Post by: samiira on March 23, 2010, 06:26:35 pm
Post by: Stroodle on March 23, 2010, 06:38:08 pm
Maybe cause I'm using a mac...
Post by: samiira on March 23, 2010, 06:39:21 pm
Post by: moekamo on March 23, 2010, 07:45:25 pm
so 3 complex roots, so c. probably didnt need to find the roots, just simplify to get z^3=-4i and you should know that this gives 3 complex roots.
2. by inspection:
so A
3.
this is just a straight line equation:
so has no stationary points or asymptotes, hence C.
4.
put into calc and see imaginary component of the reciprocal is 1/3, so must be A
5.
6.
This is perpendicular bisector of origin and
hope this is all right :S
Post by: samiira on March 23, 2010, 07:46:30 pm
Post by: samiira on March 23, 2010, 07:48:43 pm
Post by: samiira on March 23, 2010, 08:20:32 pm
i just noticed some problems..
you wrote their z^3 + 2^2i^2 + 2z^2 + 4i.. How did u get 2^2i^2 + 2z^2 .. because the question said 2iz^2 + 2z
and also in Q3. its not a straight line its x+ (b/x).. so its a straight line added to a hyperbola....
sorry for the trouble
Post by: moekamo on March 23, 2010, 08:34:59 pm
so the z bit that i made a mistake on cancels anyway... to give the same result.
also in q 3, in the pdf its x + x/b, is that a mistake??? if it is then you can exclude C as x cant equal 0, B is correct as there is asymptote at x=0 and at y=x
Post by: samiira on March 23, 2010, 09:19:19 pm
Post by: moekamo on March 23, 2010, 09:30:45 pm
so 3 complex roots, so answer is C
Post by: samiira on March 23, 2010, 10:42:28 pm
Post by: samiira on March 25, 2010, 08:03:35 pm
Post by: samiira on March 25, 2010, 09:10:41 pm
Post by: the.watchman on March 26, 2010, 06:47:44 am
Dilate
So replace
Post by: qshyrn on March 26, 2010, 03:57:22 pm
x^2 + (y/3)^2=9 expand the y bit and divide all by 9
Post by: the.watchman on March 26, 2010, 04:26:12 pm
i got option C for that its 1/3 dilation from x axis :
x^2 + (y/3)^2=9 expand the y bit and divide all by 9
Well about that, when you dilate by a factor of n from the x-axis, you replace y with y/n (in this case,
So it should be B
Post by: superflya on March 26, 2010, 04:32:25 pm
i presume u dont need help with the last question.
Post by: qshyrn on March 26, 2010, 05:07:13 pm
doesnt it go (x,y) --> (x,ny) for a dilation of factor n from x axis. so the n is put in front of the y?i got option C for that its 1/3 dilation from x axis :
x^2 + (y/3)^2=9 expand the y bit and divide all by 9
Well about that, when you dilate by a factor of n from the x-axis, you replace y with y/n (in this case,), which results in
So it should be B
Post by: qshyrn on March 26, 2010, 05:10:00 pm
for q2) restrict it between [-1,1] as the domain ofand... how do u continue ?? (tbh im having trouble with doing that as well, trying to make x the subject) i tried doing it as 2 seperate inequalities, but what i got isnt one of the answers...is [-1,1] so it looks a lil something like this
. u should be able to do the rest.
i presume u dont need help with the last question.
Post by: the.watchman on March 26, 2010, 05:15:01 pm
doesnt it go (x,y) --> (x,ny) for a dilation of factor n from x axis. so the n is put in front of the y?i got option C for that its 1/3 dilation from x axis :
x^2 + (y/3)^2=9 expand the y bit and divide all by 9
Well about that, when you dilate by a factor of n from the x-axis, you replace y with y/n (in this case,
So it should be B
Ah but y' = ny, y = y'/n (replace y with y'/n) :)
Post by: qshyrn on March 26, 2010, 05:19:59 pm
ooo yeah ur right.. i never really use those y' in transformations, but i guess theyre helpful sometimesdoesnt it go (x,y) --> (x,ny) for a dilation of factor n from x axis. so the n is put in front of the y?i got option C for that its 1/3 dilation from x axis :
x^2 + (y/3)^2=9 expand the y bit and divide all by 9
Well about that, when you dilate by a factor of n from the x-axis, you replace y with y/n (in this case,
So it should be B
Ah but y' = ny, y = y'/n (replace y with y'/n) :)
Post by: m@tty on March 26, 2010, 05:20:40 pm