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I am trying to understand concave functions, for which wikipedia gives an expression written as: enter image description here

now I am confused as how to understand this as the line over there says that this definition merely states that for every z between x and y, the point (z, f(z) ) on the graph of f is above the straight line joining the points (x, f(x) ) and (y, f(y) ). Now what I know is that the line joining two point should be $f'(u)= \frac{f(y)-f(x)}{y-x}u+c $ where f'(.) I have taken as straight line co ordinate. Now how does this equation fits the definition that f(z) should be above than f'(u)?

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    Do you know that, given an interval $[a,b]$, the point $p$ on $[a,b]$ which divides $[a,b]$ into segments $[a,p],[p,b]$ with length ratio $\alpha:(1-\alpha)$ (so $p-a:b-p=\alpha:(1-\alpha)$) is given by $\alpha b + (1-\alpha)a$? – 2017-01-26
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    Notice that when you let $\alpha$ range between 0 and 1, then $p_{\alpha}=(1-\alpha)x+\alpha y$ is a point in the segment from $x$ to $y$, $p_0=x$ $p_1=y$. So what you are saying here is that when the argument $p_{\alpha}$ is in between $x$ and $y$ then the function is greater or equal than $f_{\alpha}=(1-\alpha) f(x)+ \alpha f(y)$ which is precisely the point with $p_{\alpha}$ $x$-coordinate in the segment from $f(x)$ to $f(y)$. Does that make more sense? – 2017-01-26
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    Thanks! I got it just one more doubt. With my calculation I got$f'(u)= \frac{f(y)-f(x)}{y-x}u+c $ which gives me $f'(u)=(1-\alpha)f(x)+\alpha f(y)+c$ why we haven't taken that intercept 'c' into consideration. Will it have no effect – 2017-01-26

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