第90回全国算数・数学教育研究(福島)大会に参加して
滋賀県立栗東高等学校 岡 良夫
第90回全国算数・数学教育研究大会の1日目は郡山市民文化センターで始まった。

午前の日程は、9時から受付で9:30〜開会式、10:00〜日本数学教育学会総会、10:50〜茨城大学教授 根本博氏の記念講演「改訂学習指導要領の根柢にあるもの」、12:10〜郡山市立郡山第二中学校合唱部のアトラクション、というものであった。

合唱もすばらしかったが、その伴奏をしていたのも中学生で、第1バイオリン3人、第2バイオリン2人、ビオラ2人、チェロ2人、コントラバス1人という、本格的な弦楽器の編成であったことのほうが驚きであった。

昼食は800円の弁当で、ホール内は飲食禁止であったので、展示室でとった。ホール内で飲食禁止というわからぬわけではないが、こういう大会では初めての体験であった。
午後は校種別に別れて、高校は少し離れた郡山視聴覚センターにおいて、13:40〜15:30〜千葉大学名誉教授佐藤恒雄氏の講演、次期開催京都代表からの挨拶があった。



氏はセンター試験の問題作成にも関わっておられて、有益な話がきけたと思う。
演題は「問題解決力を育てる問題の読解と分析の方法」というもので、96年の東大の石油タンクの問題を詳しく解説された。
ただ、講演の本題ではないものの、十年ごとにころころと変わる学習指導要領について批判されたことのほうにインパクトを受けた。今にして思えば、数学T、基礎解析、代数・幾何、確率・統計、微分積分の教育課程が一番すぐれていたと言われ、聴衆の大半は同感のようであった。
第90回全国算数・数学教育研究(福島)大会の2日目は校種別の分科会で、高校は郡山萌世高校で開催された。この高校は駅前の12階のビルの上の方の階にあった。

私は数学Tの会場での5番目(昼一番)の発表だったので、午前は隣の数学Uの分科会の発表を聴かせていただいた。
@ 関数指導に関する一考察2〜「対応」と「値域」に焦点をあてて〜 福島・郡山高校 佐藤 拓 先生

ここで発表された実践は、一人の生徒からの次のような質問から始まった。
「実数x,yが19x2+6xy+11y2=1を満たしながら動くとき、x2+y2の最大値、最小値、およびそれらを与えるx,yの値を求めよ。」(2007学習院大学文学部)
その質問に来た生徒は、解説を見てもその方針がまったく理解できないというのである。その方針とは
19x2+6xy+11y2=1のもとで、x2+y2が値kをとることができる条件は、この両方を満たす実数x,yが存在すること。
これをうけて、先生は「簡単な1変数関数の値域を“グラフを利用せずに”調べる活動を通して、対応としての関数や値域の定義に目を向けさせたというのである。
このグラフのイメージを借りずに関数を考えるという実践は大いに示唆を受けた。
A 幾何ソフト Cabri を活用したベクトルの導入〜ニュートンのファインマンによる変奏曲〜 岡山・岡山後楽館高校 河合 伸昭 先生

先生は工学部出身で、ニュートンの著作プリンキピアをファインマンが初等幾何のみで解説した Feynman’s Lost Lecture をもとに、幾何ソフト Cabri を駆使してケプラーの法則を生徒に理解させようと取り組まれた実践であった。
B 加法定理の発展的指導 福島・福島西高校 村田弓子・伊藤みすず 先生

両先生は、三角関数の加法定理から導かれる2倍角・3倍角・半角の公式で終わらずに、教科書にないオリジナルの公式を生徒に作らせるという取り組みを実践された。
また、それに関連して、sin18°,sin15°,sin33°,sin42°,sin48°,sin54°などの値を、平方根を用いて表すことも生徒にさせたことはたいへん興味深いものであった。
C 指数の拡張の指導〜有理数指数の導入の指導事例報告〜 東京・武蔵野北高校 小松 真 先生

