The graphs of the elementary functions
The graphs of algebraic functions
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
The graphs of the quartic function
The graphs of algebraic and transcendental functions
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 ƒ 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.
Quartic function    y = a4x4 + a3x3 + a2x2 + a1x + a0
1)  Calculate the coordinates of translations by plugging n = 4 into
2)  To get the source quartic function we should plug the coordinates of translations (with changed signs)
into the general form of the quartic, i.e.,
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
3)  Inversely, by plugging the coordinates of translations into the source quartic
y - y0 = a4(x - x0)4 + a2(x - x0)2 + a1(x - x0),
after expanding and reducing we obtain
y = a4x4 + a3x3 + a2x2 + a1x + a0   the quartic function in the general form.
 Thus,      y = a4x4 + a3x3 + a2x2 + a1x + a0    or    y - y0 = a4(x - x0)4 + a2(x - x0)2 + a1(x - x0), by setting  x0 = 0  and  y0 = 0 we get the source quartic   y = a4x4 + a2x2 + a1x.
By setting the coefficients a2 and a1 of the source quartic to zero, interchangeably, obtained is the basic classification shown in the diagram.
There are ten types (shapes of the graphs) of quartic functions.
 type 1 y = a4x4 + a3x3 + a2x2 + a1x + a0    or    y - y0 = a4(x - x0)4,  a2 = 0 and a1 = 0.
 The zeroes or roots:
 type 2 y = a4x4 + a3x3 + a2x2 + a1x + a0    or    y - y0 = a4(x - x0)4 + a1(x - x0),  a2 = 0.
 The zeroes of the source function:
 The zeroes of the translated function we get by adding x0 to the solution of the equation   a4x4 + a1x + y0 = 0.
 type 3 y = a4x4 + a3x3 + a2x2 + a1x + a0    or    y - y0 = a4(x - x0)4 + a2(x - x0)2,  a1 = 0.
 type 3/1 a4·a2 > 0
 type 3/2 a4·a2 < 0
 type 3/1 a4·a2 > 0
T (x0, y0).
 type 3/2 a4·a2 < 0
Remaining six types of quartic polynomial satisfy the criteria shown in the diagram below.
The roots of the source quartic  y = a4x4 + a2x2 + a1x  Types, 4/1, 4/2, 4/3 and 4/4
the roots of the Types, 4/5 and 4/6
The abscissa of the turning point of the Types, 4/1, 4/2 and 4/3
the abscissas of the turning points of the Types, 4/4, 4/5 and 4/6
 The abscissas of the points of inflection of the source quartic of  Types 4/2 to 4/6,
 The roots of the translated quartic Type 4 we get by adding x0 to the solutions of the equation a4x4 + a2x2 + a1x + y0 = 0.
Intermediate algebra contents