is the answer 8, 8 and1...

this is one method.....

for it to be a limiting case let us concider an eqn. in whos graph, f(1)=1, f(0)=1 and  consequently f(1/2)= -1

substituting it in the eqn.s for ax2+bx+c we get 3 eqn.s solving which we have

a=8,b=-8 and c=1

moda=8, modb=8, mod c=1

 

or

 

we can imagine of another eqn. in which we have f(1)=-1, f(0)= -1 and consequently f(0)=1

following the same procedure for this eqn. we obtain a= -8 b=8 and c= -1

but again moda=8 modb=8 and modc=1

 

rate if useful