MATLAB: What does the value s do in the code?

asymptotescodegraphproblemvalue

clear
clf
f=@(x)((x.^2)-1)./((x.^2)-4);
xa=-10; xb=-2; s=1; 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,hold on
xlabel('x'), ylabel('y'), title('((x.^2)-1)./((x.^2)-4)')
f=@(x)((x.^2)-1)./((x.^2)-4);
xa=-2; xb=2; s=0; 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, 
f=@(x)((x.^2)-1)./((x.^2)-4);
xa=2; xb=10; s=1; ya=-5; yb=5; 
xv=linspace(xa,-s);  xh=linspace(s,xb);
%xv=linspace(xa,xb); 
plot(xv,f(xv),'blue',xh,f(xh),'blue','linewidth',2)
axis equal, axis([-5 5 ya yb]), grid on,
Can anyone explain what s does in my equation. When I tried making the graph from the helpful input from my earlier question (thank you everyone) the problem was that the graph had two blue lines where the asymptotes should be. I splitted the x-values in 3 parts and made s=1 instead of s=0 for the two graphs above y=1 and then my graph worked the way it should. This however has not made me any wiser to what s means and does…Can anyone explain?

Best Answer

The ‘s’ variable sets the upper limit of ‘xv’ and the lower limit of ‘xh’. It was in your original code, so I assumed you wanted it kept in the revised code I posted. You can set ‘s’ to be whatever you want.
The call to axis:
axis([xa xb ya yb])
sets the axis limits. It is independent of ‘s’.
The title will display as close to ‘normal’ as possible by removing the dot-operators:
title('(x^2-1)/(x^2-4)')
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