Hyperelliptic curves express by branch points In a book of algebraic geometry, the author says that equation
$$
y^2=x(x^4-1)
$$
defined a curve $S$ of genus $2$. 
My problem is this: Because the curve has genus 2, it can be expressed as follows
$$
y^2=(x-a_1)(x-a_2)...(x-a_6)
$$ 
where $a_i$, $ 1\le i\le 6$, is the branch points of the canonical map $\varphi_{K}: S \longrightarrow \mathbb{P}^1$. Thus, $y^2=x(x^4-1)$ can not be a curve of genus $2$. A curve of genus $2$ has $6$ branch points, and not $5$ as suggested by equation $y^2=x(x^4-1)$.
Am I making a mistake in my thinking?
Thank you!
 A: I see there is no checked answer, so I try to fill the gaps. Maybe it could be useful for someone.
Disclaimer: I'm studying Riemann Surfaces right now. It is possible that I'm misunderstanding something.
As Angina already stated, $\infty$ is a branch point. How to check that?
Hyperelliptic curves $S$ consist of two affine curves glued together, say $$X=V(y^2-p(x))$$ and $$Y=V(w^2-z^6p(\tfrac{1}{z}))$$.
You can see $\varphi_{K}$ in the $X$ point of view, or in the $Y$ point of view. in both cases, you can see it as first-coordinate projection on $X$ or $Y$ respectively to $\Bbb{C}$. Branch points are the projection of points where the first-coordinate projection fails to be a locally of degree 1. These are points which set the partial derivatives of the defining polynomial w.r.t. the second coordinate to zero. They happen to be determined by zeros of $p(x)$. If $p(x)$ has even degree and $0$ is not among its roots, then you can symmetrically do the same job using $Y$.
Call $j_{X}:X\rightarrow S$ the embedding of $X$ in $S$ and $j_{Y}$ similarly.
Here $j_{x}(0,0)$ is a ramification point of $\varphi_{K}$ but checking for ramification points with the "$Y$ point of view", you see  that also $j_Y(0,0)$ is a ramification point. But what is its image under projection $\varphi_{K}$? It is exactly $\infty \in \Bbb{C}^{\infty}$ or equivalently $[0:1] \in\Bbb{P}^{1}$
