Hello, the question is, is it possible to set an interval in y when plotting a graph, if the interval is already set in x?

3 件のコメント

Dyuman Joshi
Dyuman Joshi 2023 年 11 月 22 日
It's not clear to me what you want to do.
Could you elaborate on this and provide more information? An example alongside would be helpful.
Maria
Maria 2023 年 11 月 22 日
I have a task to build a graph of a trigonometric equation (cos(2*t))^(-2) - (sin(2*t))^(-2) =8/3), I build the graph as follows (I will attach part of the program) , but they told me that I need to limit the schedule by f, but I didn’t find information about this
t = 0:0.01:3*pi;
f= (cos(2*t)).^(-2) - (sin(2*t)).^(-2) - 8/3;
figure;
plot(t, f, 'LineWidth', 2);
xlabel('t');
ylabel('y');
title('График уравнения');
grid on;
Dyuman Joshi
Dyuman Joshi 2023 年 11 月 22 日
"but they told me that I need to limit the schedule by f"
I do not understand this. Can you explain a bit more about this?
Also, do you know what is the expected output? If yes, then please share.

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 採用された回答

Sam Chak
Sam Chak 2023 年 11 月 22 日

1 投票

The ylim() command can be used to set the interval on the y-axis.
x = -1:0.01:1;
y = x.^3 - 0.25*x.^2 - 0.5*x;
figure (1)
plot(x, y), grid on
figure (2)
plot(x, y), grid on, ylim([-0.4, 0.2])

2 件のコメント

Maria
Maria 2023 年 11 月 22 日
Thank you it works @Sam Chak
Sam Chak
Sam Chak 2023 年 11 月 22 日
You are welcome, @Maria. If you find the solution helpful, please consider clicking 'Accept' ✔ on the answer and voting 👍 for it.
By the way, the function has singularities (division-by-zero events). That's why the plot shows the discontinuities.
t = 0:0.01:3*pi;
f = (cos(2*t)).^(-2) - (sin(2*t)).^(-2) - 8/3;
figure;
plot(t, f, 'LineWidth', 2);
ylim([-20 20])
xlabel('t');
ylabel('y');
title('График уравнения');
grid on;

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