

Polynomial and/or Polynomial
Functions and Equations 
Polynomial functions 
Source or original polynomial function 
Translating (parallel shifting) of the polynomial function 
Coordinates of translations and their role in the polynomial
expression 
Transformations of the polynomial function applied to the
quadratic and cubic functions 





Polynomial functions 
Source
or original polynomial function 
Any
polynomial f
(x)
= a_{n}x^{n}
+ a_{n}_{}_{1}x^{n}^{}^{1}
+
.
. .
+
a_{1}x
+ a_{0}
of degree n >
1, 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 

f_{s
}(x)
= a_{n}x^{n}
+ a_{n}_{}_{2}x^{n}^{}^{2}
+
.
. .
+
a_{2}x^{2}
+ a_{1}x



has
n
 1 terms
lacking second and the constant term, since its coefficients, a_{n}_{}_{1
}=
0
and a_{0
}=
0
while
the leading coefficient a_{n},
remains unchanged. 
Therefore,
the source polynomial function passes through the
origin. 
A
coefficient a_{i
}of
the source function is expressed by the coefficients of the general
form. 
A
coefficient a_{i
}of
the source polynomial can be calculated like
coefficients of the Taylor polynomial 

where
f ^{(}^{i}^{
)}^{
}(x_{0})
denotes ith derivative at
x_{0}. 

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 x_{0}
and in the direction of the yaxis
by y_{0}. 
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 f_{s}(x),
i.e., 

y
 y_{0}
= a_{n}(x
 x_{0})^{n}
+ a_{n}_{}_{2}(x
 x_{0})^{n}^{}^{2}
+
.
. .
+
a_{2}(x
 x_{0})^{2}
+ a_{1}(x
 x_{0}) 


and
by expanding above expression we get the polynomial function in
the general form. 

Inversely, by plugging the coordinates of translations into a given polynomial function
f(x), that is expressed in the general form,
i.e., 

y
+ y_{0}
= a_{n}(x
+ x_{0})^{n}
+ a_{n}_{}_{1}(x
+ x_{0})^{n}^{}^{1}
+
.
. .
+
a_{1}(x
+ x_{0})
+ a_{0} 


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. 
Therefore,
each polynomial missing second term (a_{n}_{1}
=
0),
represents a source polynomial whose graph is translated in
the direction of the yaxis
by y_{0}
= a_{0}. 

Transformations of the polynomial function
applied to the quadratic and cubic functions 
The
application of the above theory to the quadratic and cubic
polynomial functions. 

Quadratic
function f
(x)
=
a_{2}x^{2}
+ a_{1}x
+ a_{0} 
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
=
a_{2}x^{2}
the source quadratic function 
3)
Inversely, by plugging the coordinates of translations into the source quadratic function 
y

y_{0}
= a_{2}(x

x_{0})^{2}, 

and
after
expanding and reducing we obtain 
y
=
a_{2}x^{2}
+ a_{1}x
+ a_{0} the quadratic function
in the general form. 

Cubic
function f
(x)
= a_{3}x^{3}
+
a_{2}x^{2}
+ a_{1}x
+ a_{0} 
Applying
the same method we can examine the third degree polynomial
called cubic function. 
1)
Calculate the
coordinates of translations 
substitute
n
= 3
in 



then 


2)
To
get the source cubic function we should plug the coordinates
of translations (with changed signs) 
into the general form
of the cubic,
i.e., 

after
expanding and reducing obtained is 


the source
cubic function. 

3)
Inversely, by plugging the coordinates of translations into the source
cubic 
y

y_{0}
= a_{3}(x

x_{0})^{3}
+
a_{1}(x

x_{0})^{}, 

after
expanding and reducing we obtain 
y
=
a_{3}x^{3}
+ a_{2}x^{2}
+ a_{1}x
+ a_{0} the cubic function
in the general form. 
According to mathematical induction we can examine any
ndegree polynomial function using shown method. 








College
algebra contents C




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