Cubic
Functions
NCS Mathematics
DVD Series
Outcomes for this DVD
In this DVD we will:
•
Apply Remainder and Factor Theorems.
LESSON 1
•
Finding Intercepts, Turning- and Inflection Points of Cubic
Functions.
LESSON 2
•
Sketch Cubic Functions.
LESSON 3
•
Make deductions from given Graphs of Cubic Functions.
LESSON 4
Lesson 1
Application of
Remainder and
Factor Theorems
NCS Mathematics
DVD Series
Outcomes for this Lesson
In this lesson we will :
 Formulate the Remainder and Factor Theorems.
 Determine at least one linear factor for a cubic
expression by means of the Factor Theorem.
 Determine the quadratic quotient when a cubic
expression is divided by a linear factor by means of :
Long Division or
Inspection.
 Utilize Factor Theorem and Quotient strategies
to Factorize Cubic Expressions.
Remainder by means of long Division
and Division Terminology
2x 2
x2
1
2 x3  4 x 2  x  5
2x  4x
3
2
x 5
x2
 Remainder
 3
3


Furthermore g  x   2 x3  4 x 2  x  5  2 x 2  1  x  2   3
Polynomial = Quotient × Divisor + Remainder
Remainder Theorem
Remainder Theorem
If q  x  is the quotient when a polynomial f  x 
 a
is divided by ax  b then the remainder R  f b
Proof
f  x    ax  b   q  x   R
 ax  b  0  x  b 
a

 a   a  b a  b  q  b a   R
Then f b
 a  R  0 q b a  R  0  R  R
 b  b   q b
 a
R  f b
Factor Theorem
If  x  p  is a factor of f  x  , then
Remainder  R  f  p   0
Example:
Prove that 2 x  3; x  3 and 3x  2
Solution
3
 2
From Remainder Theorem: Remainders are g   ; g  3 and g   
2
 3
are factors of g  x   6 x3  23x 2  9 x  18
 3
27
9
3
 g  2   6  8  23  4  9  2  18  0
 


 g  3   6  27  23  9  9  3  18  0

2
8 
4

 2
g 
 9      18  0
    6 
  23 
9

 27 
 3
  3
It follows from factor theorem that 2 x  3; x  3 and 3x  2
are factors of g  x   6 x3  23x 2  9 x  18
Factorization of a Cubic Expression
Factorize h  x   12 x  19 x  45x  18
3
2
 Linear factors must be linked to coefficients of first or last terms.
Linear factors will be of the format ax  b
where a can be equal to 1; 2; 3; 4; 6 or 12 and
b can be equal to  1;  2;  3;  6;  9 or  18
 Utilize the factor theorem to determine at least one linear factor.
h   1  0; h   2   0 but h  3  0   x  3 is a factor of h  x .
 Utilize Long Division or Inspection Strategy to determine
the Quadratic Quotient.  Will be discussed in the next slides 

 h  x    x  3 12 x 2  17 x 2  6

 h  x    x  3 3x  2  4 x  3
Factorize trinomial.
Finding the Quadratic Quotient
by means of Long Division
We mentioned in the previous slide that


h  x   12 x3  19 x 2  45 x  18   x  3 12 x 2  17 x  6   x  3  q  x 
In addition q  x  can be determined by means of Long Division.
12x 2 17x  6
12 x3  19 x2  45x  18
x3
12 x3  36 x2
17 x2  45x
17 x2  51x
 q  x   12 x  17 x  6
2
6 x  18
6 x  18
Finding the Quadratic Quotient
by means of Inspection Strategy
We mentioned earlier that


h  x   12 x3  19 x 2  45 x  18   x  3 12 x 2  17 x  6   x  3  q  x 
In addition q  x  can be determined by means of Inspection Strategy.
12 x  19 x  45x  18   x  3  q  x 
3
2
12 x3  19 x 2  45x  18   x  3  ax 2  bx  c 
By inspection a  12 and c  6.
12 x3  19 x 2  45x  18   x  3 12 x 2  bx  6 
By further inspection 6  3b  45  3b  51  b  17.
 q  x   12 x 2  17 x  6
Tutorial 1: Factorization of Cubic Expressions
Given f  x   2 x  x  5x  2 :
3
2
1) Determine one linear factor by means
of the Factor Theorem.
2) Determine the quadratic quotient by
means of the inspection strategy.
3) Check your answer in (3) by means
PAUSE DVD
of long division.
•
4) Factorize the cubic expression. •
Do Tutorial 1
Then View solutions
Tutorial 1 Problem 1:
Suggested Solution
Determine one linear factor for f  x   2 x3  x 2  5 x  2
by means of the factor theorem.
Possible factors :  x  1 ;  x  2 ;  2 x  1
Test a variety of the 6 possibilities.
f  1  2  1  5  2  0   x  1 is not a factor of f  x 
f 1  2  1  5  2  0   x  1 is a factor of f  x 
Tutorial 1 Problem 2:
Suggested Solution
We know that :


