Behaviour of Function
Before and After Turning Point
Given any quadratic functions, it is the behaviour of the function before and after the
turning point, and consequently that of the differential coefficient, which determines a
maximum or minimum point.
Exploration
1.
Open a new TI InterActive! Document. Title this document Turning
Points. Add your names and the date.
2.
Given a function y1 = x2 – 4x + 3, click on
Palette to find its derivative.
3.
Select
4.
In the Graph Functions Y= menu, enter both the function and the
derivative equation.
5.
Click on the Copy All tab to display the function curve and its derivative
equation on the same graph.
6.
On the Graph Window, click on Calculate Numerical Derivative tab
7.
In the Calculate Numerical Derivative tab, Enter –2 for the Derivative at:
box and Click Calculate. Record the Derivative value.
8.
Repeat the Calculate Numerical Derivative for range –2 to 6 in steps of 1.
9.
Close the Graph Section. Label the graphs of y and
10.
Select List
11.
In the Data Editor, label x in the L1 tab and gradient for the L2 tab.
and use the TI Maths
to open the Graph section.
dy
accordingly.
dx
to open the List Editor.
12.
In the Data Editor, Enter x values (from –2 to 6) in the L1 tab and the
corresponding Derivative values calculated in step 8 in the L2 tab.
13.
Close the Data Editor.
What do you observe about the values obtained from the Calculate
Numerical Derivative for range –2 to 6 and the Derivative equation?
14.
Repeat the above steps 2 to13 for function y2 = 3 + 4x – x2.
For each of both cases,
15.
Observe from the graphs, the trend of y curve and its corresponding
derivative when x < 2, x > 2 and x = 2.
16.
Compare and Contrast the characteristics of the two functions and its
derivative. Draw any Similarities and Differences.
17.
Examine the turning point in the two y functions. Which function has a
maximum point? Which one is a minimum point? What is the
corresponding value of x?
18.
Calculate the maximum and minimum value for the corresponding curves.
19.
Save your activity document as turning.tii. Print a copy of this document
for submission.
Note: Please key in your answers to the related questions together with your
graphs and corresponding tables in the same document.
Teacher Notes
Subject:
Additional Mathematics
Topic:
Stationary points
Level:
Sec. 3 Express
Duration:
2 periods
Pre-requisite:
Students are able to
(a) state definition of gradient of a curve at a
given point
(b) recognize the common notation for gradient
function
Learner Ability:
Mixed average ability
Learning Environment:
Half Computer Laboratory
Thinking Skill:
Comparing, Identifying Patterns and
relationships, Spatial visualization, Problemsolving, Deduction.
Teaching Strategies:
Think-Pair-Share, KWL.
Objectives
Students should be able to:
a) identify maximum and minimum turning
points by observing gradient of curve
b) apply the knowledge of turning points in
problem solving questions.
Applicable TI Interactive!
Functions:
TI Maths Palette
Graph
Calculate Numerical Derivative
List Editor
Solution to Exploration Activity
9.
10
2
y = x – 4x + 3
8
6
4
2
-10 -8 -6 -4 -2
2
-2
-4
4
6
8 10
dy/dx = 2x – 4
-6
-8
-10
12.
x
gradient
dy/dx
-8
-6
-4
-2
0
2
4
6
8
2
-1
0
1
2
3
4
5
6
13.
The dy/dx values obtained from the Calculate Numerical Derivative are
points on the corresponding Derivative equation.
14.
10
y = 3 + 4x –x2
8
dy/dx = 4 - 2x
6
4
2
-10 -8 -6 -4 -2
2
-2
-4
-6
-8
-10
4
6
8 10
14.
x
-2
-1
0
1
2
3
4
5
6
15.
(i)
gradient
dy/dx
8
6
4
2
0
-2
-4
-6
-8
y1 = x2 – 4x + 3
For x < 2, when the y curve is decreasing, the corresponding derivative is
increasing linearly.
For x > 2, when the y curve is increasing, the corresponding derivative is
increasing linearly.
When x = 2, the y curve is stationary, the corresponding derivative cuts
the x-axis.
(ii)
y2 = 3 + 4x – x2
For x < 2, when the y curve is increasing, the corresponding derivative is
decreasing linearly.
For x > 2, when the y curve is decreasing, the corresponding derivative is
decreasing linearly.
When x = 2, the y curve is stationary, the corresponding derivative cuts
the x-axis.
16.
Similarities
-
Both graphs have a turning point at x = 2
Both derivative graphs are linear
Both derivative graphs cut through the x - axis (ie. when dy/dx = 0)
Both the derivative graphs change sign before and after the turning
point
Differences
-
y1 = x2 – 4x + 3 graph has a minimum point
y2 = 3 + 4x – x2 graph has a maximum point
Derivative of y1 curve increases
Derivative of y2 curve decreases.
17.
Characteristics of Turning Point
-
Essential condition that dy/dx = 0
The function y decreases before and increases after turning point, or
vice versa
Sign of gradient dy/dx changes after the turning point
Turning point of y1 at x = 2 is a minimum point.
Turning point at y2 at x = 2 is a maximum point.
18.
Substituting the x value in the functions (or just read off from graphs), we
get the values of the turning points:
y1 has a minimum point of –1 when x = 2.
y2 has a maximum point of 7 when x = 2.
Rationale for questions:
-
To allow students to observe and deduce a relationship between
quadratic graph and its derivative graph.
Based on the relationship, students should be able to observe that the
turning points for quadratic graphs correspond with the points when
derivative is zero.
Students would be able to notice that a change in the sign of the
derivative would mean the presence of turning point.
Discussion to be conducted by teacher:
Tests for discrimination between maximum and minimum values of a function
may be summarized as follows:
For turning point at x = a
y or f(x)
dy
or
dx
f’(x)
Maximum
Minimum
- Increasing before x = a
- Decreasing after x = a
- Decreasing before x = a
- Increasing after x = a
- Positive before x = a
- Negative after x = a
- Equal to 0 at x = a
- Negative before x = a
- Positive after x = a
- Equal to 0 at x = a
∴decreasing
∴increasing
Additional Exercise
Question:
Divide 10 into two parts such that their product is a maximum.
Rationale:
-
To evaluate students’ knowledge of turning point
Application of students’ prior knowledge of algebra in the exercise
To test students’ application of knowledge of turning point (maximum,
minimum) in creative problem-solving
Solution:
Let part the two parts be x and (10 - x).
Therefore the product, y = x (10 - x)= 10x – x2.
For the product to be maximum, dy/dx = 0.
dy/dx = 10 - 2x
For dy/dx = 0, x = 5
Hence for the product to be maximum, we divide 10 into two parts of 5.
y = 10 x - x2
dy/dx = 10 - 2x