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| Polynomial function |
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· The
polynomial function
f (x) =
y = an
xn
+ an-1
xn-1
+ an-2
xn-2
+
.
. . +
a2
x2
+
a1 x + 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 |
|
- - - -
- - - - - - - - -
- - - - - - - -
- - - - - - - - |
|
| 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)
= an
xn
+ an-2xn-2
+
.
. . +
a2x2
+ a1
x |
|
|
| 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) =
y = an
xn
+ an-1
xn-1
+ an-2
xn-2
+
.
. . +
a2
x2
+
a1 x + 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. |
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| By
setting the coefficients a2
and a1
of the source quartic to zero interchangeably, we get the
basic classification of the quartic function. |
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