我试图创建一个带有2个不同缩放轴的根函数图,所以让我们说x轴从0到1.2,步长为0.1,y轴从0到1.4,步长为0 0.2(一个函数,2个不同缩放的轴)。我认为我的缩放比例是正确的,如果有更好的方法来编程,请纠正我。
这是我的代码:
x = linspace(0,1.2);
y = 0.5 + (0.9 * (x.^2 - 0.0432)).^(1/2);
% here I need the negative part as well: 0.5 - [...] as follows:
% y2 = 0.5 - (0.9 * (x.^2 - 0.0432)).^(1/2);
% How can I create this function and plot it?
plot(x,y)
axis([0 1.2 0 1.4])
set(gca,'xTick',0:0.1:1.2)
set(gca,'yTick',0:0.2:1.4)
grid on
我有函数的上半部分,但不是较低的部分(负数,请参见上面的代码注释)。它怎么能被创造出来?或者,如果不可能,我怎样才能从不同定义的子图中创建图形'?域以某种方式需要限制为x> = 0.206。
答案 0 :(得分:3)
你关闭了!我会做以下事情。请参阅以下代码中的评论:
n = 1000; % number of points. More points, smoother
% looking piecewise linear approx. of curve
x0 = sqrt(.0432)+eps; % Choose smallest xvalue to be at or epsilon to the right
% of the apex of the parabola
x = linspace(x0, 1.2, n)'; %' transpose so x is a column vector (more convenient)
y_pos = 0.5 + (0.9 * (x.^2 - 0.0432)).^(1/2); % positive branch of parabola
y_neg = 0.5 - (0.9 * (x.^2 - 0.0432)).^(1/2); % negative branch of parabola
plot(x,[y_pos, y_neg],'blue') % we´re graphing two series but use 'blue'
% for both so it looks like one series!
axis([0 1.2 0 1.4])
set(gca,'xTick',0:0.1:1.2)
set(gca,'yTick',0:0.2:1.4)
grid on