30 at 15:23














    • 1




      marmot, why does it have to be 90? (Side note $f'(a)$ does not exist).
      – gusbrs
      Nov 30 at 15:17












    • @gusbrs Well, that's how I interpreted "Note it is not exactly symmetric either but the head is cleaner.". I thought the OP wants to have it symmetric. (And thanks for the note!)
      – marmot
      Nov 30 at 15:20












    • @Marmot, that works but I thought I tried that. Eyes playing tricks on me. Not sure exactly what we changed Thanks Marmot!
      – MathScholar
      Nov 30 at 15:21










    • @MathScholar I guess the sign of in in draw (0.25,0.4) to [out=10,in=80] (1.5,2.5); is crucial. I have a - there.
      – marmot
      Nov 30 at 15:22










    • Well, conceptually, we could still have side derivatives (is this how this is said in english?...) reaching "a" at 80 degrees and the derivative in a would not exist. The graphic would be symmetric like this. But the question is why doesn't tikz accept it?
      – gusbrs
      Nov 30 at 15:23








    1




    1




    marmot, why does it have to be 90? (Side note $f'(a)$ does not exist).
    – gusbrs
    Nov 30 at 15:17






    marmot, why does it have to be 90? (Side note $f'(a)$ does not exist).
    – gusbrs
    Nov 30 at 15:17














    @gusbrs Well, that's how I interpreted "Note it is not exactly symmetric either but the head is cleaner.". I thought the OP wants to have it symmetric. (And thanks for the note!)
    – marmot
    Nov 30 at 15:20






    @gusbrs Well, that's how I interpreted "Note it is not exactly symmetric either but the head is cleaner.". I thought the OP wants to have it symmetric. (And thanks for the note!)
    – marmot
    Nov 30 at 15:20














    @Marmot, that works but I thought I tried that. Eyes playing tricks on me. Not sure exactly what we changed Thanks Marmot!
    – MathScholar
    Nov 30 at 15:21




    @Marmot, that works but I thought I tried that. Eyes playing tricks on me. Not sure exactly what we changed Thanks Marmot!
    – MathScholar
    Nov 30 at 15:21












    @MathScholar I guess the sign of in in draw (0.25,0.4) to [out=10,in=80] (1.5,2.5); is crucial. I have a - there.
    – marmot
    Nov 30 at 15:22




    @MathScholar I guess the sign of in in draw (0.25,0.4) to [out=10,in=80] (1.5,2.5); is crucial. I have a - there.
    – marmot
    Nov 30 at 15:22












    Well, conceptually, we could still have side derivatives (is this how this is said in english?...) reaching "a" at 80 degrees and the derivative in a would not exist. The graphic would be symmetric like this. But the question is why doesn't tikz accept it?
    – gusbrs
    Nov 30 at 15:23




    Well, conceptually, we could still have side derivatives (is this how this is said in english?...) reaching "a" at 80 degrees and the derivative in a would not exist. The graphic would be symmetric like this. But the question is why doesn't tikz accept it?
    – gusbrs
    Nov 30 at 15:23











    4














    Here is a solution using the pgfplots package, which extends tikz to include a wide variety of plotting options:



    documentclass{article}
    usepackage{pgfplots}
    begin{document}
    begin{tikzpicture}begin{axis}[xmin=0,xmax=1,ymin=0,xtick={0.5},xticklabels=$a$,axis lines=left,ymajorticks=false,xlabel=$x$,ylabel=$y$,clip=false]
    addplot [domain=0.1:0.5] {0.1+x^2} node [pos=0.2,above left] {$f$};
    addplot [domain=0.5:0.9] {0.1+(x-1)^2} node[pin={90: $f'(a)$ does not exist.} ] at (axis cs:0.5,0.35) {};
    end{axis}
    end{tikzpicture}
    end{document}


    enter image description here






    share|improve this answer





















    • @Ubiquitos Thank you for your answer. I needed a corner with a more vertical right and left tangent, but someone else will find this very useful to them. Thanks for sharing
      – MathScholar
      Nov 30 at 15:41


















    4














    Here is a solution using the pgfplots package, which extends tikz to include a wide variety of plotting options:



    documentclass{article}
    usepackage{pgfplots}
    begin{document}
    begin{tikzpicture}begin{axis}[xmin=0,xmax=1,ymin=0,xtick={0.5},xticklabels=$a$,axis lines=left,ymajorticks=false,xlabel=$x$,ylabel=$y$,clip=false]
    addplot [domain=0.1:0.5] {0.1+x^2} node [pos=0.2,above left] {$f$};
    addplot [domain=0.5:0.9] {0.1+(x-1)^2} node[pin={90: $f'(a)$ does not exist.} ] at (axis cs:0.5,0.35) {};
    end{axis}
    end{tikzpicture}
    end{document}


    enter image description here






    share|improve this answer





















    • @Ubiquitos Thank you for your answer. I needed a corner with a more vertical right and left tangent, but someone else will find this very useful to them. Thanks for sharing
      – MathScholar
      Nov 30 at 15:41
















