Using the CASIO fx-82ZA PLUS for functions in the FET band Rencia Lourens RADMASTE Centre.
Download ReportTranscript Using the CASIO fx-82ZA PLUS for functions in the FET band Rencia Lourens RADMASTE Centre.
Slide 1
Using the CASIO fx-82ZA PLUS for
functions in the FET band
Rencia Lourens
RADMASTE Centre
Slide 2
Some remarks
A calculator is a tool.
Learners should
Know where answers come from.
Understand mathematics.
Teachers should
Teach the mathematics.
Explain the reasoning behind why the calculator methods work.
BUT the calculator can (and should) become a tool to assist.
Slide 3
CAPS
Functions form 35% of the Grade 12 paper 1, 45% in Grade
11 and 30% in Grade 10 (CAPS). The calculator can be used
to support the calculations needed to draw and interpret the
graphs of the functions.
Slide 4
Intersection of two graphs
Find the points of intersection of the straight line f(x) = x –
3 and the parabola g(x) = x2 – x – 6 if 𝑥 ∈ −3; 4 .
We need to be
Be in TABLE mode.
Have the DUAL table SETUP.
Who is NOT sure?
Slide 5
So how can I as a teacher use this to enhance understanding?
Some thoughts
The meaning of simultaneous equations.
The meaning of a plotted graph.
Slide 6
Next example
Find the point(s) of intersections of the graphs 𝑓 𝑥 =
𝑥 2 − 3𝑥 − 4 and 𝑔 𝑥 = −𝑥 + 1
1
4
This example is different from the previous one because
The domain is not given
It is not a convenient example where the answer(s) are “in your
face”.
So we choose our own domain and start with 𝑥 ∈ −5; 5 .
Slide 7
x
f(x)
g(x)
-5
36
6.25
-4
24
5.25
-3
14
4.25
-2
6
3.25
-1
0
2.25
0
-4
1.25
1
-6
0.25
2
-6
-0.75
3
-4
-1.75
4
0
-2.75
5
6
-3.75
f(-2) > g(-2)
f(-1) < g(-1)
f(3) < g(3)
f(4) > g(4)
Slide 8
Hence somewhere between x = -2 and x = -1 we will have
f(x) = g(x). (We will look at the other value later on).
We keep the table as is, but change our domain to 𝑥 ∈
−2; −1 .
We also change the steps and make that 0,25.
Slide 9
x
f(x)
g(x)
-2
6
3.25
-1.75
4.3125
3
-1.5
2.75
2.75
-1.25
1.3125
2.5
-1
0
2.25
Slide 10
Also somewhere between x = 3 and x = 4 we will have f(x)
= g(x).
We keep the table as is, but change our domain to 𝑥 ∈
3; 4 .
We also change the steps and make that 0,25.
Slide 11
x
f(x)
g(x)
3
-4
-1.75
3.25
-3.1875
-2
3.5
-2.25
-2.25
3.75
-1.1875
-2.5
4
0
-2.75
Slide 12
The graphs 𝑓 𝑥 = 𝑥 2 − 3𝑥 − 4 and
1
4
− 𝑥 + 1 intersect at
(-1,5; 2,75)
(3,5; 2,25)
𝑔 𝑥 =
Slide 13
Turning point of a parabola
Find the turning point of 𝑓 𝑥 = 𝑥 2 − 4𝑥 − 1
We do not know the range so we will start with 𝑥 ∈
−5; 5 .
We do not need the second function, so we CAN disable the
second function.
Slide 14
x
f(x)
-5
44
-4
31
-3
20
-2
11
-1
4
0
-1
1
-4
2
-5
3
-4
4
-1
5
4
Slide 15
So how can I use this as a teacher to enhance understanding?
Some thoughts
The meaning of symmetry
The minimum value
The meaning of a plotted graph
The shape of a quadratic function
Just checking – the turning point is (2; -5)
Slide 16
New example
Find the turning point of 𝑓 𝑥 = 4𝑥 2 − 4𝑥 − 2
We do not have a domain so we start with 𝑥 ∈ [−5; 5].
Slide 17
x
f(x)
-5
118
-4
78
-3
46
-2
22
-1
6
0
-2
1
-2
2
6
3
22
4
46
5
78
The turning point
should be somewhere
between x = 0 and x =
1
Slide 18
So……..
We keep the table and change the domain to….
