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
Sigma notation of the polynomial
The graph of the linear function
The graph of the quadratic 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, ƒ (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.
Sigma notation of the polynomial
Coefficients of the source polynomial in the form of a recursive formula
According to mathematical induction we can examine any n-degree polynomial function using shown method.
Therefore, the polynomial   f (x) =  yanxn + an-1xn-1 + an-2xn-2 + . . .  + a2x2 + a1x + a0
we can write as
while,  for   k = 0,            an = an,
and from which, for  k = n,            a0 = f (x0) = y0.
Thus, expanded form of the above sum is
y - y0 = an(x - x0)n + an-2(x - x0)n-2 + . . . + a2(x - x0)2 + a1(x - x0)
where x0 and y0 are coordinates of translations of the graph of the source polynomial
fs(x) = anxn + an-2xn-2 + . . . + a2x2 + a1x
in the direction of the x-axis and the y-axis of a Cartesian coordinate system.
Therefore, every given polynomial written in the general form can be transformed into translatable form by calculating the coordinates of translations x0 and y and the coefficients a of its source function.
Linear function   ya1x + a0
ya1x + a0  or  ya1(x - x0)  or  y - y0a1x,
where  x0 - a0/a1  and/or   y0a0.
By setting  x0 = 0  or  y0 = 0  obtained is
 the source linear  y = a1x.
Quadratic function    y = a2x2 + a1x + a0
1)  Let calculate the coordinates of translations of quadratic function using the formulas,
 substitute n = 2 in
 then
2)  To get the source quadratic function we should plug the coordinates of translations (with changed signs)
into the general form of the quadratic, i.e.,
after expanding and reducing obtained is
y = a2x2   the source quadratic function.
3)  Inversely, by plugging the coordinates of translations into the source quadratic function
y - y0 = a2(x - x0)2,
and after expanding and reducing we obtain
y = a2x2 + a1x + a0   the quadratic function in the general form.
ƒ (x) = y = a2x2 + a1x + a0   or   y - y0a2(x + x0)2,
 where
By setting  x0 = 0  and  y0 = 0  obtained is
 the source quadratic  y = a2x2 .
 If  a2 · y0 < 0  then the roots are
The turning point  V(x0 , y0 ).
College algebra contents B