私もかねがね指数表示したものをわざわざ累乗根に直すことに疑問を抱いてきたので、先生の取り組みに対しては期待していた。
たとえば、2の3/4乗とは何を意味するのか。また、それとは別に2の0.75乗はどうか。指数関数を構築するためにはxは整数値だけではだめである。ただ、それらの有理数乗を生徒にどう納得させるかはまだまだ課題が多いと感じた。
来年度はたぶん私は2年生対象に数学Uを指導することになるだろうから、以上の各実践はそれぞれ有益であった。また、指導助言の国士舘大学教授の正田良先生、筑波大学附属駒場中・高等学校の駒野誠先生は、丁寧な資料を配布され、非常に勉強になった。
D Quarry a math method to understand perfectly quadratic function without graph〜To make mathematicsT be interesting〜 滋賀・栗東高校 岡 良夫

この私の発表は後述するので参照されたい。
E 根号を含む式の値について〜いろいろな判定法〜 岩手・花巻南高校 尻引 正光先生

大学入試問題ではよく、平方根が絡んだ値と有理数との大小を判定する必要が出てくる。先生は、これに対して4つの判定法を示された。すなわち、
(ア) 開平法による近似値を利用
(イ) 差、2乗の差などによる不等式の証明
(ウ) 整数または有理数でのはさみうち
(エ) 2次関数のグラフの利用
このうち最初の2つが一般的であるが、ケースによって使い分けるというものである。
F sin・cos・tanの意味が理解できる指導法の提案 東京・南多摩高校 森本 貴彦先生

数学Tの三角比と、数学Uの三角関数を統一的に扱うことで、生徒がいまいち理解できていないsin・cos・tanの意味をとらえさせようという取り組みは、今年の私の課題でもあるので、大いに参考になった。私はsinだけをまず詳しく学習させればと考えている。
ただ、先生が提案された「中学校で三角比を学習する」ということに関しては無理があると思う。高校の側の都合だけで中学校の内容をとやかく言うことはできないのである。
G 基礎学力定着への取り組み 福島・勿来高校 井上 修司 先生

勿来高校は福島県いわき市の最南端にある普通科高校で、20年度からは3クラス規模になるという厳しい高校である。
ただそこで奮闘されている先生方や、勉強ができないなりに必死にとりくむ生徒の姿勢の中に、ヒントになるものをキラッと感じた。たとえば、2次関数の最大・最小問題では、一人の先生が放物線のグラフをマグネットでたどり、もう一人の先生(TT制)がそれにあわせてy軸をたどるというのは、値域の理解にとっては完璧である。なんとか生徒にわかってもらおうとした苦肉の策が実は最善の方法であるかもしれないのである。
私も前任校は農業高校であったから、そういう授業の場面は容易に想像できるが、基礎学力の定着という観点からは、貴重な示唆を与えられたと感謝している。
なお、指導助言の先生が指摘されたのは、2次関数のグラフの最大・最小で、頂点のところにそれをかく教師がいたりするが、それはまちがいとのこと。座標平面上の点そのものには最大も最小のないからである。
第90回全国算数・数学教育(福島)大会の3日目は、再び全体会で、シンポジウムの形式であった。

一番左はコーディネーターで筑波大学大学院准教授の清水静海氏、シンポジストは、文部科学省初等中等局教科調査官の永田潤一郎氏、一関市教育委員会指導主事の高畑嗣人氏と、

椙山女学園大学教授の浪川幸彦氏、東京都台東区立浅草中学校副校長の増田律子氏、成城学園初等学校教諭の島田功氏であった。
題目は「これからの算数・数学教育で期待されているもの」ということで、小学校・中学校ですでに告示された新学習指導要領についての論議が中心であった。