𝑥 ∈ [0; 1]
And we make the steps…..
1
4
Slide 19
x
f(x)
0
-2
0.25
-2.75
0.5
-3
0.75
-2.75
1
-2
The turning point is
1
; −3
2
Slide 20
Next example
Find the turning point of 𝑓 𝑥 = 2𝑥 2 − 8,5𝑥 + 4.
We do not know the domain hence…
𝑥 ∈ −5; 5
Slide 21
x
f(x)
-5
96.5
-4
70
-3
47.5
-2
29
-1
14.5
0
4
1
-2.5
2
-5
3
-3.5
4
2
5
11.5
The turning point
should be somewhere
between x = 1 and x =
3
Slide 22
We will change the domain to 𝑥 ∈ 1; 3 .
The steps should be
1
.
4
Slide 23
x
f(x)
1
-2.5
1.25
-3.5
1.5
-4.25
1.75
-4.75
2
-5
2.25
-5
2.5
-4.75
2.75
-4.25
3
-3.5
The turning point
should be somewhere
between x = 2 and x =
2.25
Slide 24
We will change the domain to 𝑥 ∈ 2; 2,25 .
The steps should be
1
.
16
Slide 25
x
f(x)
2
-5
2.0625
-5.0234375
2.125
-5.03125
2.1875
-5.0234375
2.25
-5
1
1
The turning point is 2 8 ; −5 32
Slide 26
Finding the intercepts with the axes
Find the intercepts with both the axes of the graph of
𝑓 𝑥 = 𝑥 2 − 5𝑥 + 6.
Domain 𝑥 ∈ −5; 5 .
Steps of 1
Slide 27
x
f(x)
-5
56
-4
42
-3
30
-2
20
-1
0
Just checking…….
Where
12
will the turning
6 point be?
y intercept
1
2
2
0
x intercept
3
0
x intercept
4
2
5
6
Slide 28
So how can I as a teacher use this to enhance understanding?
Some thoughts
The meaning of 𝑓 𝑥 = 0 vs the meaning of 𝑓(0).
The meaning of a plotted graph
Solving of quadratic equation
Slide 29
Next example
Find the intercepts with both the axes of 𝑓 𝑥 = −𝑥 2 +
3𝑥 − 3.
Domain 𝑥 ∈ −5; 5 .
Steps of 1.
Slide 30
x
f(x)
-5
-43
-4
-31
-3
-21
-2
-13
-1
-7
0
-3
1
-1
2
-1
3
-3
4
-7
5
-13
No x-intercept?
y-intercept
Turning point
should be here
Slide 31
Seems as there are no x
intercepts.
Focus on turning point
first.
Will be between x=1 and
x=2.
The turning point is below
the x-axis.
All the graph values are
below the x-axis.
So no x-intercepts.
x
1
-1
1.25
-0.8125
1.5
-0.75
1.75
-0.8125
2
-1
Slide 32
Next example
Find the intercepts with both the axes of 𝑓 𝑥 = −4𝑥 2 +
8𝑥 + 21.
Domain 𝑥 ∈ −5; 5 .
Steps of 1.
Slide 33
x
f(x)
-5
-119
-4
-75
-3
-39
-2
-11
-1
9
0
21
1
25
2
21
3
9
4
-11
5
-38
x-intercept should
be here
y-intercept
Turning point
should be here
x-intercept should
be here
Slide 34
Somewhere between x = -2 and x = -1 the one
x-intercept
should lie and somewhere between x = 3 and x = 4 the
other x-intercept should lie.
So we are going to look at smaller domains and smaller steps.
Slide 35
x
f(x)
-2
-11
-1.75
-5.25
-1.5
0
-1.25
4.75
-1
9
x-intercept
Slide 36
x
f(x)
3
9
3.25
4.75
3.5
0
3.75
-5.25
4
-11
x-intercept
Slide 37
Looking at the reciprocal function
𝑦=
4
𝑥−1
+2
Work with domain 𝑥 ∈ −5; 5
Slide 38
x
f(x)
-5
1.333333
-4
1.2
-3
1
-2
0.6666666
-1
0
x-intercept
0
-2
y-intercept
1
ERROR
Asymptote
2
21
3
9
4
-11
5
--38
Slide 39
Finding equations of graphs
We now need to move to the STATS mode
Let us have a look at the Menu
Is everybody sure how to get into STATS mode?
