(a) The 'graphs' are sets of points satisfying the equations and, as the equations are different we should expect the graphs to be different. We know that the graph of x+ y -1=0 is a straight line through (1,0) and (0,1) with gradient -1. As (x+y-1) is a factor of (x + y - 1)(x2 + y2)=0 we see that all the points on the line x +y-1=0 satisfy this equation and lie on the graph of this relation. From the other factor we find the remaining points on the graph of the relation given by x2 + y2 = 0 and so the origin (0,0) is also on this graph and no other points.

(b) Similarly we know that the graph of x+ y = 0 is a straight line through (0,0) with gradient -1. As (x+y) is a factor of (x + y)(y2+ (x+1)2)=0 we see that all the points on the line x +y=0 satisfy this equation and lie on the graph of this relation. From the other factor we find the remaining points on the graph of the relation given by y2+ (x+1)2 = 0 and so the point (-1,0) is also on this graph and no other points.

(c) Both graphs go through the points (0, 1), (1/2, 1/2) and (1, 0) and have gradient -1 at these three points as can be seen if we find the derivative. As before, they do not have the same equations so we might expect extra points, or even extra curves (or branches) to belong to the graph of the relation. As the software does not show any extra branches we will expect to find isolated points.

Differentiating:
3x2+3x dy
dx
 +3y+3y2 dy
dx
 =0
Simplifying this expression gives:
dy
dx
 = - x2 + y
x + y2
 .
The algebraic curve F(x,y) = x3 + 3xy + y3 - 1 may have other branches and in order to find them we have to factorise the expression x3 + 3xy + y3 - 1. We expect x+y-1 to be a factor because we have seen that the graph of this relation seems to have the line x+y-1 as one of its branches. To make the algebra more manageable, assuming x+y-1 is a factor, we substitute Y=y-1 and the expression becomes:
F(x,Y) = x3 + 3x(Y+1) +(Y+1)3 - 1.
Expanding this and factorising the expression we get:
F(x,Y) = (x+Y)(x2 -xY +Y2 +3Y + 3).
We need to return to the (x,y) coordinates to locate the other points belonging to this relation. Substituting for Y=y-1 and simplifying we get:
F(x,y)=(x+y-1)(y2 +y(1-x)+(1 + x + x2)).
The remaining points (or branches) of the graph of this relation are given by the quadratic:
y2 +y(1-x) + (1 + x +x2) = 0
which has roots given by
y =
-(1-x) ±   _______________________
Ö[ 1 - 2x + x2 -4 - 4x - 4x2]
 
2
 = (x-1) ±(x+1)Ö-3
2
 .
The only real values satisfying this equation are x = -1, y = -1 so the graph consists of the straight line y=1-x and the isolated point (-1, -1).