周知のとおり、算数・数学の授業時間および授業内容が増えた今回の改訂は現場の教師からは概ね歓迎されているということであった。
ただ、前の段階に単に戻るわけではないということは強調されていた。総合的な学習の時間も数学を増やすために減らされたものの、まだ残っているし、ゆとりとか生きる力も厳然として存在しているのである。
今回新たに入ったのは「算数的活動・数学的活動」と「資料の(整理ではなく)活用」というものだが、どうも言葉遊びの域を出ていないように思う。
最近教育に登場する用語はますます漠然としてきた感がある。「生きる力」もそうであるが、「活用」と「利用」は区別していると言われてもよくわからないのが正直なところである。私が英語でレポートを出す一つの理由が、ますます加速される用語のあいまいさなのである。とても日本語を用いてまともな教育研究ができそうもないというのが実感である。
Quarry a math method to understand perfectly quadratic function without a graph
〜To make mathematicsT be systematic and interesting for 1st grade students〜
Shiga・Ritto High School Yoshio Oka
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From the open lesson of mine
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Quadratic function is the most important in mathematicsT , but it is also the most difficult part for the 1st grade students in high school.
So I intend its study to be more systematic. For example, the meaning of square root should be taught as length of the side of square.
I did the open lesson on June 11, 2008, whose theme was drawing square root with compasses & ruler and simplifying square root, i.e. √8=2√2.
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Why do I think that square root is the most important? The reason is that it is the base for not only quadratic equation but also function.
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Many teachers hurry at the point of square root because they want to reach to quadratic function as fast as they could.
But I do not think hurrying to quadratic function is the best way to the perfect understanding it. We should not skip the studying square root as the preparation for it.
So I taught to students drawing square from any rectangle at first.
Secondly I proved that √2 is irrational by reduction to absurdity.
Thirdly I introduced extraction square root i.e. √2 ≒ 1.414 because memorizing the figure is nonsense.
Factorization is important to solve quadratic equation. But its practice is not enough in the textbook. So I intend its practice to be more systematic as following:
@ finding two integers whose sum and product are given
The 1st stage
( 1 ) sum is 3, product is 2 → 1 & 2
( 2 ) sum is 4, product is 3 → 1 & 3
(36) sum is 17, product is 72 → 8 & 9
The 2nd stage
( 1 ) sum is −3, product is 2 → −1 & −2
( 2 ) sum is −4, product is 3 → −1 & −3
(36) sum is −17, product is 72 → −8 & −9
The 3rd stage
( 1 ) sum is 1, product is −2 → −1 & 2
( 2 ) sum is 2, product is −3 → −1 & 3
(36) sum is 1, product is −72 → −8 & 9
The 4th stage
( 1 ) sum is −1, product is −2 → 1 & −2
( 2 ) sum is −2, product is −3 → 1 & −3
(36) sum is −1, product is −72 → 8 & −9
A factorization of type x2+(α+β)x+αβ
The 1st stage
( 1 ) x2+3x+2 = (x+1)(x+2)
( 2 ) x2+4x+3 = (x+1)(x+3)
(36) x2+17x+72 = (x+8)(x+9)
The 2nd stage
( 1 ) x2−3x+2 = (x−1)(x−2)
( 2 ) x2−4x+3 = (x−1)(x−3)
(36) x2−17x+72 = (x−8)(x−9)
The 3rd stage
( 1 ) x2+x−2 = (x−1)(x+2)
( 2 ) x2+2x−3 = (x−1)(x+3)
(36) x2+x−72 = (x−8)(x+9)
The 4th stage
( 1 ) x2−x−2 = (x+1)(x−2)
( 2 ) x2−2x−3 = (x+1)(x−3)
(36) x2−x−72 = (x+8)(x−9)
B filling the blank at the “tasuki-gake” (cross-product)
The 1st stage
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C factorization of type ax2+bx+c
The 1st stage
(1) 2x2+5x+2
(2) 2x2+5x+3
(16) 6x2+13x+6
The 2nd stage
(1) 2x2−5x+2
(2) 2x2−5x+3
(16) 6x2−13x+6
The 3rd stage
(1) 2x2+3x−2
(2) 2x2+x−3
(16) 6x2+5x−6
The 4th stage
(1) 2x2−3x−2
(2) 2x2−x−3
(16) 6x2−5x−6
Although square root is regarded as length of diagonal line of square, I think that is not universal.
@ Transformation from rectangle to square remaining its area
Step 1: put the point E such that AE=AB

