Involutory Parabola

We will use each bit of information in turn to figure out the coefficients. (Then, we'll find we don't have quite enough information. Or do we?)

The key bits of information are: the function describes a parabola; it's an involution; and it passes through $(-3,0)$ and $(4,0)$ .

Parabola:

This tells us the discriminant of the curve's equation is zero - ie $B^2=4C$

Involution:

We can swap $x$ and $y$ in the equation without changing it. This tells us that $C=1$ and $D=E$ . The current form of the equation is then

$x^2+Bxy+y^2+D(x+y)+F=0$

Point $(-3,0)$ :

$9-3D+F=0$

Point $(4,0)$ :

$16+4D+F=0$

Solving these two simultaneous equations gives $D=-1$ and $F=-12$ .

But we still have two possibilities for $B$ , namely $\pm2$ . So the curve is either

$x^2+2xy+y^2-x-y-12=0$

or

$x^2-2xy+y^2-x-y-12=0$

Both of these are involutions passing through the given points, and both have discriminant zero. However, they are not both parabolas! We have

$x^2+2xy+y^2-x-y-12=(x+y+3)(x+y-4)=0$

and this factorisation shows that the equation describes two parallel lines (namely $y=-x-3$ and $y=-x+4$ ) - not a parabola. (This is known as a degenerate conic).

Having eliminated $B=2$ , we are left with $x^2-2xy+y^2-x-y-12=0$ , which is indeed a parabola, giving the answer $-2+1-1-1-12=\boxed{-15}$ .