
The
graphs of the
polynomial functions 
The
source
or original polynomial function 
Translating
(parallel shifting) of the polynomial function 
Coordinates of translations
and their role in the polynomial expression 
The
graph of the quadratic 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 

f_{s}
(x)
= a_{n}_{
}x^{n}
+ a_{n}_{}_{2}_{
}x^{n}^{}^{2}
+
.
. . +
a_{2}_{
}x^{2}
+ a_{1}x



has
n
 1 terms
lacking second and the constant term, since its coefficients, a_{n}_{}_{1
}=
0
and a_{0
}=
0
while
the leading coefficient a_{n},
remains unchanged. 
Therefore,
the source polynomial function passes through the
origin. 
A
coefficient a_{i
}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 xaxis
by x_{0}
and in the direction of the yaxis
by y_{0}. 
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 f_{s}(x),
i.e., 

y
 y_{0}
= a_{n}_{
}(x
 x_{0})^{n}
+ a_{n}_{}_{2}_{
}(x
 x_{0})^{n}^{}^{2}
+
.
. .
+
a_{2}_{
}(x
 x_{0})^{2}
+ a_{1}_{
}(x
 x_{0}) 


and
by expanding above expression we get the polynomial function in
the general form 
f
(x) =
y = a_{n
}x^{n}
+ a_{n}_{1
}x^{n}^{}^{1}
+ a_{n}_{}_{2}_{
}x^{n}^{}^{2}
+
.
. . +
a_{2}_{
}x^{2}
+
a_{1 }x + a_{0}. 
Inversely, by plugging the coordinates of translations into
the given polynomial f(x)
expressed in the general form,
i.e., 

y
+ y_{0}
= a_{n}_{
}(x
+ x_{0})^{n}
+ a_{n}_{}_{1}_{
}(x
+ x_{0})^{n}^{}^{1}
+
.
. .
+ a_{1}_{
}(x
+ x_{0})
+ a_{0} 


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. 

Quadratic
function y
=
a_{2}x^{2}
+
a_{1}x + a_{0} 
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
=
a_{2}x^{2}
the source quadratic function. 
3)
Inversely, by plugging the coordinates of translations into the source quadratic function 
y

y_{0}
= a_{2}(x

x_{0})^{2}, 

and
after
expanding and reducing we obtain 
y
=
a_{2}x^{2}
+ a_{1}x
+ a_{0}
the quadratic function
in the general form. 
ƒ(x)
=
y
=
a_{2}x^{2}
+
a_{1}x + a_{0}
or y

y_{0}
= a_{2}(x
+ x_{0})^{2}, 
where 


By
setting x_{0}
= 0 and y_{0}
= 0
obtained is 
the
source quadratic
y
= a_{2}x^{2}
. 

If
a_{2
}·
y_{0}
< 0
then
the roots are 


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













Calculus
contents A 



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