1
the part of curve in IIIrd quadrant is the inverted image of part of curve in Ist quadrant
15 Answers
6i cant get the ans ...someone please show it!!!
16how did u work this out...
am getting the part of 3rd quad as a mirror image in 2d quad...
1i don't think the graph exists in 2nd quadrant
there is no -ve value of x for which y is +ve so the graph can't exist in 2nd quadrant
you also confirm this from othersas i am not so good in graphs
6at x=0 it f(x) will be ∞ because curve is not defined
take
lim x+1/x
x→0
which equals infinte
if i solved correctly...:P
6i agree to the fact that it exists in 3rd and 1st quad...
my prob is
am not geting theCURVES alright...
can someone help
6oh i got it man!!thanzz
i also drew the graph..so posting it!!
1rather tanuj's post must be pinked ...very nice and precise graph
6well i never found that x=y can also be a asymptote!!
23y = x + 1x
or (y-x) = 1x
since y = 1x has asymptote y = 0 ,
u can say that the former one has y-x=0 as asymptote
