A revealing insight into the polynomial function - continue
The principle provides full control over the polynomial
Hence, by substituting coordinates of translations into the source polynomial function we can move its graph to any desired position on the coordinate plane.
Since an nth degree polynomial with real coefficients can be expressed as a product of its leading coefficient an and n linear factors of the form (x - ri), where ri denotes its real root and/or complex root
f (x) = anxn + an - 1xn - 1 + . . . + a1x + a0 = an(x - r1)(x - r2) . . . (x - rn),
the above expression shows following coefficients and roots relations called Vieta's formulas.
Therefore, if the leading coefficient an = 1 we can write for example,
the quadratic function shown by its roots
f (x) = x2 + a1x + a0 = (x - r1)(x - r2) = x2 - (r1 + r2)x + r1r2,
the cubic function expressed by roots
f (x) = x3 + a2x2 + a1x + a0 = (x - r1)(x - r2)(x - r3)    or
f (x) = x3 - (r1 + r2 + r3)x2 + (r1r2 + r1r3 + r2r3)x - r1r2r3,
the quartic function wrtten by roots
f (x) = x4 + a3x3 + a2x2 + a1x + a0 = (x - r1)(x - r2)(x - r3)(x - r4)    or
f (x) = x4 - (r1 + r2 + r3 + r4)x3 + (r1r2 + r1r3 + r1r4 + r2r3 + r2r4 + r3r4)x2 -
- (r1r2r3 + r1r2r4 + r1r3r4 + r2r3r4)x + r1r2r3r4,
and so on.
Note that the coefficients of a polynomial are expressed as the alternating sums of corresponding combinations of products of roots.
Example:   Let write the cubic polynomial whose roots are r1 = 1 and  r2,3 = 1 ± i  assuming its leading coefficient a3 = 1, and find its source function and draw their graphs.
Then, find the cubic function obtained by moving the source function by  xt = - 1 and  yt = 2, and draw the graph of the translated cubic function.
Solution:   We write the cubic polynomial given by its roots to calculate its coefficients, since
f (x) = x3 - (r1 + r2 + r3)x2 + (r1r2 + r1r3 + r2r3)x - r1r2r3,
then                  a2 = - (r1 + r2 + r3) = - [1 + (1 + i) + (1 - i)] = - 3,
a1 = r1r2 + r1r3 + r2r3 = 1 · (1 + i) + 1 · (1 - i) + (1 + i) · (1 - i) = 4,
a0 = - r1r2r3 = - 1 · (1 + i) · (1 - i) = - 2,
therefore     f (x) = x3 - 3x2 + 4x - 2    or     f (x) = (x - 1)[x - (1 + i)][x - (1 - i)].
To find the source form,   fs(x) = a3x3 + a1x  of the general cubic polynomial we calculate coordinates of translations x0 and y0,
and plug into      y + y0 = a3(x + x0)3 + a2(x + x0)2 + a1(x + x0) + a0
thus,              y = (x + 1)3 - 3(x + 1)2 + 4(x + 1) - 2 = x3 + x
or         fs(x) = x3 +   - the source cubic function of the type 2/1 since,  a3a1 > 0.
The obtained source cubic we then move to new position by plugging the given coordinates of translations,
xt = -1 and  yt = 2,      into         y - yt = a3(x - xt )3 + a1(x - xt )
thus,      y - 2 = (x + 1)3 + (x + 1)    or     y = x3 + 3x2 + 4x + 4
as shows the below figure.
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