Step 2: draw the semi-circle whose diameter is AD, and put the point P on the semi-circle such that PE⊥AE.

Step 3: draw the square PQRA, whose area equals to one of rectangle ABCD.

A Put the point of square root for any positive number which is given on the number line
Step 1:

Step 2:

Step 3:

B Meaning of simplifying square root
We usually simplify square root e.g. √12=2√3. But its meaning is not clear for students.
I was surprised when many students failed to answer 3√6 instead of 2×33 for the problem “factorize 54 to product of prime numbers.”
I wonder that they are taught factorizing to product of prime numbers only for simplifying square root in their junior high school.
So I added figure image for simplifying square root as following.
Type 1: √a=2√b

Type 2: √a=3√b

Type 3: √a=4√b

In these types a is the area of the given large square, b is that of small squares divided equally.
At the type 1, a=4b for the relation between the large square and the small one as the area.
On the other hand, √a=2√b for the relation between the large square and the small one as the length of side.
Thus it is not essential to factorize to product of prime numbers for simplifying square root. It is important whether a(number in the root signal √) is divided by square of some integers or not. For example:
8, 12, 20, 24 are divided by 4=22, so they are expressed 2√2,2√3,2√5,2√6.
18, 27, 45, 54 are divided by 9=32, so they are expressed 3√2,3√3,3√5,3√6.
32, 48, 80, 96 are divided by 16=42, so they are expressed 4√2,4√3,4√5,4√6.
Although in any mathT textbooks it is written that √2 is known to be irrational, it is not proven in that case.
On the other hand, in math A textbooks that is introduced for the example of proof by reduction to absurdity.
Fill the following blank.
@ Suppose that √2 is .
a
A Then we can think that √2= ―― ,
b
Provided that a and b are .(i.e. which have no common divider except 1)integers.
B Square both sides of A, and clear its denominator.
=a2
C From B, a2 is .
D If a is odd, then a2 is also odd. So from C, a is .
E From D, we can express that a= .(in the case, c is integer.)
F Substitute E for B, and arrange it, we get
b2= .
G By the same logic with that of C→D, we get bis .
H From D and G,a and bhave ,as the common divider which is not 1.
I This fact contradicts to the supposition that a andb is .
∴ √2 is not rational.(is irrational)
※ This is called “the proof by reduction to absurdity.”
In the textbook of mathT, it is written that √2 ≒ 1.41421356….
In some textbooks it is introduced with the pun “hito yo hito yo ni hito mi go ro”
1. 4 1 4 2 1 3 5 6
But memorizing the figure is nonsense. The way to get it (extraction) is more important.
1.
1 √2.000000
+)1 1
2 1
1. 4
1 √2.000000
+)1 1
24 1 00
+) 4 96
28 4
1. 4 1
1 √2.000000
+)1 1
24 1 00
+) 4 96
281 400
+) 1 281
282 119

