curve intersection
- Thread starter mufc802
- Start date
It's not about finding the actual co-ordinates of the intersection for these curves, it's just knowing that they actually intersect or not. The textbook specifically says "You would not be expected to solve this equation in C1".
This is the answer that "explains" what to do:
This is the answer that "explains" what to do:
LOLok, my bad, didnt realise what the question was getting at. Ok, what they're asking for is for you to figure out how many points of intersection there are and what quadrants the intersections lie in.
So, they do something similar to what Morbius described and look at some points where one function is bigger than the other, and try to look for a point where it's less than. We know that x(x-4) is 0 at x = 0. Furthermore, -1/x < 0 for all x > 0. So can we find a point of x(x-4) such that x(x-4) < -1/x?
If so, then we must have 2 intersections in the 4th quadrant! The A-Level reason for this is that in order to get from the positive side of -1/x to the negative side and back again, we have to cut it twice! So what they do is say, well let's try x = 2. They find that in fact for x = 2, x(x-4) < -1/x, so there are 2 points of intersection in the 4th quadrant!
Then for the third one you can just sort of use the same principle. At x = -0.000000001 (note for the pedants that i didnt say 0), x(x-4) < -1/x (check!). So just find a point where it's greater than -1/x and happy days are had (e.g. x = -12 or something). So we have 3 intersections, with our last one in the 2nd quadrant
So they used 2 because it's the mid-section of 0 and 4 so that would be the lowest point on the curve? If that meant that x(x-4) was less than -1/x it would mean it did intersect because it went lower on the y axis? If x(x-4) was greater than -1/x when x was 2 it would mean that it didn't go down enough on the y-axis to cut through -1/x?
What if they just touched rather than cut so would it be possible for just one intersection?
