1. Let $a$ and $b$ be unequal real numbers. If the quadratic function $f(x)=x^{2}+a x+b$ satisfies $f(a)=f(b)$, then the value of $f(2)$ is $\qquad$
- 1.4.

From the known conditions and the axial symmetry of the quadratic function graph, we get
$$
\begin{array}{l}
\frac{a+b}{2}=-\frac{a}{2} \Rightarrow 2 a+b=0 \\
\Rightarrow f(2)=4+2 a+b=4 .
\end{array}
$$