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Translated
sextic function example
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| The
graphs
of algebraic
and transcendental functions |
| Elementary
functions are, Algebraic functions
and Transcendental functions |
| Algebraic
functions |
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· 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 |
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y
=
a4x4
+
a3x3
+
a2x2
+
a1x + a0
-
Quartic
function |
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y
=
a5x5
+ a4x4
+
a3x3
+
a2x2
+
a1x + a0
-
Quintic
function |
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- - - -
- - - - - - - - -
- - - - - - - -
- - - - - - - - |
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Rational
functions - a ratio of two polynomials |
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Reciprocal function |
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Translation of the reciprocal function,
called linear rational function. |
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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. |
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| 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. |
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The source polynomial function |
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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. |
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| 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. |
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| Coordinates of translations
and their role in the polynomial expression |
| The
coordinates of translations we calculate using the formulas, |
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| 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) |
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| 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 |
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| 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. |
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| Translated
monomial (or
power) function |
| If
we set all coefficients, an-2
to a1,
in the above expanded form of the
polynomial to zero, we get |
| y
- y0
= an(x
- x0)n,
x0
= -
an-1/(
n · an)
and y0
=
f(x0). |
| 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
= anxn
is symmetric about the y-axis,
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
= anxn
is symmetric about the origin, that
is f(-x)
=
-f(x). |
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