Find the range of $a$ such $|f(x)|\le\frac{1}{4},$ $|f(x+2)|\le\frac{1}{4}$ for some $x$. 
Let $$f(x)=x^2-ax+1.$$
  Find the range of all possible $a$ so that there exist $x$ with 
  $$|f(x)|\le\dfrac{1}{4},\quad |f(x+2)|\le\dfrac{1}{4}.$$

A sketch of my thoughts: I write 
$$f(x)=\left(x-\dfrac{a}{2}\right)^2+1-\dfrac{a^2}{4}\ge 1-\dfrac{a^2}{4}$$
so if $1-\dfrac{a^2}{4}>\dfrac{1}{4}$ or $-\sqrt{3}<a<\sqrt{3}$
this case impossible
But I don't know how to prove the other case, or if this there are better ideas. 

My Work: I will approach in a very short way.we have been given
 two characteristic equations of a function and their range .
 $$f(x)=x^2-ax+1$$
 $$\text{and, we know}$$
 $$\lvert x^2-ax+1 \rvert \le \dfrac{1}{4}$$
 $$\lvert (x+2)^2-a(x+2)+1 \rvert \le \dfrac{1}{4}$$
from the absolute value,we can write-
 $$-\dfrac{1}{4}\le x^2-ax+1 \le \dfrac{1}{4}$$
 $$-\dfrac{1}{4}\le (x+2)^2-a(x+2)+1 \le \dfrac{1}{4}$$
Now, Let's see how do these two inequalities come to our help? We have to find the range
 of $a$ for which these two inequalities held.It is certain that we will find a point of a if we work on
upper limit of these two inequalities and will find another point if we work on lower limit of these two inequalities.
to make my life easy ,I am counting them as equations.[Look,I am just using the equations to determine the peak values of  "$a$"]
So,if we take upper limit,we get-
$$x^2-ax+1 = \dfrac{1}{4}.......(1)$$
$$(x+2)^2-a(x+2)+1  = \dfrac{1}{4}......(2)$$
now if we solve these two equation we will get ,$a^2=7$
Again, if we take lower limit,we get-
$$ x^2-ax+1=-\dfrac{1}{4}.........(3)$$
$$ (x+2)^2-a(x+2)+1=-\dfrac{1}{4}..........(4)$$
If we solve these two equations we will get $a^2=9$
Hence, $ 7\le a^2 \le 9$
