Need assistance using bisection method to find equation roots in Matlab function.

I'm currently taking a course using Matlab, and I am having trouble finding the error in this code I wrote. The goal is to find the zero of a quadratic equation using the bimethod approach. However, my code outputs the wrong answer everytime, and I do not see my mistake. Can anyone help me?

% input for function that we're finding the roots of
f1 = @(x) ((-0.6)*x^2)+(2.4*x)+(5.5);

xl = 5; % xl sets the inital low value for the code
xu = 10; % xu sets the initial high value for the code
i = 0; % i is the number of iterations that are to be performed

while i <= 15 % this sets the number of iterations the loop is allowed to compute
    i = i + 1;
    
% This if function is testing if the product of fxl and fxm is positive or
% negative. The sign of this value determines which bound is to be replaced
% with the midpoint, cutting the sample size in half. 

    if f1(xl) * f1(xu) > 0 
        xu = xm; % xu replaces the value of xm
        xm = (xl + xu)/2; % xm is the "midpoint" of the initial low and high values
    else
        xl = xm; % xl replaces the value of xm
        xm = (xl + xu)/2; % xm is the "midpoint" of the initial low and high values
    end
end

root = f1(xu);

error_a = (abs(f1(xu)) - f1(xl))/(f1(xu)); % approximate error function 
error_t = (abs(f1(xm)) - 5.62859)/5.62859; % true error function 

xl_inital = 5; 
xu_inital = 10; 

fprintf('The zero of the input function between %d and %d is %f.\n',xl_inital,xu_inital,root);
fprintf('The approximate error is %f and the true error is %f.\n',error_a,error_t);

This is the output:

The zero of the input function between 5 and 10 is -10.255035.

The approximate error is -1.999951 and the true error is 0.821955.

Answer 1

[Alan Stevens]

Try this

% input for function that we're finding the roots of
f1 = @(x) ((-0.6)*x^2)+(2.4*x)+(5.5);

xl = 5; % xl sets the inital low value for the code
xu = 10; % xu sets the initial high value for the code
i = 0; % i is the number of iterations that are to be performed

while i <= 15 % this sets the number of iterations the loop is allowed to compute
    i = i + 1;
    xm = (xl + xu)/2;
% This if function is testing if the product of fxl and fxm is positive or
% negative. The sign of this value determines which bound is to be replaced
% with the midpoint, cutting the sample size in half. 

    if f1(xm) * f1(xu) > 0 
        xu = xm; % xu replaces the value of xm
    else
        xl = xm; % xl replaces the value of xm
    end
end

root = f1(xu);

% you are confusing function values with x values below!!
% error_a = abs((f1(xu) - f1(xl))/f1(xu)); % approximate error function 
% error_t = abs((f1(xm) - 5.62859)/5.62859); % true error function 

error_a = abs((xu - xl)/xm);
error_t = abs((xm - 5.62859)/5.62859);

xl_inital = 5; 
xu_inital = 10;

fprintf('The zero of the input function between %d and %d is at x = %f.\n',xl_inital,xu_inital,xm);
fprintf('The approximate error is %f and the true error is %f.\n',error_a,error_t);

[Jaden Evans]

Thank you so much for helping, I see what I was doing wrong. Now I can sleep in peace lol