Recover the curve from its range and count line intersections

$-2x^2-2x+a=xy+y$ gives $2x^2+xy+2x+y-a=0$. At least one real x requires $(y+2)^2-4(2)(y-a)\ge0$, hence $y^2-4y+4+8a\ge0$. Since a≠0, x=−1 does not satisfy the rearranged equation, so this criterion respects the domain.

The supplied range y≤−2 or y≥b is equivalent to $[y-(-2)](y-b)\ge0$. Expanding gives $y^2+(2-b)y-2b\ge0$. Comparing with $y^2-4y+4+8a$ yields −4=2−b and 4+8 a=−2 b. Thus b=6 and $a=(-4-2b)/8=-2$.

For a=−2, divide −2 x²−2 x−2 by x+1. The quotient is −2 x; subtracting −2 x(x+1)=−2 x²−2 x leaves remainder −2. Thus $y=-2x-2/(x+1)$. The asymptotes are x=−1 and y=−2 x. Stationary points are (−2,6), a minimum, and (0,−2), a maximum also giving the y-intercept. The numerator −2(x²+x+1) has no real zero, so there is no x-intercept.

At x=−3, $y=p(-3)+3p+6=-3p+3p+6=6$. Thus every line y=px+3 p+6 passes through (−3,6). As in the source sketch argument, two distinct intersections with C occur precisely when p>0 or p<−2.

As a supplementary check, multiplication by x+1 gives $(p+2)x^2+(4p+8)x+3p+8=0$. Its discriminant is $4p(p+2)$, positive exactly for p>0 or p<−2. p=−2 reduces to the impossible equation 2=0; p=0 gives the double intersection x=−2. Substituting x=−1 gives 2, so no counted root is an excluded input.