Fimplicit not plotting properly

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Nethanel Benzaquen
Nethanel Benzaquen 2020-12-3
评论： Walter Roberson 2020-12-3
Hello,
Im trying to implicity plot conic sections using this deffinition in the app desinger.
fun1=@(x,y) a.*x.^2+2*h.*x.*y+b*y.^2+2*g.*x+2*f.*y+c;
after inputing
a=2 , 2*h=0, b=0, 2*g=3, 2*f=1, c=7
fimplicit is returning a blank graph although this is a perfectly fine parabola.
when plotting it with a linspace as such:
xx=-10:10;
yy=-2.*xx.^2-7.*xx-7;
plot(xx,yy)
Matlab plots the function perfectly fine. So why does't the fimplicit plot it aswell.
thanks

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采纳的回答

Stephan
Stephan 2020-12-3
编辑：Stephan 2020-12-3
fun1=@(x,y) a.*x.^2+2*h.*x.*y+b*y.^2+2*g.*x+2*f.*y+c;
a=2;
h=0;
b=0;
g=3/2;
f=1/2;
c=7;
fimplicit(fun1,[-3 2 -12 -5])
  2 个评论
Nethanel Benzaquen
Nethanel Benzaquen 2020-12-3
Thanks!
do you know why when writting the code like this
fun1=@(x,y) a.*x.^2+2*h.*x.*y+b*y.^2+2*g.*x+2*f.*y+c;
fimplicit(app.ax1,fun1);
xlim(app.ax1,'auto')
ylim(app.ax1,'auto')
the xlim and ylim didn't work. The Idea is for any value of coefficients i should see the graph. but in this specif case it didnt show it to me
Stephan
Stephan 2020-12-3
No, i also did not get it to work in an automatic way...

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更多回答（1 个）

Walter Roberson
Walter Roberson 2020-12-3
Ran in:
a=2; h=0/2; b=0; g=3/2; f=1/2; c=7;
fun1=@(x,y) a.*x.^2+2*h.*x.*y+b*y.^2+2*g.*x+2*f.*y+c;
fimplicit(fun1, [-10 10 -10 10])
Looks plausible to me.
  2 个评论
Nethanel Benzaquen
Nethanel Benzaquen 2020-12-3
Thanks!
do you know why when writting the code like this
fun1=@(x,y) a.*x.^2+2*h.*x.*y+b*y.^2+2*g.*x+2*f.*y+c;
fimplicit(app.ax1,fun1);
xlim(app.ax1,'auto')
ylim(app.ax1,'auto')
the xlim and ylim didn't work. The Idea is for any value of coefficients i should see the graph. but in this specif case it didnt show it to me
Walter Roberson
Walter Roberson 2020-12-3
Setting the xlim or ylim has no effect on the range that fimplicit uses to plot, and changing the limits to auto after the call does not trigger an already existing fimplicit to go back and analyze the function to find a range of bounds that will give an interesting plot. fimplicit only draws according to the range passed in or the default for the fimplicit call.
You are not passing in limits and the default limits happen to have no interesting content for this function.

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