f  x   2 x3  x 2  5 x  2   x  1 ax 2  bx  c   x  1  q  x 
q  x  can be determined by means of Inspection Strategy.
 2 x  x  5x  2   x  1  ax  bx  c 
3
2
2
By inspection a  2 and c  2.
 2 x3  x 2  5x  2   x  1  2 x 2  bx  2 
By further inspection  2  b  5  b  3  b  3.
 q  x   2 x  3x  2
2
Tutorial 1 Problem 3:
Suggested Solution
We know that


f  x   2 x3  x 2  5 x  2   x  1 2 x 2  3x  2   x  1  q  x 
q  x  can be determined by means of Long Division.
2x 2  3x  2
2 x3  x 2  5 x  2
x 1
2 x3  2 x 2
3x 2  5 x
3x 2  3x
 q  x   2 x  3x  2
2
2 x  2
2 x  2
Tutorial 1 Problem 4:
Suggested Solution
We know that

f  x   2 x 3  x 2  5 x  2   x  1 2 x 2  3x  2

Factorize the trinomial :
f  x   2 x  x  5 x  2   x  1  2 x  3x  2 
3
2
2
  x  1 x  2 2 x  1
Alternatively we can use the Factor Theorem
to determine two factors of f  x  :
Can show that :
 f  x   2 x3  x 2  5 x  2
f 1  0   x  1 is a factor and
  x  1 x  2  2 x  1
f  2   0   x  2  is also factor. Third factor by inspection.
Lesson 2
Finding:
 Intercepts
 Turning Points
 Inflection Points
NCS Mathematics
DVD Series
X- and Y- Intercepts
In a previous lessons on graphs of functions you have learnt that the
intercepts on the axes can be determined as follows :
The y - intercept of a function can be found where x  0.
i.e.
at the point  0, f (0)  .
The x-intercepts of a function
can be found where y  0.
y  x 2  x  6   x  3 x  2 
In the case of cubic functions (polynomial of degree 3)
the x-intercepts can be determined by using either
Factorisation or the Factor Theorem.
Example 1 : X- and Y- Intercepts
EXAMPLE : Determine the intercepts on the axes of the
following cubic function:
f ( x)  x 3  4 x 2  4 x
y - intercept :
i.e. y  f (0)  (0)3  4(0) 2  4(0)  0
x - intercepts :
i.e.
f ( x)  0  x3  4 x 2  4 x  0
Use factorisation.

x ( x 2  4 x  4)  0


x ( x  2) 2  0
x  0
or
x  2
Example 2 : X- and Y- Intercepts
EXAMPLE :
Determine the intercepts on the axes of the
following cubic function: g ( x)  x3  4 x 2  x  6
y - intercept :
i.e.
y  g (0)  (0)3  4(0)2  (0)  6  6
x - intercepts :
i.e.
g( x)  0
 x3  4 x 2  x  6  0
Use the Factor Theorem :
g (1)  (1)3  4(1)2  1  6  0
 ( x  1) is a factor.




g ( x )  ( x  1)( x 2  5 x  6)
 ( x  1)( x  2)( x  3)
( x  1)( x  2)( x  3)  0
x 1
or
x  2
or
x  3
Positive Gradients
Implies Function is Increasing
In a previous lesson on calculus you have learnt that the
gradient of the tangent to a curve of a function can be
determined by using the first derivative.
At a point where a curve is sloping upwards the gradient of
the tangent is positive and the y -value is increasing as
the x-value increases.
i.e.
If f / (a )  0 , then f ( x ) is increasing at x  a
Negative Gradients
Implies function is decreasing
We have also learnt that:
At a point where a curve is sloping downwards the gradient of
the tangent is negative and the y  value is decreasing as
the x  value increases.
i.e.
If f / (a )  0 , then f ( x) is decreasing at x  a
Zero Gradient
We have also learnt that:
The turning point is where the gradient of the tangent
is zero and the y - value is neither increasing or
decreasing i.e stationary.
i.e.
If f / (a )  0 , then f ( x) is stationary at x  a
Turning Points and Gradients
A turning point is a stationary point where the curve smoothly
turns from increasing to decreasing or from decreasing to increasing
as shown in the figures below:

