Question 3370
you mean you had trouble with, surely? :-)


{{{y = x^2 + px + q}}}


For (0,2) --> 2 = 0 + 0 + q, so q = 2
For (-3,-5) --> {{{-5 = (-3)^2 - 3p + 2}}}, which gives 
-5 = 9 - 3p + 2
3p = 16
p = 16/3


so, {{{y = x^2 + (16/3)x + 2}}}


jon