1) Ponder your parabola on a coordiante axis. You'll have to decide on an orientation and a direction.
2) It seems prudent to decide (there is nothign magic about it) that the car will be moving generally from left to right as we consider this one seciton of track.
3) Having said that, on the left, the slope is -4/3 -- It is coming down. On the right, the slopr is 2/3 -- It is going back up. It's pretty clear, then that we have a cup-shape, of opening-up parabola.
4) Now you must decide one more thing. Where does the parabolic segment start? It's doesn't much matter, but you can make matters a little wors or a little better by choosing wisely. Since we get 50 horizontal feet, it may be wise to put the vertex on the origin. Maybe not. Frankly, it seemed easier to me at first to start the parabolic section on the y axis.
Given these definitions
y(x) = ax^2 + bx + c
y'(x) = 2ax + b
"start at a height of 30 feet"
y(0) = 30
incoming track has a slope of -4/3
y'(0) = -4/3
the end of the curve is to have a slope of 2/3
y'(50) = 2/3
That should be enough.
Let me return to your opening statement. As soon as you believed that you could not do it, you lost track of your thinking process. Everything broke down ONLY when you gave up. Here's a plan. Don't do that.
Note: Really, it doesn't much matter what you assume. If we had decided to make it symmetrical about the y-axis, it would have looked like this...
y(-25) = 30
y'(-25) = -4/3
y'(25) = 2/3
It WILL produce a different solution, but the structure will be the same. Just make rational assumptions and WRITE THEM DOWN.