## Solution.

Since the graph of the function $f(x)$ has an axis of symmetry parallel to the line $x=0$, then the axis of symmetry has the equation $x=c$, where $c$ is a constant real value. Substituting $t=x-c$, we obtain that the graph of the function $f(t)$, as a function of $t$, will be symmetric with respect to the line $t=0$, i.e., the $O y$ axis. We have:

$$
\begin{gathered}
P(t)=x^{4}-8 x^{3}+14 x^{2}+a x=(t+c)^{4}-8(t+c)^{3}+14(t+c)^{2}+a(t+c)= \\
=t^{4}+4 t^{3} c+6 t^{2} c^{2}+4 t c^{3}+c^{4}-8\left(t^{3}+3 t^{2} c+3 t c^{2}+c^{3}\right)+14\left(t^{2}+2 t c+c^{2}\right)+a t+a c= \\
=t^{4}+t^{3}(4 c-8)+t^{2}\left(6 c^{2}-24 c+14\right)+t\left(4 c^{3}-24 c^{2}+28 c+a\right)+\left(c^{4}-8 c^{3}+14 c^{2}+a c\right)
\end{gathered}
$$

The graph of the function $f(t)$ is symmetric with respect to the $O y$ axis if and only if the function $f(t)$ is even, i.e., if and only if the coefficients of the function $f(t)$ corresponding to the odd powers of $t$ are equal to 0. We obtain

$$
\left\{\begin{array} { c } 
{ 4 c - 8 = 0 } \\
{ 4 c ^ { 3 } - 2 4 c ^ { 2 } + 2 8 c + a = 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { c } 
{ c = 2 } \\
{ a = - 4 c ^ { 3 } + 2 4 c ^ { 2 } - 2 8 c }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
c=2 \\
a=-4 \cdot 8+24 \cdot 4-28 \cdot 2=8
\end{array}\right.\right.\right.
$$

Final answer: $a=8$.
11.1. For what real values of the parameter a does the graph of the function $f: \square \rightarrow \square$, $f(x)=x^{4}-8 x^{3}+14 x^{2}+$ ax have an axis of symmetry parallel to the line $x=0$?