Points of Inflection for Sine Curve

In the code below, I have successfully detected the peaks of the sine curve. I have also put the peaks on the plot. May someone PLEASE help me now identify the points of inflection. It is where the graphs goes from concave up to concave down (NOT the peaks). The points of inflection should be points across the Sine curve that go across the middle of the graph. The code is below. If you could also mark the points on the graph that would be much appreciated. Thanks!

I found a solution to this problem, it stated that

***************************************************************************** "If your curve are data (not calculated by a function) I would use the gradient function to calculate the derivative. To find the zero-crossing of the derivative, you can either use a threshold test, or if appropriate for your derivative, the interp1 function." *************************************************************************

I just dont know how to code the above..but here is my code again below. Thanks again guys: Basically,inflection points in the signal is to be determined automatically by detecting zero-crossings in derivatives of the signal. I need help finding the zero-crossings of the signal. Thank you!!

x  = 1:500;
X  = x;
J  = 1; 
Fs = 499;
N  = J*Fs;
t = 0: 1/Fs : J;
Fn = 3;             % this control the number of cycles/periods
deltaJ = 0.0020;
deltax = 1;
y_sine_25HZ = sin(Fn*2*pi*t);
y = y_sine_25HZ ;
plot(x,y, 'k.')
% drvY = diff(y)/deltax;       % one unit shorter than y 
% drvY = [drvY , drvY(end)];   % making up for the missing point 
% secondDrvY = diff(drvY)/deltax;
% secondDrvY = [secondDrvY, secondDrvY(end)];
%  [~,locup]=findpeaks(y)  % (+)ive peaks                         % location up
%  [~,locdn]=findpeaks(-y) % (-)ive peaks (positive negative y)   % location down 
dy=[0 sign(diff(y))];
locdn = find((diff(dy))==2)          % location down
locup = find((diff(dy))==-2)         % location up

plot(t,y)
ylim([-1.05 1.05])
hold on
scatter(t(locup)',y(locup)',30,'r','*')
scatter(t(locdn)',y(locdn)',30,'g','d','filled')

[dpb]

You've kept all the previous code now unused plus the duplication of methods I showed you previously between using findpeaks and the search for sign changes.

Clearly you've not yet bothered to understand what any of the above actually does...if you'll follow up the suggestions I made earlier on delving into the "why" of the search for values==+/-2 it should become clear on a very much related technique for zero crossings.

[Osita Onyejekwe]

I completely understand your technique, and I understand the mathematics in solving for the zero crossing, I just have a problem putting it into code. Zero crossing is when the second derivative crosses the zero -line from positive to negative and the first derivative has a local maxima. I'm just a week new to Matlab...so its all new to me ..i just need a little help on this one and I'm good to go

[dpb]

 find(abs([0 diff(sign(y)])==2)