Personally, I always do it using x = -b/2a and then average the x-intercepts and check, 9 times out of 10, if they don't match, I've managed to catch an error whilst solving for the x-intercepts, usually bad factorising. x = -b/2a is really safe because it's easy to sub it, and yep, then just sub back in for y
- completing the square is also something which I do.
It's hard to advise on what's the best method, because when you have enough experience, you intuitively know what to do because of recognition, like "oh, i'll complete the square here cause I can already see it" - kind of thing.
In regards to 1 - 5,
1. All are fine, I usually write - "ran f = ..." - I find that interval notation is easiest to use, but again, if the question uses set notation, I'd use set notation
3. If it's negative shouldn't a = -1? But yeah, that's how I would go about it too.
4. It doesn't really matter, I just put a comma: "x = 1, x = 2"
5. Drawing a graph would be best, but again, experience will give you intuition
This is how I would approach 5.
y = (x-4)^2 for x E [0,6]
This is a positive parabola with a turning point at x = 4 i.e. (4, 0). I know that 0 is smaller than 4 and 6 is greater than 4, so that means that the minimum y value occurs at x = 4, so it is y = 0
For the maximum, I know that the ends are x = 0 and x = 6, now with the turning point at x = 4, x = 0 is "further away" from the turning point, meaning it will be "higher" so maximum y will occur at x = 0, so y = 16. Since x = 0 is inclusive, this will be a square bracket.
The range is y E [0, 16]