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Graphing a cubic
function
examples |
Graphing a cubic
function type
2/1 |
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The
graphs
of the cubic functions |
There are three types
(shapes of the graphs) of cubic
functions: |
type
1 |
y
=
a3x3
+ a2x2
+ a1x
+ a0
or y
-
y0
= a3(x
-
x0)3,
-
a22
+ 3a3a1
= 0 or a1
= 0. |
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therefore,
its source function y
=
a3x3,
and the tangent line through the point of
inflection is horizontal. |
type
2/1 |
y
=
a3x3
+ a2x2
+ a1x
+ a0
or
y
-
y0
= a3(x
-
x0)3
+
a1(x
-
x0),
where a3a1>
0 |
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whose
slope of the tangent line through the point of inflection is
positive and equals a1. |
type
2/2 |
y
=
a3x3
+ a2x2
+ a1x
+ a0
or
y
-
y0
= a3(x
-
x0)3
+
a1(x
-
x0),
where a3a1<
0 |
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whose
slope of the tangent line through the point of inflection is
negative and is equal a1. |
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Graphing
translated
cubic function type
2/1 |
type
2/1 |
y
=
a3x3
+ a2x2
+ a1x
+ a0 or
y
-
y0
= a3(x
-
x0)3
+ a1(x
-
x0),
a3
· a1
>
0, |
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I
(x0,
y0). |
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Example: Find the coordinates of
translations, the zero point, the point of inflection and draw
graphs of
the cubic function y
=
-
x3
+ 3x2
-
5x
+
6
and its source function. |
Solution:
1)
Calculate the coordinates of translations
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y0
= f(x0)
=> y0
= f(1)
=
-13
+ 3 · 12
-
5
· 1
+
6 =
3,
y0
= 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
+ y0
=
a3(x
+
x0)3
+ a2(x
+
x0)2
+ a1(x
+
x0)
+ a0 |
thus,
y
+ 3
=
-1·
(x
+
1)3
+
3 · (x
+
1)2
-
5
· (x
+
1)
+
6 => y
=
-
x3 -
2x
the source
function. |
Let
prove that the source cubic is an odd function, which means
that f(x)
=
-f(-x), |
since
f (x)
=
-
x3 -
2x
then
-f
(-x)
=
- [-
(-x)3
-
2(-x)]
=
- [
x3
+ 2x]
=
-
x3 -
2x. |
As,
a3
=
-
1 and
a1
=
-
2
then a3a1 >
0
therefore, given function is of the type 2/1. |
Since given function is symmetric to its
point of inflection, and as the y-intercept
a0
=
6, then the x-intercept
or zero of the function must be at the point (2,
0) as
shows the below figure. |
To
prove the root plug the values, a3
=
-
1,
a1
=
-
2,
x0
=
1 and
y0
=
3 into |
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so
x1
= 2, therefore
-
x3
+ 3x2
-
5x
+
6
= -1(x
-
2)(x2
-
x
+
3). |
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Pre-calculus contents
E |
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