Got Problem?....lets Discuss Here (1 Viewer)

[victorling]

hello, everone
i have posted a same thread in maths2u forum to answer people's questions

feel free!

by victorling of'03

now tutoring maths2-4u

 

[:: ck ::]

man u're really desperate to get students aren't u.................

 

[xiao1985]

well, he's takin up freakin_out's advice and start helpin pplz before tryin to get any more students i fink... ^^

 

[victorling]

Originally posted by :: ryan.cck ::
man u're really desperate to get students aren't u.................

lol

 

[Grey Council]

just hang around, when people have a problem they post it in the forums.

The problem is, you don't log on too often. Usually someone else helps.

And don't take that wrongly. I'm just saying.

 

[J0n]

You should post this thread in the 4U forum, drbuchanan will give you the Riemann Hypothosis - if you solve that, i would love to have you as a tutor

 

[lil tiger]

http://www.boredofstudies.org/community/showthread.php?s=&threadid=24470

 

[overwhelming]

u can tutor me for $-5

cost of transport = ~$5

 

[ToO LaZy ^*]

....i need help wif curve sketching..=(
can anyone plz help..??

Q:
x^2 / x^2 + 4

 

[victorling]

Originally posted by ToO LaZy ^*
....i need help wif curve sketching..=(
can anyone plz help..??

Q:
x^2 / x^2 + 4

solution: since the power of the denominator and the numerator are the same: both ^2, we need to simplify the question first

= x^2/ x^2 +4

= (x^2+4-4)/x^2+4

=(x^2+4)/(x^2+4) - (4/(x^2+4))

=1- (4/(x^2+4))

it is a graph with no vertical asymptote, one horizontal asymptote: y=1, turning point0,1)
As x aprroaches positive maximum, y approaches 1 but at a value smaller than 1
As x approaches negative maximum, y approaches 1 but at a value smaller than 1

 

[ToO LaZy ^*]

thnx buddy =)
helped a lot =)

 

[ToO LaZy ^*]

hey its me again...
still on the same topic : curve sketching

Q:
x / x^2+1

showing any TPs, inflexions or asymp.

thx buddy =)

 

[Grey Council]

passes through 0,0, inflexion there, i think
turning point:
min at -1, -0.5
max at 1, 0.5
asymptote: y=0
note, horizontal asymptotes occur at infinity, thats why curve passes through 0,0

 

[victorling]

Originally posted by GuardiaN
passes through 0,0, inflexion there, i think
turning point:
min at -1, -0.5
max at 1, 0.5
asymptote: y=0
note, horizontal asymptotes occur at infinity, thats why curve passes through 0,0

yep..sorry
i am going to post the graph tomorrow
too busy tonight

 

[ToO LaZy ^*]

ok...thx for da help =)

 

[Grey Council]

np

 

[victorling]

give me some qs , people

 

[victorling]

*questions

 

[ToO LaZy ^*]

ok here's one..
1.
i) by solving the equation z^3+1=0, find the three cube roots of -1

ii) let @ be a cube root of -1, where @ is not real. show that
@^2 = @-1

iii) hence simplify (1-@)^6 <----dunno understand this part


btw..this a 4u q

 

[victorling]

so only part3?