Recent content by life92

f(x) = 1/6 (x^2 - 4x + 24) = 1/6 ((x-2)^2+20) which means the vertex is (2, 10/3) and the axis of symmetry is x=2 Consider some z that exists in the restricted domain of f(x) and then N which doesn't belong to that domain, where f(z)=f(N). Now, by symmetry, (z+N)/2 = 2, since the axis of...

Miss Wong IS THAT YOU? I don't know if you remember me or not but you took Miss Ogara's class at Baulko back in '07 from memory. Hope everything's going good ! But anyway guys, Miss Wong a.k.a Joan is a REALLY good Jap teacher. She was heaps mad back when she was my teacher and is probably even...

OMG All this talk about DDR lol. Dyou know when they open it up? I've been there a couple of times, but it's always closed :( And so I thought that they only opened it up for like, when other people use the uni, like Animania or something... Otherwise I'd probably end up spending my breaks...

OH SHIT i totally did not see you message me O_O add me on facebook so i know what you look like XD (and viceversa...) my email is just,, life92@gmail.com :) weee japanesee

Hey, thanks for the add but.. Do I know you? x__x Sorry if I do but have forgotten, I have REALLY bad memory~~

Oh my bad. I made a very silly mistake lol. x^2 = 4ay y = x^2 / 4a y' = x/2a At some point (x1,y1) y' = x1/2a Therefore, the gradient of the normal = -2a / x1 y-y1=-2a / x1 (x-x1) Rearranging: 2ax + x1 y = x1 y1 + 2a x1 Subbing (0,a) 0 + x1 a = x1 y1 + 2a x1 0 = x1 y1 + x1 a 0 = x1 (y1 + a)...

lol wth dude. I'm just trying to help out........... But in answer to super.muppy's question about not using parameters, I tried it again and the answer doesn't seem to come out which is pretty weird. I think you could possibly think of this problem as a non-algebraic problem and try to think...

x^2=4ay Focus (0,a) y = x^2 / 4a y' = x / 2a At some point (2ap,ap^2) y' = 2ap / 2a = p Therefore, gradient of normal = -1/p Eqn of normal: y -ap^2=-1/p(x-2ap) py-ap^3=-x+2ap x+py=ap^3+2ap Through (0,a) (focus) ap = ap^3 + 2ap 0 = ap^3 + ap 0 = ap (p^2+1) Therefore, p = 0 only That means...

Sometimes even if the second derivative is 0 at a certain point it doesn't necessarily mean its an inflexion. Also Xcelz, that goes for if the first AND second derivative are 0, doesn't necessarily mean its an inflexion. For example, f(x) = x^4 f'(x) = 4x^3 and f''(x) = 12x^2 If you sub x=0...

y = ax^3 + bx^2 - x + 5 y' = 3ax^2 + 2bx - 1 y'' = 6ax + 2b Since it has an inflexion at (1,-2) then y'' = 0 at (1,-2) y'' = 6ax + 2b 0 = 6a(1) + 2b b = -3a Now sub this and (1,-2) into the original equation and you get... -2 = a(1)^3 - 3a(1)^2 - 1 + 5 -2 = -2a + 4 -2a = -6 a = 3 Therefore...

P(x)=4x^3+12x^2-15x+4 P'(x)=12x^2+24x-15 = 3 (4x^2+8x-5) = 3 (2x+5)(2x-1) P(1/2) = 0 , therefore the double root is 1/2, or the double factor is (2x-1) By inspection, the last factor must then be (x+4) Therefore the roots are 1/2, 1/2 and -4

This is actually obviously very close to the answer. Btw you are missing a pi XD But anyway, as you can see, you integrated from 1=>2, however, if you look at a diagram, there is still a space underneath y=1 in the first quadrant. This is the missing volume that you are leaving out behind...

LOL I think his jersey name was "Jeganaut" from memory... might want to correct me on that Jega :)

=o You might've saw me at the welcome O_O ,, but there were so many people there lol XD But nah, theres 3 others, but one of them didn't come to the welcome...

I think most courses only have Lectures in Week 1. BUT, I know that some courses only start in Week 2, so that means you'll only have lectures next week and then normal classes start from Week 3. Check your timetable for week 2 and 3 and compare.