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Curve Sketching (1 Viewer)
- Thread starter angeleyez
- Start date
Nelly
Member
Aiigh, this is EXTREMELY hard to describe in words. but here goes:
BTW, this is a 4U topic, so if u just do 3U, you dont have to worry.
f'(x) = whatever
graph f(x).
Now, each point on the graph f'(x) is the gradient of f(x).
So when f'(x) = 0, then f(x) is having a turning point.
If f'(x) is having a turning point, then f(x) has a point of inflexion.
You don't have to worry about where it crosses the y-axis or whatever, because you dont know f(x), because if you integrate f'(x) you get a + C, so you never know how high or low it might be.
I hope that helped, and good luck in 2 weeks
BTW, this is a 4U topic, so if u just do 3U, you dont have to worry.
f'(x) = whatever
graph f(x).
Now, each point on the graph f'(x) is the gradient of f(x).
So when f'(x) = 0, then f(x) is having a turning point.
If f'(x) is having a turning point, then f(x) has a point of inflexion.
You don't have to worry about where it crosses the y-axis or whatever, because you dont know f(x), because if you integrate f'(x) you get a + C, so you never know how high or low it might be.
I hope that helped, and good luck in 2 weeks
Jellymonsta
Member
i dont know if it is in the syllabus but i was given a 4 mk simple version of that in my 3u trials.
the teacher said it was a tough exam, so i don't know if it would actually show up in the hsc
the teacher said it was a tough exam, so i don't know if it would actually show up in the hsc
mastermind
Member
- Joined
- Oct 15, 2002
- Messages
- 37
angeleyez -
i dont know about actually being in the syllabus, BUT this type of question was asked in 2000 2U maths paper. So even if you don't do 4U, be prepared for possibly another one this year...
i dont know about actually being in the syllabus, BUT this type of question was asked in 2000 2U maths paper. So even if you don't do 4U, be prepared for possibly another one this year...
Last edited:
I did a question like that in the trials for 2U. The question was later removed because there were 2 answers and I never found out whether I was right. BUT what I did was find the integral of the gradient function and sub in a known point to find the constant.
I'll post later on cos I have math tutoring in an hour.
I'll post later on cos I have math tutoring in an hour.
underthesun
N1NJ4
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- 2010
actually, you should take a look at the 4u graphs. Just ask someone for it, and it'll help you a lot. Mind you tho, not to go to the other topics other than graph 
Dumbarse
Member
i did one of these questions it was bullcrap, and hwat the hell is a cusp, how do u know when the graph has a cusp and how do u draw the cusp??
BlackJack
Vertigo!
Think a cusp is some point formed by two curves touching ... the middle bit of this } would probably be one...
Dumbarse
Member
thanx