Gradient Zero

Local Maximum
if sign of first derivative
changes from
positive to zero to negative
around critical point
Local Minimum
if sign of first derivative
changes from
negative to zero to positive
around critical point
Turning Points and Gradients: Example

y  f  x   sin x
 90;1 is a local maximum
f   x  changes from   0  

 270; 1 is a local minimum
f   x  changes from   0  
Note
 At a local extreme point the first derivative must be zero
 The sign of the first derivative must change around this point
Concavity and
Second Derivative
y  2ax  b
y  2ax  b
y  2a  0
y  2a  0
y  ax  bx  c where a  0
y  ax 2  bx  c where a  0
 Curve is concave up
 Second derivative
always positive
 Tangents below curve
or curve above tangents
 Curve is concave down
 Second derivative
2
always negative
 Tangents above curve
or curve below tangents
Concavity and
Points of Inflection
y  x 3  y   3x 2  0
y   x3  y  3x 2  0
y  6 x
y  6 x
y  0 if x  0
 Concave Up 

y  0 if x  0
y  0 if x  0
 Concave Down 
y  0 if x  0
 Concave Up 

y  0 if x  0
y  0 if x  0
 Concave Down 
Observations at or around (0;0) :
 First and second derivatives zero at (0;0)
Inflection Point at (0;0)
 No change in sign of first derivative near (0;0)
 Second derivative zero at this point
 Changes in sign of second derivative near (0;0)
 Change in concavity around this point
Concavity and Points of Inflection: Example
EXAMPLE : Given: f ( x)  x3  3x 2  7
Determine : a) the interval on which the curve is concave up or down
b) the coordinates of the inflection point.
a) Firstly determine f ( x) :
f ( x)  3x 2  6 x  f ( x)  6 x  6
b) Inflection point when x  1:
 1; f 1   1;5 is inflection point
Then determine where f ( x)  0 :
f ( x)  0 when 6 x  6  0  x  1
Next determine the sign of
f ( x) near x  1:
f   x  : 
0 

 f is concave down  f is concave up
when x  1
1
when x  1

Concavity and Inflection Points:
Example from Trigonometry
180; 0  is a point of inflection
f   x  changes from   0  

y  f  x   sin x
Enrichment :
y  f   x   cos x
y  f   x    sin x
 0  180 k ;0  where k  are all points of inflection
f   x  changes from   0   or from   0  
 f  180   sin(180)  0
Note :
 At an inflection point the second derivative must be zero
 The sign of the second derivative must change around this point
Tutorial 2: Turning and Inflection Points
A function f is defined by f ( x)  2 x 3  6 x.
Determine :
(1) the turning points (maximum or minimum) and
(2) the inflection point.
PAUSE DVD
• Do Tutorial 2
• Then View solutions
Tutorial 2 Problem 1:
Suggested Solution
1)
f ( x)  2 x3  6 x
For the turning points we first find f ( x):
 f ( x)  6 x 2  6  6( x 2  1)  6( x  1)( x  1)
hence f ( x)  0 at x  1 or x  1
Next we determine the signs of f ( x) :
f  x :  0
1

0 
1
Local maximum at x  1
with y  2(1)3  6(1)  4
 maximum turning point at (  1 ; 4)
Local minimum at x  1
with y  2(1)  6(1)  4
 minimum turning point at (1 ;  4)
3
Tutorial 2 Problem 2:
Suggested Solution
f ( x)  2 x 3  6 x  f   x   6 x 2  6
(2)
For the inflection point we find f ( x):
 f ( x)  12 x
Possible inflection point where f   x   0 :
 f ( x)  12 x  0 at x  0
Only an inflection point if there is a change in concavity:
 We discuss the signs of f   x  around x  0 :
f   x  :
 0 
0
 Curve change from concave down to concave up around x  0.
Thus  0; f  0     0;0  is the inflection point.
Lesson 3
Sketching
Cubic Functions
NCS Mathematics
DVD Series
Sketching Cubic Functions
In order to sketch a cubic function, one should:
STEP 1: Find intercepts of the curve and the axes
STEP 2 : Determine the sign of the function over
the domain. (This determine the position of the curve)
STEP 3 : Find the turning point(s) of the function
STEP 4 : Find the inflection point(s) of the function
STEP 5 : Sketch the curve of the function
Sketching cubic functions defined by y  ax3  c
Example: Sketch the function defined by y  f ( x)  x3  8.
Step 1: y  intercept at  0;8 
x  intercept(s) where x3  8  0 or at  2;0 
Step 2: f  x   0 when x  2 (Curve lies above x  axis) f  x  :
f  x   0 when x  2 (Curve lies below x  axis)
Step 3: f   x   3x  0  x  0
2
If x  0 then f   x   0 and if x  0 then f   x   0
 There is no turning point(s)
Step 4: f   x   6 x  0  x  0
If x  0 then f   x   0 and if x  0 then f   x   0
  0; f  0     0;8 is the point of inflection.
 0 
2
Step 5: Sketch
Sketching cubic functions defined by y  ax3  bx2
Example: Sketch function defined by y  f ( x)  x3  3x 2
y  intercept: Where y  f (0)  (0)3  3(0) 2  0
x  intercepts: Where x3  3x 2  0  x 2  x  3  0  x  0 (equal) or x  3
Signs of f  x  :

