How can I draw the slope of the given values ​​on the chart?

x=[0.5, 1.0, 1.10, 1.20, 1.30, 1.40, 1.50, 2.0, 2.5, 3.0, 3.5, 4.0, 4.10, 4.20, 4.30, 4.40, 4.5, 5 ];

y=[4.91, 4.91, 4.9190, 4.9187, 4.9165, 4.9099, 4.89, 4.58, 3.86, 2.85, 1.65, 1.06, 1.0153, 0.9751, 0.9422, 0.9072, 0.87, 0.77];

plot (x,y);

hold on;

slope=(diff(y)./diff(x));

for the given values how can ı draw and calculate slope on the graph.

Answer 1

[Star Strider]

I would use the gradient function rather than diff for this, since gradient produces a numerical derivative the same size as the argument:

slope = (gradient(y)./gradient(x));

plot (x,y);
hold on
plot(x, slope)
hold off

Experiment to get the result you want.

[Mert Polat]

My chart must be the same as in this picture.

[Star Strider]

If you simply want the region where both slopes are -1, that is relatively straightforward:

x=[0.5, 1.0, 1.10, 1.20, 1.30, 1.40, 1.50, 2.0, 2.5, 3.0, 3.5, 4.0, 4.10, 4.20, 4.30, 4.40, 4.5, 5 ];
y=[4.91, 4.91, 4.9190, 4.9187, 4.9165, 4.9099, 4.89, 4.58, 3.86, 2.85, 1.65, 1.06, 1.0153, 0.9751, 0.9422, 0.9072, 0.87, 0.77];

zci = @(v) find(v(:).*circshift(v(:), [-1 0]) <= 0);                    % Returns Approximate Zero-Crossing Indices Of Argument Vector

slope = (gradient(y)./gradient(x));
neg1 = zci(slope+1);
for k1 = 1:numel(neg1)
    idx = neg1(k1) + [-1 1];
    xneg1(k1) = interp1(slope(idx), x(idx), -1);
    yatx(k1) = interp1(x, y, xneg1(k1));
end

plot (x,y);
hold on
for k1 = 1:numel(xneg1)
    xv = xneg1(k1)+[-1 1]*0.5;
    yv = -xv + (xneg1(k1)+yatx(k1));
    plot(xv, yv, 'LineWidth',1.5)
    plot([0 xneg1(k1)], [1 1]*yatx(k1), '--k')
    plot([1 1 ]*xneg1(k1), [0 yatx(k1)], '--k')
end
hold off
text(0, yatx(2), 'V_{OL(max)}', 'HorizontalAlignment','right', 'VerticalAlignment','middle')
grid

I leave the other annotations and the reference lines to you. This should get you started on them.

Experiment to get the result you want.

[Mert Polat]

Thank you soo much :)

[Star Strider]

My pleasure!

If my Answer helped you solve your problem, please Accept it!