

ALGEBRA


::
Quartic
function, the fourth degree polynomial f
(x) = a_{4}x^{4}
+ a_{3}x^{3} + a_{2}x^{2}
+ a_{1}x + a_{0}

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
+ y_{0}
= 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},

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 n^{th}
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


y
= a_{4}x^{4}
+ a_{3}x^{3}
+
a_{2}x^{2}
+
a_{1}x + a_{0}_{
}or_{ }y

y_{0}
=
a_{4}(x

x_{0})^{4},
a_{2}
= 0 and a_{1}
=
0. 


If
a_{4
}·
y_{0}_{
}<
0
then
the roots are 

The turning point T(x_{0}
, y_{0}
).




:: 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
 y_{0}
= a_{n}(x
 x_{0})^{n},

where x_{0}
= 
a_{n}_{
}_{}_{
}_{1}/(n·a_{n}),
y_{0}
=
f
(x_{0})
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
(x_{0},
y_{0}). 
The real
roots, 

as
shows the right figure (where a_{n}
> 0 ). 




For
n
= 2m + 1, m Î
N the odd power function 
has,
the point of inflection I
(x_{0},
y_{0}),
a_{t}
=
0. 
The real
root, 

as
shows the right figure (where a_{n}
> 0 ). 





:: Drawing translated power or monomial
function example

Example:
Given sextic
polynomial 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 function. 
Solution:
1)
Calculate the coordinates of translations,

x_{0}
= 
a_{n}_{
}_{}_{
}_{1}/(n·a_{n})
= 
a_{5}/(6·a_{6})
= 4, x_{0}
= 4
and y_{0}_{
}= f
(x_{0})_{
}= 
16. 
2)
To get the source function, plug x_{0}
and y_{0}_{ }
with changed signs, into the general form of the sextic 
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} 
after
expanding and reducing we get, y
= (1/4)
x^{6 }its
source function. 
3)
Inversely, by plugging the coordinates of translations into
the source function 
y
 y_{0}
= a_{6}(x
 x_{0})^{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
a_{6}
· y_{0}
<
0 then, 

The only turning point is
T(x_{0}, y_{0})
or T(
4, 
16). 















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