<s>[INST] 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. 
 [/INST] Considering the problem geometrically, then
$$
f(x) = x^{\,2}  - ax + 1
$$
is a vertical parabola, convex (upward concave), with axis $x=a/2$ and vertex in $V=(a/2,1-a^2/4)$.
Thus the parameter $a$ is then just a translation parameter.

So, from the above and with the help of the sketch it is clear that, to satisfy the requirements we shall have
$$
\eqalign{
  & \exists \,x\;:\;\left\{ \matrix{
  \left| {f(x)} \right| \le 1/4 \hfill \cr 
  \left| {f(x + 2)} \right| \le 1/4 \hfill \cr}  \right.\quad  \Leftrightarrow   \cr 
  &  \Leftrightarrow \quad \left\{ \matrix{
  2 \le \overline {A_{\,1} A_{\,2} }  \hfill \cr 
   - 1/4 \le V_{\,y}  \le 1/4 \hfill \cr}  \right.\quad  \cup \quad \left\{ \matrix{
  2 \le \overline {A_{\,1} A_{\,2} }  \hfill \cr 
  V_{\,y}  \le  - 1/4 \hfill \cr 
  \overline {B_{\,1} B_{\,2} }  \le 2 \hfill \cr}  \right. \cr} 
$$
which means
$$
\left\{ \matrix{
  2 \le \overline {A_{\,1} A_{\,2} }  \hfill \cr 
  \overline {B_{\,1} B_{\,2} }  = \emptyset \quad  \vee \quad \overline {B_{\,1} B_{\,2} }  \le 2 \hfill \cr}  \right.
$$
i.e.
$$
\left\{ \matrix{
  2 \le \sqrt {a^{\,2}  - 3}  \hfill \cr 
  a^{\,2}  - 5 \le 0\quad  \vee \quad \sqrt {a^{\,2}  - 5}  \le 2 \hfill \cr}  \right.
$$
and finally
$$ \bbox[lightyellow] {  
\left\{ \matrix{
  7 \le a^{\,2}  \hfill \cr 
  a^{\,2}  \le 9 \hfill \cr}  \right.\quad  \Rightarrow \quad \sqrt 7  \le \left| a \right| \le 3
}$$
</s>