Given a point $P = (-1,6)$. $2P$ is found to be $(3,-2)$.
But how? I have used the standard formula and can never get this
With equation $y^2 = x^3 - 15x + 22$
$$m' = (3(x_o)^2+a)/2y_o,\\ x_1 = (m')^2 - 2x_o ,\\y_1 = y_o + m'(x_1-x_o).$$ Then $2P = (x_1 -y_1).$ No P was given it is a torsion subgroup question.
Reposted with more of the information I missed.
If the elliptic curve is $y^2=x^3-15x+22,$ then, differentiating, $$\frac{dy}{dx}=\frac{3x^2-15}{2y}.$$ So the slope at $(-1,6)$ is $\frac{-12}{12}=-1.$
Then you want the other point on the curve on the line $y=-x+5$ that goes through $P$ and is tangent to the curve.
So you are trying to solve:
$$(-x+5)^2= x^3-15x+22$$
This will give a cubic equation with repeated roots $x=-1.$ since the constant term of the cubic is $-3,$ the third root is $x_1=3,$ and thus $y_1=-x_1+5=2.$
But then $(x_1,-y_1)=(3,-2)=2P.$
The multiples of a point on an elliptic curve $y^2=x^3+Ax+B$ can be computed using the division polynomials. See https://en.wikipedia.org/wiki/Division_polynomials.
For example, if $P=(x,y)$ is not of order $2$ (i.e. $y\neq 0$) we have
$$2P=\left(\frac{x^4-2Ax^2-8Bx+A^2}{4y^2}, \frac{x^6+5Ax^4+20Bx^3-5A^2x^2-4ABx-8B^2-A^3}{8y^3} \right).$$
Of course, the $4y^2$ can be replaced by $4(x^3+Ax+B).$ Plugging in $P=(-1,6),A=-15,B=22$ verifies that $2P=(3,-2)$. On the Wikipedia page, you can find a formula for $3P$ as well.
As a bonus, another useful formula is the following, valid for $P=(x_P,y_P), Q=(x_Q,y_Q)$ such that $P\neq \pm Q$. $$x(P\pm Q)=\frac{x_P^2x_Q+x_Px_Q^2+A(x_P+x_Q)\mp 2y_Py_Q+2B}{(x_P-x_Q)^2}.$$ Setting $Q=2P$ yields a formula for $x(3P)$.
