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yikes this is the answer i got.. so how would i find the turning points and nature of this y'' = (x-4)^1/2.1/4 (x-3)(4-x)^-1/5 - 1/2 (x-3)(4-x)^-1/2 . 1/2(x-4)^-1/2

My name is Homer Jr , you can call me Hoju i do 2u Maths , IPT , Software Design , Business studies and Advanced English i suck , i wish i had done more units ..... but oh well ill have to kill these subjects then:chainsaw:

yea i know how to use it .. its just for those questions... i wasnt getting a right answer on point (0,1) for that first question and the 2nd one was how on earth do i differentiate this (1st and 2nd) y=(x-3)(4-x)^1/2

hmm i sort of understand that , but i really want to know how to do it via differentation method thx anyway :)

ok i have a few questions , i dont seem to be getting these right my first question is Find the turning points on the curve y=(4x^2 - 1)^4 okay so we use the chain rule yes? first derivatve y' = 4(4x^2 - 1)3 x 8x so y' = 32x(4x^2 - 1) put that to 0 to find points 32x(4x^2 - 1)^3...

argh... anyone doin exactly 10 units get over 90 ?

far out , memorizing definitions from my book is actually quite easy , but im having problems memorizing these case studies!!!

hmm so when are we actually gonna be told about our Major assignment? or do we just have to automatically start it whenever? cause my teacher has told me NOTHING about it yet oh well only one way to find out

5th that (err probably C)

hmm thats odd it says the Maximum value is 31 when x = - 4 and how did u draw up that curve if u cant factorize it oh while im here might as well ask this to say with an equation x^2 = 8y Why is the axis of symmetry x = 0? why is it not y= 0 and how do i find axis of symmetry for...

err well what i do i basically make myself memorize notes liek a parrot and i say them to myself over and over again , i do this on the weekend as well only i go for basically the whole day lmao and i spent proportions of my time working on assessments but i do this on weekdays as well...

Okay i have 2 more questions if anyones there consider 15-8x-x^2 in form a - (x+b)^2 Once u put that in that form its 31 - (x+4)^2 but then it says find the maximum value of 15-8x-x^2 and the value for x for which it occurs.. i dont understand this.. also the following equation 8y...

heh .. sue the school for not being ergonomically friendly?

uhh well actually see i try to revise my work for ALL of my subjects daily ive been trying this , sure it worked at the start but as i get more homework it gets harder to actually do homework/assessments for the required subjects and then study my notes for all the subjects so im...

me too

yep me too

True but cant an imaginative journey also be percieved as putting yourself in the position of a person enduring an experience??? and imagining what he/she went through?

actually i did use the distance formula For that Particular example i gave above Find the equation of the locus of a point that moves so its equidistant from the point (0,-3) and the line y = 3 i ended up gettin just 12y answer says x^2 + 12y im not sure what i did wrong after...

Hey ppl , with locus is there an easier way to do the quadratic type problems e.g Find the equation of the locus of a point that moves so its equidistant from the point (0,-3) and the line y = 3 I know you can use the distance formula , but someone told me you can use a parabola...

yea alrite i see , but what about old topics??? i mean daily you cant revise every topic you've done in each course.. once we do a topic... does that mean to forget about it until the exams???