Extend a line of best fit

I have the following code which gives a plot and a line of best fit.

ARI_Weibull = ones(n,1)./P_Weibull;% Average Return Interval 1/P
figure
semilogx((ARI_Weibull), Q, 'r');
p_1 = polyfit(log(ARI_Weibull),Q,1); % Linear best fit
f_1 = polyval(p_1,log(ARI_Weibull));
hold on
semilogx(ARI_Weibull,f_1,'--r')
Q_100yr = polyval(p_1, log(100)) % Q value for 1 in 100 year
xline(100) ;

I now want to extend the line of best fit which I calculated to show its intersection with x = 100, which should correspond to a y value of Q_100yr calculated above.

How do I go about extending this line.

Thanks in advance,

Brian

Answer 1

[KSSV]

You have used polyfit, so you have slope and y-intercept in your hand. For a given value of x, you can find respective y value using polyval.

y_100 = polyval(p_1,100);

If you want to extend to certain range of values [xmin,xmax], also you can do using polyval.

m = 100 ;
xi = linspace(xmin,xmax,m) ;  % give your values for xmin, xmax 
yi =  polyval(p_1,xi);

[Brian Robinson]

Ok well I have y_100 already as Q_100yr

My original figure looks like this:

Now I have done as you say:

m = 100;
xi = linspace(1,1000,100);
yi = polyval(p_1,xi);

Now what is the command to plot this? Because I tried semilogx(xi,yi) and that isn't right.

[KSSV]

yi = polyval(p_1,log(xi));

Answer 2

[Brian Robinson]

Ok thanks for that.