Rationalizing denominator is important to calculate square roots.
But students cannot understand its meaning even though they can do it.
So I started teaching it with the reciprocal of a number containing square root.
(1) Calculate the following multiplication.
√2
@ √2 × ―――
2
√3
A 2√3 × ―――
6
The all answers are 1, so two numbers of multiplication are reciprocal numbers each other.
This fact means that e.g.
1 √2
――― = ―――
√2 2
(2) Calculate the following multiplication.
@ (√3+√2) × (√3−√2)
A (2+√3) × (2−√3)
The all answers are 1, so two numbers of multiplication are reciprocal numbers each other.
This fact means that e.g.
1
―――――― = √3−√2
√3+√2
(3) Calculate the following multiplication.
√3−1
@ (√3+1) × ―――――
2
√6−√2
A (√6+√2) × ――――――
4
This fact means that e.g.
1 √6−√2
―――――― = ――――――
√6+√2 4
(4) Rationalize denominator of the following fractions. 【type T】
3
@ ―――
√2
1
A ――――
2√5
(5) Rationalize denominator of the following fractions. 【type U】
1
B ――――――
√5+√2
√3+1
C ―――――
√3−1
(6) Rationalize denominator of the following fractions. 【type V】
3+√3
D ―――――――
√6(1+√3)
You had better device for the type V.
Quadratic equation is important for the prelude of quadratic function.
So I tried students to think the easiest form of quadratic equation to solve.
(1) The following eleven quadratic equations have the common two solutions even though their forms are different. Find the two solutions. And find the easiest form of quadratic equation to solve.
@ x2 = 6x−5
A x2+5 = 6x
B x2−6x = −5
C x2−6x+9 = 4
D (x−3)2 = 4
E (x−2)(x−4) = 3
The (common) two solutions: &
(the easiest) form of quadratic equation:
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(2) The following eleven quadratic equations have the common two solutions even though their forms are different. Find the two solutions. And find the easiest form of quadratic equation to solve.
@ x2 = 5x−6
A x2+6 = 5x
B x2−5x = −6
25 1
C x2−5x+ ―― = ――
4 4
5 1
D (x− ―― )2 = ――
2 4
The (common) two solutions: &
(the easiest) form of quadratic equation:
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(3) Meaning of the expression as (x−a)2=b
What do you think the following form of quadratic equation means?
(x−2)2 = 3
The form of ( )2 means the shape of square, so this equation expresses that the area of the left square whose side is x−2 length is 3.
As the length of side is √3,
x−2=√3 that is x=2+√3

But we can also think that the length of the left square is 2−x because x is unknown.
In this case 2−x=√3 i.e. x=2−√3.