0
0

0

3
Signs of f   x  : f ( x)  3x 2  6 x  3x( x  2)
f ( x )  0  x  0 or
x2
  0; f  0     0;0  is a maximum turning point
and  2; f  2     2; 4  is a minimum turning point.
Signs of f   x  : f ( x)  6 x  6  6( x  1)
f ( x)  0  x  1
f   x  :
 0 
1
1; f 1   1; 2  is an inflection point.
f  x : 
0
 0 
0
2
Sketching cubic functions defined by y  ax3  bx2  cx
Example: Sketch function defined by y  f ( x)  2 x3  3x 2  12 x
y  intercept: Where y  f (0)  0
x  intercepts: Where x  2 x 2  3x  12   0  x  0 or x 
Signs of f  x  :
 0 0  0 
1,8
0
3,3
3  105
 x  0 or x  3,3 or x  1,8
4
f  x : 
Signs of f   x  : f ( x)  6 x 2  6 x  12  6( x  2)  x  1
f ( x )  0  x  2 or
x  1
  1; f  1    1;7  is a maximum turning point
and  2; f  2     2; 20  is a minimum turning point.
Signs of f   x  : f ( x)  12 x  6  6(2 x  1)
f ( x )  0  x 
f   x  :
 0 
1
2
0,5
 0,5; f  0,5    0,5; 6,5 is an inflection point.
0  0 
1
2
Sketching cubic functions defined by y  ax3  bx2  cx  d
Example: Sketch function defined by y  f ( x)  x3  6 x 2  9 x  4
y  intercept: Where y  f (0)  4
x  intercepts: Where  x  1  x 2  5 x  4    x  1  x  4   0  x  1 or x  4
2
Signs of f  x  :
 0 0 
1
f  x : 
4
Signs of f   x  : f ( x)  3x 2  12 x  9  3( x  1)  x  3
f ( x )  0  x  1 or
x3
 1; f 1   1;0  is a maximum turning point
and  3; f  3    3; 4  is a minimum turning point.
Signs of f   x  : f ( x)  6 x  12  6( x  2)
f ( x)  0  x  2
f   x  :
 0 
2
 2; f  2    2; 2  is an inflection point.
0  0 
1
3
Tutorial 3: Sketching Cubic Functions
For each of the functions, calculate the intercepts
with the axes, the signs of the function, the turning
points and point of inflection. Then draw a sketch
graph of the function:
PAUSE DVD
1)
f ( x)  x  2 x  4 x  8
2)
g ( x)  ( x  1)( x  3)( x  7)
3)
h( x)  2 x 3  x 2  13 x  6
3
2
• Do Tutorial 3
• Then View Solutions
Tutorial 3 Problem 1: Suggested Solutions
Sketch function defined by y  f ( x)  x3  2 x 2  4 x  8
y  intercept: Where y  f (0)  8
x  intercepts: Where x 2  x  2   4  x  2    x  2   x 2  4    x  2 
2
 x  2  0
 x  2 (Equal) or x  2
Signs of f  x  :
 0 0 
2
2
Signs of f   x  : f ( x)  3x 2  4 x  4  (3x  2)  x  2 
f ( x )  0  x 
2
3
or
f  x : 
2
x  2
  2; f  2     2;0  is a maximum turning point
 2  2   2
13 
and  ; f      ; 9  is a minimum turning point.
27 
 3  3   3
Signs of f   x  : f ( x)  6 x  4  2(3 x  2)
f ( x )  0  x  
2
3
f   x  :
 2  2   2
20 
   ; f        ; 4  is an inflection point.
27 
 3  3   3
0  0 
0 

