Example: Let
write the cubic
polynomial whose roots are r_{1}
= 1 and r_{2,3}
= 1 ±
i assuming its
leading coefficient a_{3}
= 1,
and find its source function
and draw their graphs.

Then, find the cubic
obtained by moving the source function by x_{t}
= 
1 and y_{t}
= 2, and draw the graph of
the translated cubic.

Solution:
We write the cubic polynomial
given by its roots to calculate its coefficients, since

f
(x)
= x^{3}
 (r_{1}
+ r_{2}
+ r_{3})x^{2}
+ (r_{1}r_{2}
+ r_{1}r_{3}
+ r_{2}r_{3})x
 r_{1}r_{2}r_{3}, 
then
a_{2}
=
 (r_{1}
+ r_{2}
+ r_{3})
=
 [1
+ (1 + i)
+ (1  i)]
=
 3, 
a_{1}
=
r_{1}r_{2}
+ r_{1}r_{3}
+ r_{2}r_{3}
=
1 ·
(1 + i)
+ 1
·
(1  i)
+ (1 + i)
·
(1  i)
=
4, 
a_{0}
=

r_{1}r_{2}r_{3}
=

1 ·
(1 + i)
·
(1  i)
=
 2, 
therefore
f (x)
= x^{3}
 3x^{2}
+ 4x
 2
or f
(x)
= (x

1)[x
 (1 + i)][x
 (1
 i)]. 
To
find the source form, f_{s}(x)
=
a_{3}x^{3}
+ a_{1}x 
of the general cubic polynomial
we calculate coordinates
of translations x_{0}
and y_{0}, 

and plug them into
y
+ y_{0}
= a_{3}(x
+ x_{0})^{3}
+
a_{2}(x
+ x_{0})^{2}
+
a_{1}(x
+ x_{0})
+
a_{0} 
thus,
y
= (x
+
1)^{3}
 3(x
+ 1)^{2}
+ 4(x
+ 1)
 2
= x^{3}
+ x 
or
f_{s}(x)
= x^{3}
+ x
where a
= 1 thus,
it is the source cubic of the type 2/1 since, a_{3}a_{1}
> 0. 
The source
cubic we then move to new position by
plugging the given coordinates of translations,

x_{t}
= 1 and y_{t}
= 2 into
y

y_{t}
=
a_{3}(x

x_{t })^{3}
+
a_{1}(x

x_{t
}) 
so
we get, y

2
=
(x
+
1)^{3}
+
(x
+
1)
or y
=
x^{3}
+
3x^{2}
+ 4x
+ 4 
as shows the below figure. 
