How can I find local maxima of this quadratic function?
$$f(x) = \sum _{i=1}^n -\frac{(z_i - x)_+^2}{2} - \left\{((\frac{(z_i - x)_+^2}{2})-(\frac{(y_i - x)_+^2}{2}) ) * c_i\right\} $$
which + means If $z_i$ is bigger than u its equal $z_i - x$ and else it's equal zero.
Also $z_i$, $y_i$ and $c_i$ is a vector of values with same lengths.$z_i$ and $y_i$ is an interval.
$$ z = (0.04124633, 0.05404163, 0.30939022, 0.57690859, 0.70211949, 0.73597262, 0.92130814,1.05226194, 1.68227033, 1.91068738)$$
$$y = (0.2112338, 0.4172289, 0.3991616, 0.8755037, 1.2448935, 1.3493179, 1.6522002, 2.4177490, 4.7770777, 2.3840472)$$
$$ c = (5.337958, 2.275704, 10.64689, 2.873863, 1.293402, 1.03867, 0.6372883, 0.1032198, 0.000045619, 0.5670724)$$
I plot this quadratic function in terms of x which can be change between max finite value of $z_i$ and $y_i$ and see that theres is one local maxima but I do not know how can I find it matematically?
Since the function is everywhere differentiable, it can have local extrema only at the points where $f'=0$. The formula for the derivative is $$\begin{split} f'(x) &= \sum _{i=1}^n (z_i-x)_+ + c_i \{(z_i - x)_+- (y_i - x)_+ \} \\ &= \sum_{i=1}^n (1+c_i)(z_i-x)_+ - \sum_{i=1}^n c_i (y_i - x)_+ \end{split} \tag 1 $$ if I parsed your formula correctly (it appears to have a few unnecessary parentheses). This function is piecewise linear; its graph has corners at the points $z_i$ and $y_i$.
I don't think there is a very clever way to deal with such piecewise linear functions. I would list the numbers $z_i,y_i$ in increasing order, plug each into $f'$, and noted the signs. The maximum will be found within an interval where $f'$ is positive at the left endpoint and is negative at the right endpoint. The precise location is found by solving linear equation (since $f'$ is linear within the interval of interest).
For example, on the interval $(1.68227033, 1.91068738 )$ we have $$\sum_{i=1}^n (1+c_i)(z_i-x)_+ = (1+0.5670724) (1.91068738-x)$$ and $$\sum_{i=1}^n c_i (y_i - x)_+ = 0.1032198 (2.4177490-x)+0.000045619 (4.7770777-x) + 0.5670724(2.3840472-x)$$ So, you can simplify $f'(x)=\sum_{i=1}^n (1+c_i)(z_i-x)_+ - \sum_{i=1}^n c_i (y_i - x)_+$ by expanding the parentheses, and check if $f'$ turns to zero or not.