According this fact, the solutions of the quadratic equation (x−2)2 = 3 are
x=2+√3,2−√3, i.e. x=2±√3
(4) Complete square for the quadratic equation
x2−6x+7=0
@ Transpose the constant term from the left side to the right one.
x2−6x = −7
A Find the half of the coefficient of linear term in the left side.
−6÷2=−3
B Find the square of the number in A.
(−3)2=9
C Add the number in B to both sides in @.
x2−6x+9 = −7+9
D Transform the left side to ( )2, and arrange the right side.
(x−3)2 = 2
(5) Complete square for the quadratic equation
3x2−12x−9=0
@ Divide the both sides by 3.
x2−4x−3 = 0
A Transpose the constant term from the left side to the right one.
x2−4x = 3
B Find the half of the coefficient of linear term in the left side.
−4÷2=−2
C Find the square of the number in B.
(−2)2=4
D Add the number in B to both sides in C.
x2−4x+4 = 3+4
E Transform the left side to ( )2, and arrange the right side.
(x−2)2 = 7
(6) Complete square for the general form of quadratic equation, and introduce formula of solution.
ax2+bx+c=0
(a≠0, and a,b,c are constant)
@ Divide the both sides by a.
A Transpose the constant term from the left side to the right one.
B Find the half of the coefficient of linear term in the left side.
C Find the square of the number in B.
D Add the number in B to both sides in C.
E Transform the left side to ( )2, and arrange the right side.
Then we can get completion of square root, and going on the process:
F Take off ( )2 in the left side, get the square root in the right side.
G Transpose the constant term from the left side to the right one.
※ So we can get the solution for general form of quadratic equation, and arranging its right side we get the formula of solution.
−b ± √ b2 − 4ac
x = ――――――――――――――
2a
The concept of function is very difficult for students. Its reason of difficulty is caused by generality of function. We should think individual concrete function each other.
For example, the textbook say that the linear function is the function which is written by y=ax+b, the linear expression of x.
But this explanation is hard for students to understand the point of linear function because they do not know why the linear expression is given at first. So I want to give the following table instead of the expression y=4x+1.
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x |
−3 |
−2 |
−1 |
0 |
1 |
2 |
3 |
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y |
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1 |
5 |
9 |
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Increase decrease |
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4 |
4 |
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We usually put the following 3 steps to understand function:
Step 1: given the expression of the function
Step 2: filling the blank in the table of x and y by calculating with the expression
Step 3: drawing the graph with the table
I think drawing the graph is insufficient to understand function. Maybe students are satisfied with looking at the shape of graph. But it is necessary for them calculating a lot to know the meaning of function.
Although filling the blank in the table of x and y is merely thought as preparation for drawing the graph, I want to treat it as the main method in the study of function.
(1) Linear function
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x |
−3 |
−2 |
−1 |
0 |
1 |
2 |
3 |
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y |
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1 |
5 |
9 |
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Increase decrease |
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4 |
4 |
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Look at the above table. It is only known that the function is linear; i.e. increase or decrease, e.g. velocity, is constant.
On the other hand, if function is quadratic then the increase or decrease of the increase or decrease, e.g. acceleration, is constant.
(2) Quadratic function
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x |
−3 |
−2 |
−1 |
0 |
1 |
2 |
3 |
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y |
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1 |
4 |
9 |
16 |
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Increase decrease |
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3 |
5 |
7 |
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In(de)crease of in(de)crease |
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2 |
2 |
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tAlthough two given points, (0,1),(2,9), are the same, two function are different according to whether the third point is (1,5) or (1,4).
Moreover we can define linear & quadratic function as the following:
Linear function is the function whose in(de)crease are constant.
Quadratic function is the function whose in(de)crease of in(de)crease are constant.
In textbooks there are four types at the introduction of quadratic function.
But I don’t use the graph at this time because the graph is not expressed numerically. I prefer table of correspondence of x and y.
(1) Quadratic function type T
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x |
−3 |
−2 |
−1 |
0 |
1 |
2 |
3 |
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y |
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8 |
2 |
0 |
2 |
8 |
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Increase decrease |
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−2 |
−2 |
2 |
6 |
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In(de)crease of in(de)crease |
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4 |
4 |
4 |
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At the type T they are symmetry for the y axis and vertex is the origin (0,0). Our students have studied only this type in junior high schools.
(2) Quadratic function type U
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x |
−3 |
−2 |
−1 |
0 |
1 |
2 |
3 |
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y |
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9 |
3 |
1 |
3 |
9 |
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Increase decrease |
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−6 |
−2 |
2 |
6 |
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In(de)crease of in(de)crease |
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4 |
4 |
4 |
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At the type U they are symmetry for the y axis and vertex is the point (0,1) on y axis.
(3) Quadratic function type V
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x |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
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y |
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8 |
2 |
0 |
2 |
8 |
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Increase decrease |
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−6 |
−2 |
2 |
6 |
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In(de)crease of in(de)crease |
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4 |
4 |
4 |
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At the type V they are symmetry for the line x=3 which is parallel to y axis and vertex is the point (3,0) on x axis.
(4) Quadratic function type W
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x |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
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y |
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9 |
3 |
1 |
3 |
9 |
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Increase decrease |
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−6 |
−2 |
2 |
6 |
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In(de)crease of in(de)crease |
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4 |
4 |
4 |
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At the type W they are symmetry for the line x=3 which is parallel to y axis and vertex is the point (3,1) on neither y axis nor x axis.
I will do open lesson on November 6 , 2008. Quadratic function will be the theme of it.