Graphing the polynomial functions
The source or original polynomial function
Translating (parallel shifting) of the polynomial function
Coordinates of translations and their role in the polynomial expression
Translated power function
Translated sextic function example
Graphing the polynomial functions
Translating (parallel shifting) of the polynomial function
Thus, to obtain the graph of a given polynomial function f (x) we translate (parallel shift) the graph of its source function in the direction of the x-axis by x0 and in the direction of the y-axis by y0.
Inversely, to put a given graph of the polynomial function beck to the origin, we translate it in the opposite direction, by taking the values of the coordinates of translations with opposite sign.
Coordinates of translations and their role in the polynomial expression
The coordinates of translations we calculate using the formulas,
Hence, by plugging the coordinates of translations into the source polynomial function fs(x), i.e.,
 y - y0 = an(x - x0)n + an-2(x - x0)n-2 + . . .  + a2(x - x0)2 + a1(x - x0)
and by expanding above expression we get the polynomial function in the general form
f (x) =  yanxn + an-1xn-1 + an-2xn-2 + . . . + a2x2 + a1x + a0.
Inversely, by plugging the coordinates of translations into the given polynomial f (x) expressed in the general form, i.e.,
 y + y0 = an(x + x0)n + an-1(x + x0)n-1 + . . .  + a1(x + x0) + a0
and after expanding and reducing above expression we get its source polynomial function.
Note that in the above expression the signs of the coordinates of translations are already changed.
Translated monomial (or power) function
If we set all coefficients, an-2  to a1, in the above expanded form of the polynomial to zero, we get
y - y0 = an(x - x0)n,   x0- an-1/( n · an) and  y0 = f (x0).
the translated power (or monomial) function, the exponent of which is an odd or an even positive integer.
When the exponent is even, i.e., of the form n = 2m,  m Î N, the graph of the source power function
y = anxn is symmetric about the y-axis, that is  f (-x) = f (x).
When the exponent is odd, i.e., of the form n = 2m + 1,  m Î N, the graph of the source power function
y = anxn  is symmetric about the origin, that is  f (-x) = -f (x).
Example:  Given is sextic  y = (1/4)x6 - 6x5 + 60x4 - 320x3 + 960x2 - 1536x + 1008, find its source or original function and calculate the coordinates of translations, the zero points and the turning point.
Draw graphs of the source and the given sextic.
Solution:  1)  Calculate the coordinates of translations
y0 = f (4) = (1/4) · 46 - 6 · 45 + 60 · 44 - 320 · 43 + 960 · 42 - 1536 · 4 + 1008,     y0 = - 16.
2)  To get the source sextic, plug the coordinates of translations into the general form of the given sextic, to draw its graph back to the origin,
y + y0 = a6(x + x0)6 + a5(x + x0)5 + a4(x + x0)4 + a3(x + x0)3 + a2(x + x0)2 + a1(x + x0) + a0  or
y - 16 = (1/4)(x - 4)6 - 6(x - 4)5 + 60(x - 4)4 - 320(x - 4)3 + 960(x - 4)2 - 1536(x - 4) + 1008,
after expanding and reducing above expression obtained is
y = (1/4) x6   - the source sextic.
Since all the coefficients, a4, a3, a2 and a1, of the source sextic y = a6x6 + a4x4 + a3x3 + a2x2 + a1x, are zero then, the only turning point is T(x0, y0)  or  T( 4, - 16).
3)  Inversely, by plugging the coordinates of translations into the source sextic
y - y0 = a6(x - x0)6    or     y - 16 = (1/4)(x - 4)6
what after expanding yields
y = (1/4) x6 - 6x5 + 60x4 - 320x3 + 960x2 - 1536x + 1008
the given sextic. Therefore, the given sextic is translated monomial or power function.
As the translated monomial or power function has zeros if  a6 ·  y0 < 0 then.
Calculus contents A