2
3
2
3
Tutorial 3 Problem 2: Suggested Solutions
Sketch function defined by y  g ( x)    x  1 x  3 x  7 
y  intercept: Where y  g (0)  1 3 7   21
x  intercepts: Where   x  1 x  3 x  7   0  x  1 or x  3 or x  7
Signs of f  x  :
 0  0 0 
7
1
g  x :
3
Signs of g   x  : g  x    x3  3x 2  25 x  21  g ( x)  3x 2  6 x  25
g ( x)  0  x 
6  336
 x  4,1 or x  2,1
6
 Max TP at approx.  2,1 ; 9 
and Min TP at approx.  4,1 ;  105
Signs of g   x  : g ( x)  6 x  6  6( x  1)
g ( x)  0  x  1
 IP at  1; g  1    1; 48
g   x  :  0
1

 0  0 
4,1
2,1
Tutorial 3 Problem 3 : Suggested Solutions
Sketch function defined by y  h( x)  2 x3  x 2  13x  6
y  intercept: Where y  h(0)  6
x  intercepts: Where  x  2  2 x  1 x  3  0  x  2 or x  
Signs of h  x  :
 0  0 0 
2
0,5 3
h  x  :
 0  0 
Signs of h  x  : h  x   6 x 2  2 x  13
h( x )  0  x 
1,3
2  316
 x  1, 6 or x  1,3
12
 Max TP at approx.  1, 3 ; 4,8
and Min TP at approx. 1,6 ;  21,1
Signs of h  x  : h( x)  12 x  2  2(6 x  1)
h( x )  0  x 
1
6
 1  1   1
5  h  x  :
 IP at  ; h      ; 8 
27 
 6  6   6
 0 
1
6
1
or x  3
2
1, 6
Lesson 4
Deductions
from given
Cubic Graphs
NCS Mathematics
DVD Series
Interpretations of Cubic Graphs
In this section we are going to determine equations of cubic
functions as well as some interpretations from given graphs.
We will approach this section by using some examples.
NOTE :
The equation of a cubic function is given by
y  ax3  bx 2  cx  d
If a  0 the shape is :
OR
or
y  a ( x  p )( x  q )( x  r )
If a  0 the shape is :
OR
Deductions from Graphs: Example 1
EXAMPLE 1 :
The diagram below shows the sketch graph of f ( x)  x3  ax 2  11x  30.
A(  1; 36) and B are the turning points and C is an inflection point of f .
A
C
Determine:
a)
the value of a.
b)
the coordinates of B
c)
the coordinates of C
B
d)
the values of k for which f  x   k will have three roots?
e)
the coordinates of the turning points of g if g ( x )  f ( x  2).
Solution 1 (a)
a)
/
find f ( x)
f ( x)  x3  ax 2  11x  30.
f ( x )  3 x  2ax  11
/
2
But A is a turning point at x  1

f / ( 1)  0
 3( 1)2  2a( 1)  11  0
  2a  8
 a  4
f
Solution 1 (b)
b)
using a  4 we have
f ( x )  3 x  8 x  11
/
2
f ( x)  0
/
f
gives
 3 x 2  8 x  11  0
 (3 x  11)( x  1)  0
 x  11
3

or
x  1
22
y  ( )  4( )  11( 11
)

30


14
3
27
11 3
3
11 2
3
22
 B ( 11
;

14
)
3
27
f ( x)  x3  ax 2  11x  30.
Solution 1 (c)
c)
//
Find f ( x )
f ( x)  x3  ax 2  11x  30.
f ( x)  6 x  8
//
 f ( x)  6 x  8  0
//
 x  34
 y  ( 34 )3  4( 34 )2  11( 34 )  30
 10 16
27
 The point C( 34 ; 10 16
)
27
f
Solution 1 (d)
d)
22 
 11
A  1;36  and B  ; 14
27 
 3
f ( x )  k will have 2 roots when k  36 or k  14
22
27
y  k  36

y  k  36

y  k  14
y  k  14
22
27
22
27
22
 f ( x)  k will have 3 roots when k  36 and k  14
27
Solution 1 (e)
e) Turning points of g ( x )  f ( x  2)
translation of 2 units to the right:

( 1  2 ;36)

(1;36)
22
and ( 11

2
;

