The graphs of 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
         Coefficients of the source polynomial function are related to its derivative at x0
Elementary functions are,   Algebraic functions and Transcendental functions
Algebraic functions
  · The polynomial function   f (x) =  yanxn + an-1xn-1 + an-2xn-2 + . . . + a2x2 + a1x + a0
                                                    y a1x + a0                                                   - Linear function 
                                                    y = a2x2 + a1x + a0                                                      - Quadratic function 
                                                    y = a3x3 + a2x2 + a1x + a0                                       - Cubic function
                                                    y = a4x4 + a3x3 + a2x2 + a1x + a0                        - Quartic function
                                                    y = a5x5 + a4x4 + a3x3 + a2x2 + a1x + a0         - Quintic function
                                                    -  -  -  -  -  -  -  -  -  -  -  -  -  -  -  -  -  -  -  -  -              -  -  -  -  -  -  -  -  
 · Rational functions - a ratio of two polynomials  
- Reciprocal function
  - Translation of the reciprocal function,     called linear rational function.  
The graphs of the polynomial functions
The graph of a function  f is drawing on the Cartesian plane, plotted with respect to coordinate axes, that shows functional relationship between variables. The points (x, f (x)) lying on the curve satisfy this relation.
The source or original polynomial function
Any polynomial  f (x) of degree n > 1 in the general form, consisting of n + 1 terms, shown graphically, represents translation of its source (original) function in the direction of the coordinate axes.
The source polynomial function
fs (x) = anxn + an-2xn-2 + . . . + a2x2 + a1x
has n - 1 terms lacking second and the constant term, since its coefficients, an-1 = 0 and a0 = 0 while the leading coefficient an, remains unchanged.
Therefore, the source polynomial function passes through the origin.
A coefficient ai of the source function is expressed by the coefficients of the general form.
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.
Coefficients of the source polynomial function are related to its derivative at x0
The coefficients of the source polynomial are related to corresponding value of its derivative at x0 like the coefficients of the Taylor polynomial in Taylor's or Maclaurin's formula, thus
Such for example, the coefficient a1 of the source cubic of   f (x) = a3x3 + a2x2 + a1x + a0 
since   f ' (x) = 3a3x2 + 2a2x + a1   and  x0 = - a2/(3a3)  then 
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