## Solving a cubic graphically

Following [Adler, p.250] the graphical solution of the cubic equation
x3 + ax2 + bx + c = 0,
can be reduced to the location of the intersection points of the two conics
y = x2,
yx + ay + bx + c =0.
The first is always the same well known parabola. The second is a rectangular hyperbola discussed in detail in Rectangular_Hype_From_Line.html .
Its center is at (-a, -b) and its asymptotes are parallel to the axes. The coefficients (a,b) are in reverse order from those in the above reference.
The figure below illustrates the method:
1) First draw the line  bx+ay+c=0 and find its intersection points {A,B} with the axes.
2) Then find their symmetrics {A',B'} with respect to (-a,-b).
3) Find the orthocenter H of one of the triangles formed by the parallelogram ABB'A' and its diagonals (here BAB').
4) Pass a conic throught the five points {A,B,A',B',H}. This coincides with  yx+ay+bx+c=0.
5) Draw the parabola y=x2.
6) Find the intersection point(s)  (sx, sy) of the two curves. Number sx is a root of the original equation.