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| 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
|
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|
| 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 = anxn
+ 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) =
y = anxn
+ 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 y0
and the coefficients a
of its source function. |
|
| Linear
function y
= a1x
+ a0 |
| y
= a1x
+ a0
or y
= a1(x
-
x0)
or y
-
y0
= a1x, |
| where
x0
=
-
a0/a1
and/or y0
= a0. |
| 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
-
y0
= a2(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
).
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