What does the value s do in my code??

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Britney
Britney 2014년 10월 16일
댓글: Britney 2014년 10월 16일
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?
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Britney
Britney 2014년 10월 16일
Also how do you plot the
title('((x.^2)-1)./((x.^2)-4)')
so that it does not have the matlab structure to it. I want the title to look like you write the equation if anyone understand me.

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채택된 답변

Star Strider
Star Strider 2014년 10월 16일
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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Britney
Britney 2014년 10월 16일
haha

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추가 답변(2개)

Britney
Britney 2014년 10월 16일
Is that it goes from -10 to -1 from
xv=linspace(xa, -s)
?
But why not from -10 to -2? That has to be equated to axis([xa xb ya yb])?

Matt Tearle
Matt Tearle 2014년 10월 16일
This code is... a bit muddled. At the end of the day, the s is not really necessary. The problem you were having before with the line across the asymptotes is that MATLAB simply calculates the values in the x vector and joins the points up, so on one side of the asymptote (x = -2.mumble) y is 10^23 (or whatever) and the next x value on the other side (x = -1.9mumble) y = -10^17. The simplest way to fix that is just split the x vector into three pieces ([-5,-2], [-2,2], [2,5]).
Your code kinda does that in a roundabout way, but you can make it much cleaner by just making three x vectors and plotting the corresponding y values for each:
f=@(x)((x.^2)-1)./((x.^2)-4);
xa=-5; xb=-2;
x1=linspace(xa,xb);
x2=linspace(xb,-xb);
x3=linspace(-xb,-xa);
plot(x1,f(x1),'blue',x2,f(x2),'blue',x3,f(x3),'blue','linewidth',2)
Regarding your second question, the input to title is just a string, so you can put anything you want there:
title('(x^2-1)/(x^2-4)')
MATLAB will interpret the carats as exponent markup, which is nice. But you can also go completely nuts if you know LaTeX:
title('$$\frac{x^2-1}{x^2-4}$$','Interpreter','latex')
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Britney
Britney 2014년 10월 16일
Thank you very much for answering. You didn't tell much about the s variable but I really appreciated your input about my function code.
p.s cute kiddo on avatar

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