Finding slope and y intercept

Hello, I was wondering if there is an easy way to find the slope and intercept of a line using MATLAB, like how it is so easy with Excel where you just plot the data and add a trendline, so then it will tell you the slope and intercept. Here is my code

tau = [15, 38, 100, 300, 1200];
CA = [1.5 1.25 1 0.75 0.5];
CA0 = 2;
dCdt = log((CA-CA0)./tau);
plot(log(CA),log((CA-CA0)./tau))
xlabel('ln(C_{A})')
ylabel('ln(C_{A}-C_{A0}/ \tau)')

I have a theory that says ln((CA-CA0)/tau) = ln(k) + alpha(ln(CA)), and I want to find alpha and ln k, which is my slope and intercept, respectively.

Thank you

Answer 1

[Sean de Wolski]

Have you tried the Curve Fitting App (Curve Fitting Toolbox, req'd)

>>cftool

[Rick]

The curve fitting toolbox is something that you have to pay extra for, right? By the way, I guess just using polyfit will do the trick.

I have the students edition, and I dont know if it comes with the curve fitting toolbox. If it does I really want to know how to access it!!

[Andrew Reibold]

If you just use polyfit to get linear data, just take two points from it and do the elementary calculations.

[Sean de Wolski]

Student Version usually does come with CFT. Try calling:

>>cftool

To see if you do.

Then you can fit arbitrary functions and you don't have to worry about linearizing them to play with polyfit.

Answer 2

[Torsten]

Did you look at the x-value where your "intercept" is between 0.32 and 0.33 ?

Best wishes

Torsten.

[Venkata]

x-intercept: 3.85; y-intercept: 0.328

Is this what you are asking?

[Torsten]

The p2-value always refers to x=0.

Thus to get the computed value of 0.3046, you must look at the intercept at x=0, not at x=3.85.

Best wishes

Torsten.

[Venkata]

Now it makes sense.

Thanks a ton, Torsten.

Answer 3

[Venkata]

I've used 'cftool' for my data. The intercept is in between 0.32 and 0.33 as can be seen from the figure.

However, the 'p2' value is 0.3046, with 95% confidence bounds.

Please explain me this.