Let $f(x) =ax^2 +bx+c,$ where $a, b ,c\in\mathbb{R}.$ If $f(-1), f(0) , f(1)\in [-1,1]. $ Then prove that $|f(x) |\le\frac{3}{2}\forall x\in [-1,1]$.
My method was involving graph : first of all we can observe that curve of any quadratic is symmetric about some x = k line and that if quadratic has value at f(-1) to be more negative then it will make the maximum reach more value because of greater slope intially . (a>0 case i am considering) from these ideas i first made three blocks of regions 2 of those regions the constraint points are marked so that it will lie inside that region only . Now the third region which goes from $x=1 to x=2$ (i made that because it was rather easy to observe the symmetric nature of the curve about some $x=k$ axis in this way ) By observation from graph i concluded maximum will be when the $f(1)= 1,f(-1)= -1 and f(0)=1$ which gives the $f(x) = -x^2 +x +1$ curve , whose maximum in that range is $5/4 <3/2$ , so did i finished correctly or something is missing in this graph proof?
