How to prove $-P = (x, -a_1x - a_3 - y)$ for an Elliptic Curve of the General Weirstrass equation for P not the identity? Let $P = (x,y) \ne \{\infty\}$. Then $-P$ is the other finite point of intersection of the curve and the vertical line through $P$.
General Weirstrass equation: E:  $a_1y^2+a_3xy+a_5y = x^3+a_2x^2+a_4x+a_6$

PROBLEM: For the general Weirstrass equation of an elliptic curve show that $-P = (x, -a_1x - a_3 - y)$ for all $P$ on the the elliptic curve $E$.

We know P + (-P) = {$\infty$}. 
For Weirstrass equation $y^2 = x^3 + Ax + B$, a special case, -P = (x,-y).
 A: Let $E$ be given by a Weierstrass equation
$$y^2+a_1xy+a_3y = x^3+a_2x^2+a_4x+a_6.$$
If the characteristic of the field is not $2$, then we can complete the square in $y$ to obtain
$$(y+(a_1x+a_3)/2)^2 = x^3+a_2x^2+a_4x+a_6+(a_1x+a_3)^2/4.$$
You can write the last equation as $E':Y^2=f(x)$. In this model, it is clear that if $P'=(x_0,Y_0)$, then $-P'$ (the second point of intersection of the curve with the vertical line going through $P'$) is given by $(x_0,-Y_0)$. Since $Y=y+(a_1x+a_3)/2$, it follows that if $P=(x_0,y_0)$ is a point in the original model, then $P'=(x_0,y_0+(a_1x_0+a_3)/2)$ and $-P'=(x_0,-y_0-(a_1x_0+a_3)/2)$ and therefore $-P$ has coordinates $(x_1,y_1)$ with 
$$x_1=x_0, \text{ and } y_1+(a_1x_1+a_3)/2 = -y_0-(a_1x_0+a_3)/2.$$
Hence,
$$x_1=x_0, \text{ and } y_1 = -y_0-a_1x_0-a_3,$$
i.e., $-P=(x_0,-y_0-a_1x_0-a_3)$.
Now that we have a "candidate" formula for $-P$, let us see if this also works in characteristic $2$. Let $E$ be given by a general Weierstrass equation over a field $F$ of characteristic $2$, and let $P=(x_0,y_0)$ be a point on $E$. Let $Q=(x_1,y_1)=(x_0,-y_0-a_1x_0-a_3)$. It suffices to check that $Q$ is a point on $E$. Since the vertical line that passes through $P$ also passes through $Q$, if $Q\in E$, then $Q=-P$. Here we go:
\begin{align*}
  y_1^2+a_1x_1y_1+a_3y_1 & = y_1^2+(a_1x_0+a_3)y_1\\
& = y_1(y_1+a_1x_0+a_3)\\
  & = (-y_0-a_1x_0-a_3)(-y_0-a_1x_0-a_3+(a_1x_0+a_3))\\
 & = (-y_0-a_1x_0-a_3)(-y_0)\\
 & = y_0(y_0+a_1x_0+a_3)\\
 & = x_0^3+a_2x_0^2+a_4x_0+a_6.
\end{align*}
Thus, $Q\in E$, and $Q=-P$, also in characteristic $2$.