    4












    4








    4






    Here is a solution using the pgfplots package, which extends tikz to include a wide variety of plotting options:



    documentclass{article}
    usepackage{pgfplots}
    begin{document}
    begin{tikzpicture}begin{axis}[xmin=0,xmax=1,ymin=0,xtick={0.5},xticklabels=$a$,axis lines=left,ymajorticks=false,xlabel=$x$,ylabel=$y$,clip=false]
    addplot [domain=0.1:0.5] {0.1+x^2} node [pos=0.2,above left] {$f$};
    addplot [domain=0.5:0.9] {0.1+(x-1)^2} node[pin={90: $f'(a)$ does not exist.} ] at (axis cs:0.5,0.35) {};
    end{axis}
    end{tikzpicture}
    end{document}


    enter image description here






    share|improve this answer












    Here is a solution using the pgfplots package, which extends tikz to include a wide variety of plotting options:



    documentclass{article}
    usepackage{pgfplots}
    begin{document}
    begin{tikzpicture}begin{axis}[xmin=0,xmax=1,ymin=0,xtick={0.5},xticklabels=$a$,axis lines=left,ymajorticks=false,xlabel=$x$,ylabel=$y$,clip=false]
    addplot [domain=0.1:0.5] {0.1+x^2} node [pos=0.2,above left] {$f$};
    addplot [domain=0.5:0.9] {0.1+(x-1)^2} node[pin={90: $f'(a)$ does not exist.} ] at (axis cs:0.5,0.35) {};
    end{axis}
    end{tikzpicture}
    end{document}


    enter image description here







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Nov 30 at 15:32









    Ubiquitous

    1,6511020




    1,6511020












    • @Ubiquitos Thank you for your answer. I needed a corner with a more vertical right and left tangent, but someone else will find this very useful to them. Thanks for sharing
      – MathScholar
      Nov 30 at 15:41




















    • @Ubiquitos Thank you for your answer. I needed a corner with a more vertical right and left tangent, but someone else will find this very useful to them. Thanks for sharing
      – MathScholar
      Nov 30 at 15:41


















    @Ubiquitos Thank you for your answer. I needed a corner with a more vertical right and left tangent, but someone else will find this very useful to them. Thanks for sharing
    – MathScholar
    Nov 30 at 15:41






    @Ubiquitos Thank you for your answer. I needed a corner with a more vertical right and left tangent, but someone else will find this very useful to them. Thanks for sharing
    – MathScholar
    Nov 30 at 15:41













    2














    A PSTricks solution just for comparison purposes.



    documentclass[pstricks]{standalone}
    usepackage{pst-plot,pst-plot}
    deff#1{(x-#1)^2+1}
    begin{document}
    begin{pspicture}[algebraic,ticks=none,labels=none](-.4,-.5)(4.5,3.5)
    psaxes{->}(0,0)(-.3,-.4)(4,3)[$x$,0][$y$,90]
    psplot{1}{2}{f{1}}
    psplot{2}{3}{f{3}}
    psxTick[labelsep=1pt](2){a}
    rput(2.2,3){rnode[b]{a}{$f'(a)$ does not exist}}
    pcline[nodesep=2pt]{->}(a)(*2 {f{1}})
    end{pspicture}
    end{document}


    enter image description here






    share|improve this answer


























      2














      A PSTricks solution just for comparison purposes.



      documentclass[pstricks]{standalone}
      usepackage{pst-plot,pst-plot}
      deff#1{(x-#1)^2+1}
      begin{document}
      begin{pspicture}[algebraic,ticks=none,labels=none](-.4,-.5)(4.5,3.5)
      psaxes{->}(0,0)(-.3,-.4)(4,3)[$x$,0][$y$,90]
      psplot{1}{2}{f{1}}
      psplot{2}{3}{f{3}}
      psxTick[labelsep=1pt](2){a}
      rput(2.2,3){rnode[b]{a}{$f'(a)$ does not exist}}
      pcline[nodesep=2pt]{->}(a)(*2 {f{1}})
      end{pspicture}
      end{document}


      enter image description here






      share|improve this answer
























        2












        2








        2






        A PSTricks solution just for comparison purposes.



        documentclass[pstricks]{standalone}
        usepackage{pst-plot,pst-plot}
        deff#1{(x-#1)^2+1}
        begin{document}
        begin{pspicture}[algebraic,ticks=none,labels=none](-.4,-.5)(4.5,3.5)
        psaxes{->}(0,0)(-.3,-.4)(4,3)[$x$,0][$y$,90]
        psplot{1}{2}{f{1}}
        psplot{2}{3}{f{3}}
        psxTick[labelsep=1pt](2){a}
        rput(2.2,3){rnode[b]{a}{$f'(a)$ does not exist}}
        pcline[nodesep=2pt]{->}(a)(*2 {f{1}})
        end{pspicture}
        end{document}


        enter image description here






        share|improve this answer












        A PSTricks solution just for comparison purposes.



        documentclass[pstricks]{standalone}
        usepackage{pst-plot,pst-plot}
        deff#1{(x-#1)^2+1}
        begin{document}
        begin{pspicture}[algebraic,ticks=none,labels=none](-.4,-.5)(4.5,3.5)
        psaxes{->}(0,0)(-.3,-.4)(4,3)[$x$,0][$y$,90]
        psplot{1}{2}{f{1}}
        psplot{2}{3}{f{3}}
        psxTick[labelsep=1pt](2){a}
        rput(2.2,3){rnode[b]{a}{$f'(a)$ does not exist}}
        pcline[nodesep=2pt]{->}(a)(*2 {f{1}})
        end{pspicture}
        end{document}


        enter image description here







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 30 at 16:54









        Artificial Stupidity

        5,29511039




        5,29511039






























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