In particular, we can draw the parabola y = x^{2}.
Consider the circle that is centered at (a,b) and goes
through the origin. This circle has the equation
(x  a)^{2
}+ (y  b)^{2 }= a^{2 }+ b^{2}.
If a point (x, y), other than the origin, also lies on
the parabola , y = x^{2 }then
(x  a)^{2
}+ (y  b)^{2 }= a^{2 }+ b^{2},
x^{2 } 2ax
+ x^{4 } 2bx^{2 }= 0,
x^{3 }+ (1  2b)x  2a = 0. 
(1)

The regular 7gon has a solid construction, a fact that was known to
Archimedes. To construct it, we must construct x = e^{2pi/7}, which satisfies the equation
x^{7 } 1 =
(x  1)(x^{6 }+ x^{5 }+ x^{4
}+ x^{3 }+ x^{2 }+ x + 1) = 0.
Let w = x + x^{1 }= 2cos(2p/7). Then
w^{3 }= x^{3
}+ 3x + 3x^{1 }+ x^{3},
w^{2 }= x^{2
}+ 2 +x^{2},
so
w^{3}+ w^{2
} 2w  1 = 0.
To exploit (1), we complete the cube. That is, we make the
substitution z = w + 1/3, so that
(z  1/3)^{3 }+
(z  1/3)^{2 } 2(z  1/3)  1 = 0,
which simplifies to
z^{3} (7/3)z
 7/27 = 0.
We therefore choose (a,b) = (7/54,5/3) and proceed as
follows (following along in the figure): We construct the circle
centered at (7/54,5/3) that goes through the origin. We drop a
perpendicular from a point of intersection of this circle with the
parabola y = x^{2}, to find the point (w +
1/3,0). We find the point (w/2,0) and the
perpendicular to the xaxis through this point. This
perpendicular intersects the unit circle at two of the seven points of
a regular
7gon. We use those points to find the rest of the vertices.