The transition of the cubic
function expression from the general to the translated form |
|
|
|
|
|
Elementary
functions are, Algebraic functions
and Transcendental functions |
Algebraic
functions |
· The
polynomial function
f (x) =
y = anxn
+ an-1xn-1
+ an-2xn-2
+
.
. . +
a2x2
+
a1x + 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 |
- - - -
- - - - - - - - -
- - - - - - - -
- - - - - - - - |
·
Rational
functions - a ratio of two polynomials |
|
|
-
Reciprocal function |
|
|
-
Translation of the reciprocal function,
called linear rational function. |
|
|
|
The
graphs of the
polynomial functions |
The
graph of a function f
is drawing on the Cartesian plane, plotted with respect to
coordinate axes, that shows functional relationship between
variables. The points (x,
f
(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
|
|
|
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. |
|
Cubic
function
y
=
a3x3
+
a2x2
+
a1x + a0 |
The
transition of the cubic
function expression from the general to the translated form |
Applying
the same method we can examine the third degree polynomial
called cubic function. |
1)
Calculate the
coordinates of translations |
substitute
n
= 3
in |
|
and |
|
|
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., |
y
+ y0
= a3(x
+ x0)3
+
a2(x
+ x0)2
+
a1(x
+ x0)
+
a0, |
|
after
expanding and reducing obtained is |
|
|
the source
cubic function. |
|
3)
Inversely, by plugging the coordinates of translations into the source
cubic |
y
-
y0
= a3(x
-
x0)3
+
a1(x
-
x0), |
|
after
expanding and reducing we obtain
y
=
a3x3
+ a2x2
+ a1x
+ a0 the cubic function
in the general form. |
Thus,
y
=
a3x3
+ a2x2
+ a1x
+ a0
or
y
-
y0
= a3(x
-
x0)3
+
a1(x
-
x0), |
|
by
setting x0
=
0 and y0
= 0 we get
the source cubic function
y
= a3x3
+
a1x
where a1=
tanat
. |
|
|
|
|