An algebra problem by Aaron Jerry Ninan

a,b,c are positive real numbers. $a\neq 0$ $f(x)=ax^{2}+bx+c$ 1) When x is real $f(x-4)=f(2-x)$ $f(x)\geq x$ 2) When $0< x< 2$ $f(x)\leq \frac{(x+1)^{2}}{4}$ 3) The minimum value of the function on "R" is "0". Q) Find the maximal "m" (m >1) such that there exists a real "t" , $f(x+t)\leq x$ holds so as long as $1\leq x\leq m$