Say $K=\mathbf{Q}(\sqrt{p^{*}})$ is the unique quadratic field in which $p$ is the only ramified finite prime.  That is, $p^*=(-1)^{(p-1)/2}p$.  There are two $L$-functions one can concoct out of $K$, and quadratic reciprocity manifests in the fact that the $L$-functions coincide.  

First, there is the $L$-function arising from the Galois character $\chi$ which cuts out $K$.  This is a one-dimensional Artin $L$-function.  Its Euler factor at a prime $q$ is $(1\pm p^{-s})^{-1}$, where you place $-$ whenever $q$ splits in $K$, and $+$ whenever $q$ is inert in $K$.

The other $L$-function is a Dirichlet $L$-function.  Let $\varepsilon$ be the unique character of $(\mathbf{Z}/p\mathbf{Z})^\times$ of order exactly two.  This extends to a Dirichlet character $\varepsilon$ on $\mathbf{Z}$, and the Dirichlet $L$-function is $\sum_{n\geq 1} \varepsilon(n)n^{-s}$.  

Let us see that if the two $L$-functions are equal, then quadratic reciprocity holds.  Indeed, look at the coefficient of $q^{-s}$.  The coefficient in the first $L$-function is $\left(\frac{p^*}{q}\right)$.  The coefficient in the second $L$-function is $\left(\frac{q}{p}\right)$.  (This requires some explanation:  $x\mapsto \left(\frac{x}{p}\right)$ is a character of $(\mathbf{Z}/p\mathbf{Z})^\times$ of order two, so it must equal $\varepsilon$ by uniqueness.)  We find

$$\left(\frac{q}{p}\right)=\left(\frac{p^{*}}{q}\right)=\left(\frac{(-1)^{(p-1)/2}}{q}\right)\left(\frac{p}{q}\right) = (-1)^{\frac{p-1}{2}\frac{q-1}{2}}\left(\frac{p}{q}\right).$$

Of course you didn't need $L$-functions to state any of this, but they are nonetheless very important.  For instance, you can't prove that the Artin $L$-function has an analytic continuation and functional equation without knowing that it equals a Dirichlet $L$-function.  The generalization of this coincidence to higher-degree Artin $L$-functions is still quite conjectural!