edit below.
I am having trouble with understanding the proof of III. 4.2 a) (why $[m]$ (multiplication by $m$) isogeny is nonconstant) in Silverman's "The Arithmetic of Elliptic Curves". He uses the duplication formula (for calculating $2P$) for a Weierstrass curve which computes the $x$ coordinate of a point $2P$. The formula is: $$ x([2]P) = \frac{x^4-b_4x^2-2b_6x-b_8}{4x^3+b_2x^2+2b_4x+b_6} = \frac{f(x)}{g(x)} $$ where $b_i$'s $\in K$ (field) are parameters of the curve established before and are not really relevant to my question.
He says that if a point $P=(x_0,y_0)$ on a curve is of order $2$ then it must satisfy that $g(x_0)=0$ i.e. $4x_0^3+b_2x_0^2+2b_4x_0+b_6 = 0$.
My question is why? Where does it come from? The point at infinity is $[0:1:0]$ in projective coordinates and how is it related to this affine formula? Is it clear when you calculate the projective fomulae for the duplication formula and it becomes equivalent with the condition that the $z$ coordinate is $0$?
He then also states that $f,g$ are cooprime iff the curve is smooth and from that establishes that there is a root (in $\bar{K}$) of $g(x)$ of order higher than of $f(x)$. And that root is a $x$ coordinate of a point of order $2$. What does the order of the roots have to do again with the formula? Is it again clearer from the projective formula or am I missing something?
EDIT: I now understand why the point of order has to have $f(x_0) = 0$ since it comes from the observation that $P=(x_0,y_0)$ of order has to safisfy $P= -P \iff y_0 = -y_0-a_1x_0-a_3$ and substituting $y$ in the Weierstrass equation gives the condition $g(x_0) = 0$. Now I don't really see the reasoning about orders of roots of $f,g$. Don't I need to show that $g$ has at least two distinct roots? Then there is always going to be a point of order $2$.