Slide 40
Example – linear function
We are going to work with the linear regression.
Find the equation of the straight line through
(-1; -1) and
(2; 5).
Type in the two points in the table (data).
Find the coefficients remembering that in stats the linear
equation is 𝑦 = 𝑎 + 𝑏𝑥.
The equation is 𝑦 = 1 + 2𝑥.
Or as we know it 𝑦 = 2𝑥 + 1.
Slide 41
Example – Quadratic function with
intercepts given
Typing error on page 10 (first bullet please change to
Quadratic and not linear).
Enter the three points in the table (data).
Find the coefficients remembering that in stats the linear
equation is 𝑦 = 𝑎 + 𝑏𝑥 + 𝑐𝑥 2 .
The equation is 𝑦 = 8 + 6𝑥 − 2𝑥 2 .
Or as we know it 𝑦 = −2𝑥 2 + 6𝑥 + 8.
Slide 42
Example – Quadratic function with any
three points.
Enter the three points in the table (data).
Find the coefficients remembering that in stats the linear
equation is 𝑦 = 𝑎 + 𝑏𝑥 + 𝑐𝑥 2 .
The equation is 𝑦 = −1 + 2𝑥 + 𝑥 2 .
Or as we know it 𝑦 = 𝑥 2 + 2𝑥 − 1.
Slide 43
Example – Exponential function* with
any two points.
Enter the two points in the table (data).
Find the coefficients remembering that in stats the linear
equation is 𝑦 = 𝑎. 𝑏 𝑥 .
The equation is 𝑦 = 1. 2𝑥 .
*The CASIO fx-82ZA PLUS calculator can only do the
exponential graph of the form 𝑦 = 𝑎. 𝑏 𝑥 .
Slide 44
Example – Quadratic function with the
turning point and another point.
We need to find ANOTHER point.
From the turning point we know the axis of symmetry is at x
= -1
The point symmetrical to (0;5) will be (-2; 5).
Enter the three points.
Find the coefficients remembering that in stats the linear
equation is 𝑦 = 𝑎 + 𝑏𝑥 + 𝑐𝑥 2 .
The equation is 𝑦 = 5 + 2𝑥 + 𝑥 2 .
Or as we know it 𝑦 = 𝑥 2 + 2𝑥 + 5.
Using the CASIO fx-82ZA PLUS for
functions in the FET band
Rencia Lourens
RADMASTE Centre
Slide 2
Some remarks
A calculator is a tool.
Learners should
Know where answers come from.
Understand mathematics.
Teachers should
Teach the mathematics.
Explain the reasoning behind why the calculator methods work.
BUT the calculator can (and should) become a tool to assist.
Slide 3
CAPS
Functions form 35% of the Grade 12 paper 1, 45% in Grade
11 and 30% in Grade 10 (CAPS). The calculator can be used
to support the calculations needed to draw and interpret the
graphs of the functions.
Slide 4
Intersection of two graphs
Find the points of intersection of the straight line f(x) = x –
3 and the parabola g(x) = x2 – x – 6 if 𝑥 ∈ −3; 4 .
We need to be
Be in TABLE mode.
Have the DUAL table SETUP.
Who is NOT sure?
Slide 5
So how can I as a teacher use this to enhance understanding?
Some thoughts
The meaning of simultaneous equations.
The meaning of a plotted graph.
Slide 6
Next example
Find the point(s) of intersections of the graphs 𝑓 𝑥 =
𝑥 2 − 3𝑥 − 4 and 𝑔 𝑥 = −𝑥 + 1
1
4
This example is different from the previous one because
The domain is not given
It is not a convenient example where the answer(s) are “in your
face”.
So we choose our own domain and start with 𝑥 ∈ −5; 5 .
Slide 7
x
f(x)
g(x)
-5
36
6.25
-4
24
5.25
-3
14
4.25
-2
6
3.25
-1
0
2.25
0
-4
1.25
1
-6
0.25
2
-6
-0.75
3
-4
-1.75
4
0
-2.75
5
6
-3.75
f(-2) > g(-2)
f(-1) < g(-1)
f(3) < g(3)
f(4) > g(4)
Slide 8
Hence somewhere between x = -2 and x = -1 we will have
f(x) = g(x). (We will look at the other value later on).
We keep the table as is, but change our domain to 𝑥 ∈
−2; −1 .
