
Polynomial and/or Polynomial
Functions and Equations 
Cubic functions 
Graphs
of cubic functions 
Graphing a cubic function,
example 





Cubic
functions 
Using the same method we can analyze the third degree polynomial or cubic functions. 
y
=
a_{3}x^{3}
+ a_{2}x^{2}
+ a_{1}x
+ a_{0 }or_{ }
y

y_{0}
= a_{3}(x

x_{0})^{3}
+
a_{1}(x

x_{0}), 

By
setting x_{0}
=
0 and y_{0}
= 0 we get
the source cubic function
y
= a_{3}x^{3}
+
a_{1}x
where a_{1}=
tana_{t} 
Coordinates
of the point of inflection coincide with the coordinates of
translations, i.e., I
(x_{0},
y_{0}).

The source cubic functions are
odd functions. 
Graphs of odd functions are
symmetric about the origin that is, such functions change
the sign but not absolute value when the sign of the independent variable is
changed, so that f
(x)
=

f (x). 
That
is, change of the sign of the independent variable of a function
reflects the graph of the function about the yaxis,
while change of the sign of a function reflects the graph of the
function about the xaxis. 
The
graphs of
the translated cubic functions are symmetric about its
point of inflection. 

As
shows the figure below there are three types of cubic
functions: 
type
1 
y
=
a_{3}x^{3}
+ a_{2}x^{2}
+ a_{1}x
+ a_{0 }or_{ }
y

y_{0}
= a_{3}(x

x_{0})^{3
}where
a_{1}
= 0 


therefore,
its source function y
=
a_{3}x^{3},^{
}and the tangent line
through the point of
inflection is horizontal. 
type
2/1 
y
=
a_{3}x^{3}
+ a_{2}x^{2}
+ a_{1}x
+ a_{0 }or_{ }
y

y_{0}
= a_{3}(x

x_{0})^{3}
+
a_{1}(x

x_{0}),
where a_{3}a_{1}>
0 

whose
slope of the tangent line through the point of inflection is
positive and equal a_{1}. 
type
2/2 
y
=
a_{3}x^{3}
+ a_{2}x^{2}
+ a_{1}x
+ a_{0 }or_{ }
y

y_{0}
= a_{3}(x

x_{0})^{3}
+
a_{1}(x

x_{0}),
where a_{3}a_{1}<
0 

whose
slope of the tangent line through the point of inflection is
negative and is equal a_{1}. 
The
graph of its source function has three zeros and two turning
points at 



Graphs
of cubic functions 


Graphing a cubic function,
example 
Example:
Find the coordinates of
translations, the zero point, the point of inflection and draw
graphs of
the 
cubic function
y
=

x^{3}
+ 3x^{2
}
5x
+
6
and its source function. 
Solution:
1)
Calculate the coordinates of translations


y_{0}
= f
(x_{0})
=> y_{0}
= f(1)
=
1^{3}
+ 3 · 1^{2}

5
· 1
+
6 =
3,
y_{0}
= 3 
Therefore,
the point of inflection I
(1,
3). 
2)
To
get the source cubic function, plug the coordinates
of translations into the general
form
of the cubic, 
y
+ y_{0}
=
a_{3}(x
+
x_{0})^{3
}
+ a_{2}(x
+
x_{0})^{2}
+ a_{1}(x
+
x_{0})
+ a_{0} 
thus,
y
+ 3
=
1·
(x
+
1)^{3
}+
3 · (x
+
1)^{2}

5
· (x
+
1)
+
6 => y
=

x^{3 }
2x
the source
function. 
Let
prove that the source cubic is an odd function, which means that
f (x)
=

f
(
x), 
Since
f (x) =

x^{3 }
2x
then

f (x)
=
 [
(x)^{3
}^{ }
2(x)]
=
 [
x^{3}^{
}
+ 2x]
=

x^{3 }
2x. 
As
a_{3}a_{1 }> 0
given function is of the type 2/1. 
Since given function is symmetric to its
point of inflection, and as the yintercept
a_{0}
=
6, then the xintercept
or zero of the function must be at the point (2,
0). 
Therefore,

x^{3}
+ 3x^{2
}
5x
+
6
=  (x

2)(x^{2
}

x
+
3). 









College
algebra contents C




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