Find Y of the equation

Hey everyone,

I have a problem finding the image (y) of the equations for an specific x. I build the following "wave" that it is build with f parts of 6m per each. The first part it has two equaitons while the rest have three different funcitons per each. Every function follows teh expression a*x^2+b*x+c and I have a vector of nx3 were every column corresponds to the terms of the parabolic expression. I have performed the code for the first and last sections, but the ones in the middle I have trouble. The distribution is as follows:

-The first curve of the section, starts at the previous section, prolongates until x=L(previous section)+0.1*L(current section)

-Second curve of the current section reaches the point 0.9*L(current section)/2

-Third curve of the current section starts at 0.9*L(current section)/2 and ends at 0.9*L(current section)

The main objective is to determine the y for every x. We will enter an x, lets say for every 2m, and we have to determine the y of the equation for every 2m.

The code to determine the y I have started is the following:

y_plot=[0.246913580246914	-1.33333333333333	2.10000000000000;
-1.66666666666667	20	-57.3000000000000;
0.408163265306122	-7.10204081632653	31.1938775510204;
0.246913580246914	-4.29629629629630	18.9888888888889;
-1.66666666666667	40	-237.300000000000;
0.408163265306122	-12	88.5000000000000;
0.246913580246914	-7.25925925925926	53.6555555555556;
-1.66666666666667	60	-537.300000000000;
0.408163265306122	-16.8979591836735	175.193877551020;
0.246913580246914	-10.2222222222222	106.100000000000;
-1.66666666666667	80	-957.300000000000;
0.408163265306122	-21.7959183673469	291.275510204082;
0.246913580246914	-13.1851851851852	176.322222222222;
-1.66666666666667	100	-1497.30000000000] %coeficients of a (first column), b(second column) and c(third column)
x_div=2 %find y every 2 meters
L=[6,6,6,6,6] %beam formed by 5 sections of 6meters each
position=L(:,1);+0.1*L(:,curve)
contador_vector=1
points_x=[]
points_y=[]
for curve=2:1:(n-1) %for everysection except the first one and the last one
    
    for position=position:x_div:position+(0.9*L(:,curve)/2) % first equation of the current section
        if position<sum(L)
            eq=y_plot(curve*2+contador_vector-2,1)*x^2+y_plot(curve*2+contador_vector-2,2)*x+y_plot(curve*2+contador_vector-2,3);
            points_x=[points_x,position];
            points_y=double([points_y,subs(eq,x,position)]);
            
        else
        end
    end
    for position=position:x_div:position+0.9*L(:,curve)
        if position<sum(L)
            eq=y_plot(curve*2+contador_vector-1,1)*x^2+y_plot(curve*2+contador_vector-1,2)*x+y_plot(curve*2+contador_vector-1,3);
            points_x=[points_x,position];
            points_y=double([points_y,subs(eq,x,position)]);
            
        else
        end
    end
    
    for position=position:x_div:(position+L(:,curve)+0.1*L(:,curve)-0.1*(L(:,curve)-L(:,curve+1)))
        if position<sum(L)
            eq=y_plot(curve*2+contador_vector,1)*x^2+y_plot(curve*2+contador_vector,2)*x+y_plot(curve*2+contador_vector,3);
            points_x=[points_x,position];
            points_y=double([points_y,subs(eq,x,position)]);
            
        else
        end
    end
    
    position=position+x_div;
    contador_vector=contador_vector+1;
end

%
points_x'
points_y'

[Marc Martinez Maestre]

So, the loop showld find x points between 6 and 24, but it reaches 74. The main problem is how to tell matlab that:

For every one of the sections I want that the first curve that it goes between x and x2 find the y for the current position. While this position increases by 2. And if the position is bigger than x2, check in the next function. And all this funcitons are inside a vector taht contains the coefficients of the paraolic expresssion.

[Image Analyst]

Not sure what "the image (y) of the equations" is. Can you post that image?