We know that any hyperelliptic curve over a field with characteristic not equal to $2$ has an affine model given by $y^2 = f(x)$, with $deg(f) = 2g+1$ or $2g+2$.
Can we always find a model such that $deg(f)=2g+1$?
We know that any hyperelliptic curve over a field with characteristic not equal to $2$ has an affine model given by $y^2 = f(x)$, with $deg(f) = 2g+1$ or $2g+2$. Can we always find a model such that $deg(f)=2g+1$? 


Over an algebraically closed field, yes. Let the roots of $f$ in $\mathbb{P}^1$ be $r_1$, $r_2$, ..., $r_{2g+2}$, and apply a Mobius transformation taking $r_{2g+2}$ to $\infty$ as Qiaochu says. Over a nonalgebraically closed field, not necessarily. Let $g \geq 2$ for simplicity, so the hyperelliptic involution $\tau: (x,y) \mapsto (x, y)$ is an intrinsic property of the curve. Then the curve has such a model if and only if at least one of the fixed points of $\tau$ is defined over the ground field. 

