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ALGEBRA
:: Quartic function, the fourth degree polynomial  f (x) = a4x4 + a3x3 + a2x2 + a1x + a0
Transformation of the quartic polynomial from the general to the source form
To get the source quartic function we plug the coordinates of translations, (with changed signs) into general form of the quartic polynomial,
y + y0 = a4(x + x0)4 + a3(x + x0)3 + a2(x + x0)2 + a1(x + x0) + a0,
after expanding and reducing obtained is the source quartic function The basic classification criteria applied to the source quartic polynomial shows the diagram Thus, there are ten types (different shapes of graphs) of quartic functions. Applying additional criteria defined are the conditions remaining six types of the quartic polynomial functions to appear.

Observe that the basic criteria of the classification separates even and odd nth degree polynomials called the power functions or monomials as the first type, since all coefficients a of the source function vanish, (see the above diagram).

Therefore, the first type of the qurtic polynomial
 type 1
y = a4x4 + a3x3 + a2x2 + a1x + a0    or    y - y0 = a4(x - x0)4a2 = 0 and a1 = 0. If  a4 · y0 < 0  then the roots are The turning point  T(x0 , y0 ).
:: Graph of the power function - translated power or monomial functions
Thus, we write the translated power function or the polynomial of the first type,
y - y0 = an(x - x0)n,
where  x0 - an - 1/(n·an),   y0 = f (x0)  and n is an even or an odd positive integer.
 For n = 2m, m Î N the even power function has the turning point  T (x0, y0). The real roots, as shows the right figure (where an > 0 ). For  n = 2m + 1, m Î N the odd power function has, the point of inflection I (x0, y0),  at = 0. The real root, as shows the right figure (where an > 0 ). :: Drawing translated power or monomial function example
 Example:  Given sextic polynomial  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 function. Solution:  1)  Calculate the coordinates of translations, x0 = - an - 1/(n·an) = - a5/(6·a6) = 4,   x0 = 4   and   y0 =  f (x0) = - 16. 2)  To get the source function, plug  x0  and  y0  with changed signs, into the general form of the sextic 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 after expanding and reducing we get,   y = (1/4) x6   its source function. 3)  Inversely, by plugging the coordinates of translations into the source function y - y0 = a6(x - x0)6     or     y - 16 = (1/4)(x - 4)6 that, after expanding yields given sextic polynomial. As the translated monomial or power function has zeros or roots,  if  a6 ·  y0 < 0 then, The only turning point is  T(x0, y0)  or  T( 4, - 16).    Contents A 