We also change the steps and make that 0,25.
Slide 9
x
f(x)
g(x)
-2
6
3.25
-1.75
4.3125
3
-1.5
2.75
2.75
-1.25
1.3125
2.5
-1
0
2.25
Slide 10
Also somewhere between x = 3 and x = 4 we will have f(x)
= g(x).
We keep the table as is, but change our domain to 𝑥 ∈
3; 4 .
We also change the steps and make that 0,25.
Slide 11
x
f(x)
g(x)
3
-4
-1.75
3.25
-3.1875
-2
3.5
-2.25
-2.25
3.75
-1.1875
-2.5
4
0
-2.75
Slide 12
The graphs 𝑓 𝑥 = 𝑥 2 − 3𝑥 − 4 and
1
4
− 𝑥 + 1 intersect at
(-1,5; 2,75)
(3,5; 2,25)
𝑔 𝑥 =
Slide 13
Turning point of a parabola
Find the turning point of 𝑓 𝑥 = 𝑥 2 − 4𝑥 − 1
We do not know the range so we will start with 𝑥 ∈
−5; 5 .
We do not need the second function, so we CAN disable the
second function.
Slide 14
x
f(x)
-5
44
-4
31
-3
20
-2
11
-1
4
0
-1
1
-4
2
-5
3
-4
4
-1
5
4
Slide 15
So how can I use this as a teacher to enhance understanding?
Some thoughts
The meaning of symmetry
The minimum value
The meaning of a plotted graph
The shape of a quadratic function
Just checking – the turning point is (2; -5)
Slide 16
New example
Find the turning point of 𝑓 𝑥 = 4𝑥 2 − 4𝑥 − 2
We do not have a domain so we start with 𝑥 ∈ [−5; 5].
Slide 17
x
f(x)
-5
118
-4
78
-3
46
-2
22
-1
6
0
-2
1
-2
2
6
3
22
4
46
5
78
The turning point
should be somewhere
between x = 0 and x =
1
Slide 18
So……..
We keep the table and change the domain to….
𝑥 ∈ [0; 1]
And we make the steps…..
1
4
Slide 19
x
f(x)
0
-2
0.25
-2.75
0.5
-3
0.75
-2.75
1
-2
The turning point is
1
; −3
2
Slide 20
Next example
Find the turning point of 𝑓 𝑥 = 2𝑥 2 − 8,5𝑥 + 4.
We do not know the domain hence…
𝑥 ∈ −5; 5
Slide 21
x
f(x)
-5
96.5
-4
70
-3
47.5
-2
29
-1
14.5
0
4
1
-2.5
2
-5
3
-3.5
4
2
5
11.5
The turning point
should be somewhere
between x = 1 and x =
3
Slide 22
We will change the domain to 𝑥 ∈ 1; 3 .
The steps should be
1
.
4
Slide 23
x
f(x)
1
-2.5
1.25
-3.5
1.5
-4.25
1.75
-4.75
2
-5
2.25
-5
2.5
-4.75
2.75
-4.25
3
-3.5
The turning point
should be somewhere
between x = 2 and x =
2.25
Slide 24
We will change the domain to 𝑥 ∈ 2; 2,25 .
The steps should be
1
.
16
Slide 25
x
f(x)
2
-5
2.0625
-5.0234375
2.125
-5.03125
2.1875
-5.0234375
2.25
-5
1
1
The turning point is 2 8 ; −5 32
Slide 26
Finding the intercepts with the axes
Find the intercepts with both the axes of the graph of
𝑓 𝑥 = 𝑥 2 − 5𝑥 + 6.
Domain 𝑥 ∈ −5; 5 .
Steps of 1
Slide 27
x
f(x)
-5
56
-4
42
-3
30
-2
20
-1
0
Just checking…….
Where
12
will the turning
6 point be?
y intercept
1
2
2
0
x intercept
3
0
x intercept
4
2
5
6
Slide 28
So how can I as a teacher use this to enhance understanding?
Some thoughts
The meaning of 𝑓 𝑥 = 0 vs the meaning of 𝑓(0).
The meaning of a plotted graph
Solving of quadratic equation
Slide 29
Next example
Find the intercepts with both the axes of 𝑓 𝑥 = −𝑥 2 +
3𝑥 − 3.
Domain 𝑥 ∈ −5; 5 .
