Sort transformed array

https://discuss.leetcode.com/topic/128/sort-transformed-array
Given a sorted array, and integer values A, B and C, return a new array after applying the equation x' = Ax^2+Bx+C to each element in the array.
The returned array should be sorted.
Expected time complexity : O(n)

I draw a picture that describe a little bit math, then will propose an algorithm for this problem. I must excuse myself that I am terrible painter, so don't mock at me.
As you see from the picture we have three different cases of the square function depending on the value of A.
The extreme point of the parabola in case 1(minimum) and 2 (maximum) is x* = -B /2A and y = Ax*^+Bx*+C

1. A > 0
   You notice from the draft that when we have monotonous decreasing function for all x ,where x < x* and vice verse monotonous decreasing function for all x > x*, We can use this fact and split input array in two subarrays a and b :
   a = [x0,x1,x2,x3....xk.<=x*]
   b = [x* < xk+1, ....xn]
   We calculate f(a) = [f(x0,f(x1)....f(xk)] and f[b] = [f(xk+1,f(xk+1)....f(xn)] , both arrays are sorted but in different order, one ascending, the other descending. We can revert the order of the descendin array and to merge them.
2. A < 0, the same as case 1, see the picture
3. A = 0, function is monotonous increasing , linear function, no need to merge

Based on A being -ve or positive, we can go ahead in two ways.

1. If A is +ve, it's upright parabola. Keep two pointers at both ends. Get x` = Ax^2 + Bx + C for both values, then put the maximum of the values to the end of the new array. Decrement or increment the pointers based on which element was selected.
2. If A is -ve, it's inverted parabola. This time, whichever element is lesser, put it to the start of the array