Why is $\infty$ a ramification point of a hyperelliptic curve when the defining equation has odd degree? We can work over $\mathbb{C}$ for simplicity. Let $X$ be defined by the equation $y^2 = h(x)$ where $h(x)$ has odd degree and distinct roots. Let $\pi: X \to \mathbb{P}^1$ be the ramified double covering.
Just to make sure, the roots of $h(x)$ are ramification points because if $y = 0$, then we get each root having multiplicity two, correct? But this argument doesn't really shed light on why there is another ramification point outside of the roots when $h$ has odd degree.
Which point in $X$ maps to the point at $\infty$, and why is it a ramification point? Why isn't this a ramification point when $h(x)$ has even degree?
Edit: I would like an explanation without using the Riemann-Hurwitz formula since the text I am reading counts the ramification points to compute the genus of the curve.
 A: Write $h(x) = a_{2g+1}x^{2g+1}+\ldots+a_1 x + a_0$. Remember that $X$ is really a projective curve, so let us switch to another affine chart with coordinates $(u,v) = (\frac{1}{x}, \frac{y}{x^{g+1}})$ where $X$ is represented by the equation
\begin{align*}
v^2 = \frac{y^2}{x^{2(g+1)}} &= \frac{a_{2g+1} x^{2g+1} + \ldots + a_1 x + a_0}{x^{2(g+1)}}\\
&= a_{2g+1} \frac{1}{x} + \ldots + a_1 \frac{1}{x^{2g}} + a_0 \frac{1}{x^{2g+1}} \\
&= a_{2g+1} u + \ldots + a_1 u^{2g} + a_0 u^{2g+1}.
\end{align*}
The projection map $X \to \mathbb{P}^1$ on this affine chart is given by $(u , v) \to u$.
From this, we see that $X$ has a ramification point above $u=0$ (indeed, the projection map $X \to \mathbb{P}^1$ has a ramification point wherever it fails to be a double cover). This point $u =0$ corresponds to $x=\infty$ in the other chart. 
What changes if $h(x)$ had even degree? If $h(x) = a_{2g+2} x^{2g+2} + \ldots + a_1 x + a_0$, then in the new coordinates $(u,v)$, the curve is represented by the equation
$$
v^2 = a_{2g+2} + a_{2g+1} u + \ldots + a_1 u^{2g+1} + a_0 u^{2g}.
$$
This polynomial in $u$ (on the right hand side of the above equation) does not have a root at $u=0$, hence the projection map will not have a ramification point at $x=\infty$.
Thanks to @Jyrki Lahtonen for pointing out the error in my answer.
A: Let's clarify your definition of $X$.  $y^2 = h(x)$ defines a nonsingular affine curve (we'll assume $h(0)\neq 0$), but the natural projectivization may be singular at infinity.
By convention, when we present a hyperelliptic curve as $y^2=h(x)$, we are resolving any singularities.  Since birational nonsingular curves are isomorphic, this resolution, $\hat{X}$, is unique.  Furthermore, the projection $(x,y)\mapsto y$ from $X$ onto $\mathbb{A}^1$ lifts uniquely to a map $\hat{X} \to \mathbb{P}^1$.
Now we look at the projectivization.  If $h(x)$ has degree $k\geq 3$, then the projectivization has defining equation $Y^2 Z^{k-2} - h(X,Z)=0$, where $h(X,Z) = \prod_i (X - \alpha_i Z)$, $\alpha_i$ ranging over the roots of $h(x)$.  The map to $\mathbb{P}^1$ is given by $[a:b:c] \mapsto [b:c]$.
Plugging in $Y=1$ gives us an affine curve $Z^{k-2} = h(X,Z)$, where the points mapping to $\infty$ are just those with $Z=0$, i.e. just the point $[0:1:0]$.
But this might not actually be a point of ramification, since we haven't resolved the singularity yet.  If it resolves to one point, there is ramification.  If it resolves to two points, there is no ramification.
I am not sure of the most direct way to count the points over a singularity (without invoking more advanced theory), but this can be done with a series of blow-ups, i.e. we can write the equation as $(Z/X)^{k-2} = X^2 h(Z/X)$, then as $(Z/X)^{k-4} = (X^2/Z)^2 h(Z/X)$, $(Z/X)^{k-6} = (X^3/Z^2)^2 h(Z/X)$, etc.
Finally, we either arrive at an equation $S = T^2 h(S)$ or $S^2 = T^2 h(S)$, depending on whether $k$ is odd or even.  In the first case, we have resolved the singularity to a single point ($S=T=0$), so there is ramification.  In the second case, we blow-up again to get the nonsingular $U^2 = h(S)$, and then $S=0$ has two solutions, so there is no ramification.
