

Graphing 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 
Translated
power function 
Translated
sextic function example







Graphing the
polynomial functions 
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 xaxis
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 
f
(x) =
y = a_{n}x^{n}
+ a_{n}_{1}x^{n}^{}^{1}
+ a_{n}_{}_{2}x^{n}^{}^{2}
+
.
. . +
a_{2}x^{2}
+
a_{1}x + a_{0}. 
Inversely, by plugging the coordinates of translations into
the given polynomial f
(x)
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. 

Translated
monomial (or
power) function 
If
we set all coefficients, a_{n}_{}_{2}
to a_{1},
in the above expanded form of the
polynomial to zero, we get 
y
 y_{0}
= a_{n}(x
 x_{0})^{n},
x_{0}
= 
a_{n}_{1}/(
n · a_{n})
and y_{0}
=
f
(x_{0}). 
the
translated
power (or monomial) function, the exponent of which is an odd or
an even positive integer. 
When
the exponent is even, i.e., of the form n
= 2m,
m Î N,
the graph of the source power function 
^{}y
= a_{n}x^{n}
is symmetric about the yaxis,
that is f (x)
=
f (x). 
When
the exponent is odd, i.e., of the form n
= 2m
+ 1,
m Î N,
the graph of the source power function 
y
= a_{n}x^{n}
is symmetric about the origin, that
is f (x)
=
f
(x). 


Example:
Given is sextic y
= (1/4)x^{6} 
6x^{5}
+
60x^{4}

320x^{3}
+
960x^{2}

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. 
Solution:
1)
Calculate the coordinates of translations 

y_{0}
= f (4)
= (1/4)
·
4^{6}

6 ·
4^{5}
+
60
·
4^{4}

320 ·
4^{3}
+
960
·
4^{2}

1536 ·
4
+
1008,
y_{0}
= 
16. 
2)
To get the source sextic, plug the coordinates of translations
into the general form of the given sextic, to draw
its graph back to the origin, 
y
+
y_{0}
= a_{6}(x
+
x_{0})^{6}
+
a_{5}(x
+
x_{0})^{5}
+
a_{4}(x
+
x_{0})^{4}
+
a_{3}(x
+ x_{0})^{3}
+
a_{2}(x
+ x_{0})^{2}
+
a_{1}(x
+ x_{0})
+
a_{0}
or 
y

16 = (1/4)(x

4)^{6} 
6(x

4)^{5}
+
60(x

4)^{4} 
320(x

4)^{3}
+
960(x

4)^{2}

1536(x

4)
+
1008, 
after
expanding and reducing above expression obtained is 
y
= (1/4)
x^{6 } the
source sextic. 
Since
all the coefficients, a_{4},
a_{3},
a_{2}
and a_{1},
of the source sextic y
=
a_{6}x^{6}
+ a_{4}x^{4}
+ a_{3}x^{3}
+ a_{2}x^{2}
+ a_{1}x,
are
zero then, the only turning point is T(x_{0}, y_{0})
or T(
4, 
16). 
3)
Inversely, by plugging the coordinates of translations into the source
sextic 
y
 y_{0}
= a_{6}(x
 x_{0})^{6}
or
y
 16
= (1/4)(x 
4)^{6} 
what
after expanding yields 
y
= (1/4)
x^{6} 
6x^{5}
+
60x^{4}

320x^{3}
+
960x^{2}

1536x
+
1008 
the
given sextic. Therefore, the given sextic is translated monomial or
power function. 
As
the translated monomial or power function has zeros if a_{6}
· y_{0}
< 0 then. 










Calculus
contents A 



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