Finite morphisms from $\mathbb{P}_k^1$ to a curve? I use the following definition for a curve over an algebraically closed field $k$ : a scheme $X$ over $k$ is a curve if it is integral of dimension $1$, and separated and of finite type over $k$.
Now, it is a well known fact of the theory of curves that any nonsingular curve $X$ can be embedded in $\mathbb{P}^3$. However, can we say anything about the existence of finite maps of the form $\mathbb{P}^1\to X$? If $f:\mathbb{P}^1\to X$ is finite, then can we conclude that it is an isomorphism and therefore that $X$ is rational?
 A: The existence of a finite morphism $\mathbb{P}^1 \to X$ implies, by pulling back to $\mathbb{P}^1$ the rational functions on $X$, the existence of a finite field extension $k(X) \subseteq k(\mathbb{P}^1)$. 
Now $k(\mathbb{P}^1) \simeq k(t)$, the pure transcendental extension of degree $1$ of $k$, so $k(X)$ is an intermediate extension between $k$ and $k(t)$. Then, by Lüroth's theorem, there exists a rational function $\phi(t) \in k(t)$ such that $k(X) = k(\phi(t))$. 
In other words, $k(X)$ is a simple transcendental extension of $k$, and this means that $X$ is birational, and hence isomorphic, to $\mathbb{P}^1$.    
A: If $\mathbb{P}^1 \rightarrow X$ is finite and $X$ is integral, then $X$ is complete since $\mathbb{P}^1$ is and the finite map is surjective. Then since $\mathbb{P}^1$ is smooth (hence normal) we have the factorization $\mathbb{P}^1\rightarrow \tilde{X} \rightarrow X$ where $\tilde{X}\rightarrow X$ is the normalization of $X$ (the identity if $X$ is already smooth). The normalization $\tilde{X}$ is a smooth curve and it follows from Hurwitz's formula that the genus of $\tilde{X}$ must be 0. So it must be $\mathbb{P}^1$ (by Riemann-Roch).
So with $X$ as you described, a finite map $\mathbb{P}^1 \rightarrow X$ is a finite branched covering of $\mathbb{P}^1$ by itself, composed with the normalization map of $X$ (the latter being a birational morphism). Such branched coverings $\mathbb{P}^1 \rightarrow \mathbb{P}^1$ of course exist for any degree $d$.
