
Re: Help Needed in math
Posted:
May 2, 2014 5:40 PM


On Friday, May 2, 2014 10:36:59 PM UTC+2, quasi wrote: > snmp wrote: > > > > >There is an easier way to look at this (even though it uses > > >similar principles). > > > > > >Axis of symmetry for ax^2 + bx +c = 0 is x = b/2a = 5/2a > > > > You need to consider the special case a = 0. > > > > >Value of function at axis symmetry (gives either maxima when > > >a < 0 (curve is concave down) and minima when a > 0 (curve is > > >concave up)): a(5/2a)^2 5(5/2a)  3 = 0 > a = 25/12 > > > > > >So when a = 25/12 the maxima has value 0 and so largest > > >integer for maxima to be less than 0 is 3. > > > > Yes. > > > > In my first reply in this thread, I gave hints for some > > approaches, one of which was along the lines you used above. > > > > Precalculus level methods based on properties of quadratic > > functions make sense for this problem. > > > > konyberg's solution is just an alternate method. > > > > For some reason, William Elliot couldn't make sense of > > konyberg's method, so in my reply, I tried to clarify how > > and why konyberg's approach works. > > > > quasi
Why is Elloit like this?

