function [xo,fo] = Opt_Quadratic(f,x,TolX,TolFun,MaxIter)
% 用二次插值求f(x)最優解
if nargin < 5
    MaxIter =100;
end
if nargin < 4
    TolFun= 1e-8;
end
if nargin < 3
    TolX = 1e-5;
end
%%%根據輸入確定三點的初始值，并求出對應的函數值
if length(x) > 2 
    x012 = x(1:3);
else
    if length(x) == 2
        a = x(1);
        b = x(2);
    else
        a = x-10; b = x+10;
    end
    x012 = [a (a + b)/2 b];
end
f012 = f(x012);
%%%%初始化三點值
x0 = x012(1);
x1 = x012(2);
x2 = x012(3);
f0 = f012(1);
f1 = f012(2);
f2 = f012(3);
%%%%%求出插值二項式的取最小值處的x3以及相應的最小值f3
nd = [f0-f2 f1-f0 f2-f1]*[x1*x1 x2*x2 x0*x0; x1 x2 x0]';
x3 = nd(1)/2/nd(2);
f3 = feval(f,x3); 
%%%%判斷是否停止迭代
if MaxIter<= 0 | abs(x3 - x1) < TolX | abs(f3 - f1) < TolFun
    xo = x3;
    fo = f3;
else
%%%%%根據求出的多項式值與區間中三點值的大小比較確定相應的新的三點
    if x3 < x1
        if f3 < f1
            x012 = [x0 x3 x1];
            f012 = [f0 f3 f1];
        else
            x012 = [x3 x1 x2];
            f012 = [f3 f1 f2];
        end
    else
        if f3 <= f1
            x012 = [x1 x3 x2];
            f012 = [f1 f3 f2];
        else
            x012 = [x0 x1 x3];
            f012 = [f0 f1 f3];
        end
    end
%%%%%進行新一輪迭代
    [xo,fo] = Opt_Quadratic(f,x012,TolX,TolFun,MaxIter-1);
end