How to interpolate a set of data with the cubic method?

I tried looking it up, and it comes out that i should use the spline command, but when i use it nothing comes out. I need to plot the data and the interpolant. I also need to find x for when y= 600

x = [74 29 21 12 8 5.7 4.4 3.6 2.1 1.8 1.5 1.0 0.7];
y = [80 131 189 270 320 407 450 530 620 686 740 900 1095];
s = spline(x, y, 600);

Answer 1

[Hiro Yoshino]

I would use Symbolic Math Toolbox to solve the problem analytically:

x = [74 29 21 12 8 5.7 4.4 3.6 2.1 1.8 1.5 1.0 0.7];
y = [80 131 189 270 320 407 450 530 620 686 740 900 1095];
s = spline(x, y) % create pp object

s = struct with fields:
      form: 'pp'
    breaks: [0.7000 1 1.5000 1.8000 2.1000 3.6000 4.4000 5.7000 8 12 21 29 74]
     coefs: [12×4 double]
    pieces: 12
     order: 4
       dim: 1

xq = min(x):max(x);
s1 = spline(x,y,xq); % just for plot
plot(xq,s1)
hold on;
yline(600); % serch solution roughly
ax = gca;
ax.XTick = s.breaks;
xline(s.breaks,':');
hold off;
xlim([0.82 7.35])
ylim([478 718])

The solution seems to lie in the 5th section.

The coefficients of the function corresponding to the 5th sections are

s.breaks

ans = 1×13

0.7000 1.0000 1.5000 1.8000 2.1000 3.6000 4.4000 5.7000 8.0000 12.0000 21.0000 29.0000 74.0000

s.coefs(5,:)

ans = 1×4

-64.8873 190.9875 -200.4848 620.0000

Extract coefficients

x1 = s.breaks(5)

x1 = 2.1000

x2 = s.breaks(6)

x2 = 3.6000

a = s.coefs(5,1)

a = -64.8873

b = s.coefs(5,2)

b = 190.9875

c = s.coefs(5,3)

c = -200.4848

d = s.coefs(5,4)

d = 620

Use Symbolic math to analytically solve the problem

syms x
f(x) = a*(x-x1)^3+b*(x-x1)^2 + c*(x-x1) + d

f(x) = 

sol = vpasolve(f==600,x,x1) % solutions at y = 600

sol = 

x = 2.211 or something.