interpolation methods to calculate the value of y

use linear, spline and cubic interpolation methods to calculate the value of y. Plot data points and graphs of three interpolation methods.Estimate the value of y for x = 7.5 using three interpolation methods. displayyour results along with text which technique was used to produce each value.

x = [10 20 30 40 50];
y = [0.9530 0.7221 0.4510 0.6132 0.5518];

this is what I got, but I do not know if it any correct

x = [10 20 30 40 50];
y = [0.9530 0.7221 0.4510 0.6132 0.5518];
plot(x,y,'ob')
xi=1:0.1:10;
method=char('linear','cubic','spline');
col={'--r',':g','-k'};
for k=1:3
    yi{k}=interp1(x,y,xi,method(k,:));
    hold on;
    plot(xi,yi{k},col{k})
end
legend('linear','cubic','spline', 'Location', 'best');
title(sprintf('x = 7.5 y is %.2f ',polyval(fit,7.5)))
y1=yi{1};
y2=yi{2};
y3=yi{3};

[darova]

You forgot data

legend('data','linear','cubic','spline', 'Location', 'best');

To obtain y at x = 7.5:

xp = 7.5;
yp = interp1(x,y,xp,method);
plot(xp,yp,'or')
text(xp,yp,method)

Answer 1

[KSSV]

x = [10 20 30 40 50];
y = [0.9530 0.7221 0.4510 0.6132 0.5518];
plot(x,y,'ob')
xi=min(x):0.1:max(x);
method=char('linear','cubic','spline');
col={'--r',':g','-k'};
yi = zeros(3,length(xi)) ; 
for k=1:3
    yi(k,:)=interp1(x,y,xi,method(k,:));       
end
plot(xi,yi) ; 
legend('linear','cubic','spline', 'Location', 'best');
% title(sprintf('x = 7.5 y is %.2f ',polyval(fit,7.5)))