The family of curves y = Ax + Bx² has two constants…

Step 1. Derivatives. $y' = A + 2Bx$ and $y'' = 2B$. So $B = \dfrac{y''}{2}$ and $A = y' - xy''$.

Step 2. Eliminate. $y = x(y' - xy'') + x^2\dfrac{y''}{2} = xy' - \dfrac{x^2y''}{2}$. Multiply by $2$ and rearrange: $x^2y'' - 2xy' + 2y = 0$. So $k = 2$.

Step 3. The member. For $A = 3$ and $B = -2$ we have $y'' = 2B = -4$, and $y' = 3 - 4x$. At $x = 2$ the slope is $3 - 8 = -5$.