On $y^2=x^3+5x^2+4x$, the slope through $P_1=(-1,0)$ and $P_2=(-2,2)$ is $-2$. The <elliptic-curve addition formula> gives
$$
x(P_1+P_2)=(-2)^2-5-(-1)-(-2)=2,
\qquad y(P_1+P_2)=(-2)(-1-2)=6.
$$
For doubling $P_2$, the tangent slope is
$$
\lambda=\frac{3x^2+10x+4}{2y}\bigg|_{(-2,2)}=-1,
$$
and hence
$$
\boxed{P_1+P_2=(2,6),\qquad 2P_2=(0,0).}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 125 / 3 / a / Solution

On $y^2=x^3+a_2x^2+a_4x+a_6$, let $\lambda$ be the secant slope through distinct points $P,Q$, or the tangent slope when $P=Q$. Then
$$
x(P+Q)=\lambda^2-a_2-x(P)-x(Q),
\qquad
y(P+Q)=-y(P)+\lambda\bigl(x(P)-x(P+Q)\bigr).
$$


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 Incoming links: Elliptic-curve addition formula