14
)
3
27
22
and ( 173 ;  14 27
)
f ( x)  x3  ax 2  11x  30.
f
Problem 2
EXAMPLE 2:
The diagram below shows the sketch graph of
g ( x )  x 3  bx 2  9 x  4. P and Q are the turning
points and R( 2 ;  2) is an inflection point of g.
y
P
x
R(2 ;  2)
g
Q
Determine:
a)
the value of b.
b)
the coordinates of Q
c)
the coordinates of the inflection point of h if h( x )   g ( x ).
Solution 2 (a)
a)
Find g / / ( x ) :
g ( x )  x 3  bx 2  9 x  4  g / ( x )  3 x 2  2bx  9

g / / ( x )  6 x  2b
But R is an inflection point at x  2
 f (2)  0
 6(2)  2b  0
 2b  12
 b  6
//
y
P
g
x
R(2 ;  2)
Q
Solution 2 (b), (c)
b)
using b  6 we have f / ( x )  3 x 2  12 x  9
f / ( x)  0
gives
y
 3( x  4 x  3)  0
 3( x  1)( x  3)  0
 x  1 or x  3
g
2
P
x
R(2 ;  2)
Q
 y  (3)3  6(3)2  9(3)  4  4
 Q (3 ;  4)
h  x  g  x
c)
Inflection point of h is the reflection in
the x-axis of the inflection point of g
i.e.
(2 ;  2)  (2 ; 2)
TUTORIAL 4: Part 1
Problem 1:
The diagram below shows the sketch graph of the cubic function g ( x )
with roots at x  3 ; x  1 and x  0,5.
The y -intercept is at (0 ;  3).
PAUSE DVD
y
g
• Do Tutorial 4: Part 1
• Then View Solutions
x
3
1
0,5
(0, 3)
a) Find the equation of the cubic function.
b) Determine the inflection point of the function.
Tutorial 4 Problem 1(a):
Suggested Solution
PROBLEM 1:
a) Form the cubic equation g ( x)  a( x  p )( x  q )( x  r )
 g ( x )  a( x  0,5)( x  1)( x  3)
but (0 ;  3) is the y -intercept
 a(0  0,5)(0  1)(0  3)  3
  1,5a  3
 a2
 g ( x)  2( x  0,5)( x  1)( x  3)  (2 x  1)( x 2  4 x  3)
 2 x3  7 x 2  2 x  3
Tutorial 4 Problem 1(b):
Suggested Solution
g ( x )  2 x3  7 x 2  2 x  3
b) g ( x )  6 x  14 x  2
/
2
 g / / ( x )  12 x  14
 12 x  14  0
 x   76

y  2(  )  7(  )  2(  76 )  3  1 541
7 3
6
7 2
6
 inflection point
7
1

;1
 6 54 
TUTORIAL 4: Part 2
Problem 2 :
The diagram below shows the sketch graph of the cubic function
f ( x )  x3  sx 2  tx  2
The A(1 ; 2) and B are turning points with C(2 ; 0) the inflection point of f .
Pause DVD
• Do Tutorial 4: Part 2
• Then View Solutions
a)
b)
c)
Find the values of s and t.
Determine the coordinates of point B.
Determine inflection point and
turning points of h if h( x )  2 f ( x ).
Tutorial 4 Problem 2:
Suggested Solutions
PROBLEM 2 :
b) f / ( x )  3 x 2  12 x  9  3( x 2  4 x  3)
f / ( x )  3 x 2  2 sx  t


f / ( x )  3( x  1)( x  3)  0
x  1 or x  3
 f / / ( x)  6 x  2s
but (2 ; 0) is the inflection point


y  (3)3  6(3)2  9(3)  2  2
B (3 ;  2)
a) f ( x )  x 3  sx 2  tx  2

 f / / (2)  6(2)  2 s  0
 s  6
Also (1; 2) is a turning point.
 f / (1)  3(1)2  2( 6)(1)  t  0
 t 9
c) vertical stretching by factor 2
x value remains the same
y value multiplied by 2
Inflection point of h is (2 ; 2  0)  (2 ; 0)
Max TP at 1;-2  and Min TP at 3;-4 
a) Find s and t
b) Coordinates of point B?
c) IP and TP's of h if h( x)  2 f ( x)
End of the DVD on
Cubic Functions
REMEMBER!
• Consult text-books for additional examples.
• Attempt as many as possible other similar examples
on your own.
• Compare your methods with those that were
discussed in the DVD.
• Repeat this procedure until you are confident.
• Do not forget:
Practice makes perfect!