Suppose i have points ( 0.9, 79, 76,23, 1, 3, 4.3,89), what is the slope of the line formed by this points

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SOURANGSU
SOURANGSU 2013-12-6
回答： Image Analyst 2013-12-7
slope of line formed by arbitary points

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回答（4 个）

Azzi Abdelmalek
Azzi Abdelmalek 2013-12-6
These points don't form one line. Maybe you need to use a curve fitting toolbox to fit a function ax+b
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John D'Errico
John D'Errico 2013-12-6
ax+b IS the equation of a straight line. How will the curve fitting toolbox fit a straight line any better than any other fitting tool? Magic?

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Wayne King
Wayne King 2013-12-6
编辑：Wayne King 2013-12-6
You have not told us what the "x"-values are. If we assume that they are
y = [0.9, 79, 76,23, 1, 3, 4.3,89];
x = 1:length(y); % 1 to 8
You can only fit a least-squares line to this data in order to measure the slope.
X = ones(length(y),2);
X(:,2) = 1:length(y);
y = y(:);
X\y
The intercept is 34.4071 and the slope is 0.0262, but that is a very poor approximation to the data.
Jos (10584)
Jos (10584) 2013-12-6
编辑：Jos (10584) 2013-12-6
help polyfit
Note that points are usually defined by coordinates and not by single numbers ...
Image Analyst
Image Analyst 2013-12-7
Try polyfit() to fit (regress) a line through some (x,y) coordinates:
coefficients = polyfit(x, y, 1);
slope = coefficients(1);

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