Steps of 1.
Slide 30
x
f(x)
-5
-43
-4
-31
-3
-21
-2
-13
-1
-7
0
-3
1
-1
2
-1
3
-3
4
-7
5
-13
No x-intercept?
y-intercept
Turning point
should be here
Slide 31
Seems as there are no x
intercepts.
Focus on turning point
first.
Will be between x=1 and
x=2.
The turning point is below
the x-axis.
All the graph values are
below the x-axis.
So no x-intercepts.
x
1
-1
1.25
-0.8125
1.5
-0.75
1.75
-0.8125
2
-1
Slide 32
Next example
Find the intercepts with both the axes of 𝑓 𝑥 = −4𝑥 2 +
8𝑥 + 21.
Domain 𝑥 ∈ −5; 5 .
Steps of 1.
Slide 33
x
f(x)
-5
-119
-4
-75
-3
-39
-2
-11
-1
9
0
21
1
25
2
21
3
9
4
-11
5
-38
x-intercept should
be here
y-intercept
Turning point
should be here
x-intercept should
be here
Slide 34
Somewhere between x = -2 and x = -1 the one
x-intercept
should lie and somewhere between x = 3 and x = 4 the
other x-intercept should lie.
So we are going to look at smaller domains and smaller steps.
Slide 35
x
f(x)
-2
-11
-1.75
-5.25
-1.5
0
-1.25
4.75
-1
9
x-intercept
Slide 36
x
f(x)
3
9
3.25
4.75
3.5
0
3.75
-5.25
4
-11
x-intercept
Slide 37
Looking at the reciprocal function
𝑦=
4
𝑥−1
+2
Work with domain 𝑥 ∈ −5; 5
Slide 38
x
f(x)
-5
1.333333
-4
1.2
-3
1
-2
0.6666666
-1
0
x-intercept
0
-2
y-intercept
1
ERROR
Asymptote
2
21
3
9
4
-11
5
--38
Slide 39
Finding equations of graphs
We now need to move to the STATS mode
Let us have a look at the Menu
Is everybody sure how to get into STATS mode?
Slide 40
Example – linear function
We are going to work with the linear regression.
Find the equation of the straight line through
(-1; -1) and
(2; 5).
Type in the two points in the table (data).
Find the coefficients remembering that in stats the linear
equation is 𝑦 = 𝑎 + 𝑏𝑥.
The equation is 𝑦 = 1 + 2𝑥.
Or as we know it 𝑦 = 2𝑥 + 1.
Slide 41
Example – Quadratic function with
intercepts given
Typing error on page 10 (first bullet please change to
Quadratic and not linear).
Enter the three points in the table (data).
Find the coefficients remembering that in stats the linear
equation is 𝑦 = 𝑎 + 𝑏𝑥 + 𝑐𝑥 2 .
The equation is 𝑦 = 8 + 6𝑥 − 2𝑥 2 .
Or as we know it 𝑦 = −2𝑥 2 + 6𝑥 + 8.
Slide 42
Example – Quadratic function with any
three points.
Enter the three points in the table (data).
Find the coefficients remembering that in stats the linear
equation is 𝑦 = 𝑎 + 𝑏𝑥 + 𝑐𝑥 2 .
The equation is 𝑦 = −1 + 2𝑥 + 𝑥 2 .
Or as we know it 𝑦 = 𝑥 2 + 2𝑥 − 1.
Slide 43
Example – Exponential function* with
any two points.
Enter the two points in the table (data).
Find the coefficients remembering that in stats the linear
equation is 𝑦 = 𝑎. 𝑏 𝑥 .
The equation is 𝑦 = 1. 2𝑥 .
*The CASIO fx-82ZA PLUS calculator can only do the
exponential graph of the form 𝑦 = 𝑎. 𝑏 𝑥 .
Slide 44
Example – Quadratic function with the
turning point and another point.
We need to find ANOTHER point.
From the turning point we know the axis of symmetry is at x
= -1
The point symmetrical to (0;5) will be (-2; 5).
Enter the three points.
Find the coefficients remembering that in stats the linear
equation is 𝑦 = 𝑎 + 𝑏𝑥 + 𝑐𝑥 2 .
The equation is 𝑦 = 5 + 2𝑥 + 𝑥 2 .
Or as we know it 𝑦 = 𝑥 2 + 2𝑥 + 5.