plotting a linear equation

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Andy
Andy 2012 年 7 月 25 日
編集済み: DGM 2024 年 11 月 12 日
How do I plot a linear equation y=mx+b?
So let's say I have this:
b0= 3 where b0 is the y-intercept b1= 4 where b1 is the x coefficient
Then:
Y= b0-b1*X
How do I plot this?

回答 (4 件)

per isakson
per isakson 2012 年 7 月 25 日
Try:
b0 = 3;
b1 = 4;
f = @(x) b0-b1*x;
ezplot( f, 0, 5 )
  1 件のコメント
Sergio E. Obando
Sergio E. Obando 2024 年 11 月 12 日
Just a note that as of R2016a, ezplot is no longer recommended. If anyone is reading this now, the code would still work or you can update to:
b0 = 3;
b1 = 4;
f = @(x) b0-b1*x;
fplot( f,[0,5])
See fplot for more info.

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Elizabeth
Elizabeth 2012 年 7 月 25 日
編集済み: DGM 2024 年 11 月 12 日
Or:
By all means, the solution method above will work. However, as your function increases in complexity, that command becomes more and more expensive. Try defining your domain x, then, as a vector:
b0=3; b1=4;
x= linspace(xmin,xmax, n); % Adapt n for resolution of graph
y= b0-b1*x;
plot(x,y)
  4 件のコメント
Nicholas Copsey
Nicholas Copsey 2020 年 3 月 28 日
xmin, xmax, and n are things you can change in the code for various views of the graph
DGM
DGM 2024 年 11 月 12 日
To clarify:
% curve parameters
b0 = 3; % y-intercept
b1 = 4; % (negative) slope
% define x
xmin = 0; % pick these as needed
xmax = 1;
n = 100; % number of samples
x = linspace(xmin,xmax, n);
% calculate y from x
y = b0 - b1*x;
plot(x,y)

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Ben Le
Ben Le 2015 年 12 月 9 日
編集済み: DGM 2024 年 11 月 12 日
b0 = 3;
b1 = 4;
% You can choose any range of x. In this case I chose x is from -50 to +50
x = -50:50;
y = b0 - b1*x;
plot(x,y)

Andy
Andy 2012 年 7 月 25 日
How do I get that into the same graphs? I have points for my data and the equation above is the linear regression. Ezplot gave me two graphs. :(. I want both the points and the linear equation graphed in one figure.
  1 件のコメント
Elizabeth
Elizabeth 2012 年 7 月 25 日
use the 'hold on' command

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