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Drawing a graph from a rational function. What is the problem in my code?

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I want to draw the graph of function f(x)=x^2-1/x^2-4
This is my code:
clear
clf
f=@(x)((x.^2)-1)./((x.^2)-4);
xa=-10; xb=10; s=0.05; ya=-5; yb=5;
xv=linspace(xa,-s); xh=linspace(s,xb);
plot(xv,f(xv),'blue',xh,f(xh),'blue','linewidth',2)
axis equal, axis([xa xb ya yb]), grid on
xlabel('x'), ylabel('y'), title('function')
hold on
However when draw it, it comes out wrong. It has two blue lines where the x asymptotes should be.

Accepted Answer

Star Strider
Star Strider on 10 Oct 2014
It’s necessary to not plot the singularities, then (if you want to), plot the asymptotes there instead. I had to add six lines to replace the singularities with ‘NaN’ values, move the hold line to just below the first plot call, and another two to plot the asymptotes:
f=@(x)((x.^2)-1)./((x.^2)-4);
xa=-10; xb=10; s=0.05; ya=-5; yb=5;
xv=linspace(xa,-s); xh=linspace(s,xb);
yv = f(xv); % Evaluate Function
yh = f(xh);
[mv,vi] = max(yv); % Find Maxima
[mh,hi] = max(yh);
yv(vi) = NaN; % Don’t Connect Singularities
yh(hi) = NaN;
plot(xv,yv,'blue',xh,yh,'blue','linewidth',2)
hold on
plot(xv(vi)*[1 1]', [ya yb]', '--r') % Plot X-Asymptote
plot(xh(hi)*[1 1]', [ya yb]', '--r') % Plot X-Asymptote
axis equal, axis([xa xb ya yb]), grid on
xlabel('x'), ylabel('y'), title('function')
This looks like it does what you want. If you don’t want the asymptotes, don’t include those two lines.
  9 Comments
Star Strider
Star Strider on 13 Oct 2014
They replaced phlogiston with something called ‘radium’. Never could get comfortable with all that stuff glowing by itself.

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More Answers (2)

SK
SK on 10 Oct 2014
Edited: SK on 10 Oct 2014
Since it is a discrete sequence of points, Matlab just connects the last point before -2 with the first point after -2. It has no idea that you want the graph plotted on the entire real line. Same for the asymptote at +2.
The linspace() function takes 100 equally spaced points in the specified interval. So in your case, the function is evaluated at [-10.0000, -9.9005, -9.8010, -9.7015 ...]. The closest it gets to -2 is at -2.0400 and -1.9405.
If you want it to try to hit -2 exactly try the linspace() function with the third argument, N, which tells it how many points you want. See what happens.
  1 Comment
Britney
Britney on 11 Oct 2014
Thank you for that explanation it was very insightful. I didn't know that that was how Matlab "thinks".

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SK
SK on 10 Oct 2014
Edited: SK on 10 Oct 2014
Star Strider has a point too. However with the values you have, linspace does not actually pass very close to -2 or 2. (Try linspace(-10, -0.05)). At least on my system there are no out of bounds values involved. However if I use:
f(-10 : 0.1 : -0.05) and f(0.05 : 0.1 : 10)
then I just get two straight lines where the asymptotes are. The reason is that f(-c) and f(c) where c is very-very-very close to 2, is a huge number (but not infinity) so all the other values look very-very-very small -effectively 0 on the graph which explains why the graph looks like that.
Try printing the values f(-10 : 0.1 : -0.05) and you will see that none of them are really 0 or infinity.

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