I am having trouble getting the Best fit line through the data points in these two codes can you please help me out where I am messing up the code?

%PS253 Lab 7 Analysis

close all

clear all

clc

figure(1)

%Analysis Stretch

x = [.015 .018 .025 .027 .033 .04 .044 .049 .055 .058 .065 .07 .075 .081 .087];

y = [.637 .686 .882 .98 1.176 1.3906 1.47 1.66502 1.86004 1.95804 2.15404 2.35004 2.44608 2.64306 2.83808];

err = .025*ones(size(y));

errorbar(x,y,err,'.')

p = polyfit(x,y,1);

%Find the slope in th graph

slope = (15*sum(x.*y)-sum(x).*sum(y))./(15*sum(x).^2.-(sum(x)).^2);

hold on

Q = slope.*x;

plot (x,Q,'r-')

legend('Data Points','Best Fit')

title ('Stretch')

xlabel ('Length (m)')

ylabel('Force (N)')

hold off

figure

%Analysis Part 2 Oscillations

M = [.14990 .19980 .24970 .26970 .2897 .2996 .31950 .33950 .34950 .36950 .38950 .39950 .4094 .41940 .54950];

T = [24.0 31.8 32.8 42 46 47.6 50.7 53.8 55.4 58.5 61.7 63.3 64.8 66.4 86.8];

err = .08*ones(size(M));

errorbar(M,T,err,'.')

p = polyfit(M,T,1);

%Find the slope in th graph

slope = (15*sum(M.*T)-sum(M).*sum(T))/(15*sum(M).^2.-(sum(M)).^2)

Q = slope.*M;

plot (M,Q,'r-')

hold on

legend('Data Points','Best Fit')

title ('Expected T')

xlabel ('Mass (kg)')

ylabel('T (s)')

hold off

%PS253 Lab 9 Analysis

% T equals change of temp. J equals change of L K equals original L. L

% equals 1L/2L.

close all

clear all

clc

figure(1)

%Analysis Brass

T = [52 51 49 45 38 26 0];

J = [.1 .15 .16 .19 .26 .35 .7];

K = 518.0;

L = J./K;

err = .0005*ones(size(L));

errorbar (T,L,err,'.')

slope = (sum(T.*L)./sum(T.^2));

dslope = (1./(sqrt(sum(T.^2))));

hold on

Q = slope.*L;

R = dslope.*L;

plot (T,Q,'r-')

plot (T,R,'b-')

legend('Data Points','Equation 4','Equation 5')

title ('Steel Thermal Expansion')

xlabel ('Temp Change Celcius')

ylabel('Length')

hold off

figure

clc

%Analysis Steel

T = [0 20.0 35.0 44.0 48.0 51.0 52.0];

J = [.340 .220 .160 .110 .090 .060 .040];

K = 518.0;

L = J./K;

err = .0005*ones(size(L));

errorbar(T,L,err,'.')

%Find the slope in th graph

slope = (sum(T.*L)./sum(T.^2));

dslope = (1./(sqrt(sum(T.^2))));

hold on

Q = slope.*L;

R = dslope.*L;

plot (T,Q,'r-')

plot (T,R,'b-')

legend('Data Points','Equation 4','Equation 5')

title ('Steel Thermal Expansion')

xlabel ('Temp Change Celcius')

ylabel('Length')

hold off

Answer 1

[KSSV]

Use polyval to evaluate the data points of fit curve.

figure(1)
%Analysis Stretch
x = [.015 .018 .025 .027 .033 .04 .044 .049 .055 .058 .065 .07 .075 .081 .087];
y = [.637 .686 .882 .98 1.176 1.3906 1.47 1.66502 1.86004 1.95804 2.15404 2.35004 2.44608 2.64306 2.83808];
err = .025*ones(size(y));
errorbar(x,y,err,'.')
p = polyfit(x,y,1);
%Find the slope in th graph
slope = (15*sum(x.*y)-sum(x).*sum(y))./(15*sum(x).^2.-(sum(x)).^2);
hold on
% Q = slope.*x;
Q = polyval(p,x) ; 
plot (x,Q,'r-')
legend('Data Points','Best Fit')
title ('Stretch')
xlabel ('Length (m)')
ylabel('Force (N)')
hold off