F4的數學原理速成班

梓

1.一元二次方程
2.函數及其圖像
3.續多項式
4.續方程

由於阿Sir教得太快,所以不大明白點做呢d數/.\←唔識做=星期二的數學考試唔合格/.\

請各人兄幫手教教我/.\
我只需原理就行了..../.\

能幫到我的會加分(!?)

Nidoq

有沒有英文名~_~?

梓

原帖由 Nidoq 於 7/12/2007 05:04 PM 發表

                               
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有沒有英文名~_~?

1.一元二次方程=Quadratic equation with one unknown
2.函數及其圖像=Function
3.續多項式=?,(8X^3-4X^2+3X-2)+(3X^2-X+1)之類
4.續方程=Equation(?),X^3-2=0之類

inthomaspy

亂來的
For Functions
-features of the graph of y=ax^2+bx+c
- Axis of Symmetry
   the parabola is symmertrical about a line x=h
- Vertex
  truning point (h,k) of a parabola
- Direction of opening
   if a>0, the parabola opens upward (that means the vertex is the minimum point of the parabola)
   if a<0, the parabola opens downward (that means the vertex is the maximum point of the parabola)
-y-intercept and x-intercept
   y-intercept=the intersection point of the parabola and y-axis
   x-intercept=the intersection point of the parabola and x-axis
   consider graph of y=x^2-3x-4
   the y-intercept: when x=0, y=0^2-3(0)-4
                                                   =-4
   so the coordinates of y-intercept of the graph is (0, -4)
   the x-intercept: when y=0, x^2-3x-4=0
                                                 (x-2)(x-3)=0
                                                 x=2 or 3
conclusion:
consider quadratic graph y=ax^2+bx+c
c is the y-intercept of the graph
the opening direction of the graph depends on the value of a, is it bigger or smaller than 0

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Added
consider y=a(x-h)^2+k
for a>0, parabola opens upwards
and becuase (x-h)^2>=0 and a>=0
so a(x-h)^2>=0
so, minimum value of a(x-h)^2=0
so, y=a(x-h)^2+k has minimum value of kwhen x=h (that's means a[x-h]^2=a[h-h]^2=0)
for a<0, parabola opens downwards
because (x-h)^2>=0 and a<0 (negative value)
so maximum value of a(x-h)^2=0
so the vertex (h, k)is the highest point of parabola

conclusion
consider y=a(x-h)^2+k
the value of a means the opening direction of parabola
h (NOT -h)means the x-coordinates of the vertex while k means the y coordinates of the vertex / maximum or minimum point of the vertex.

method to find maximum or minimum pointof quadractic function
method 1: using formula: -b/2a
consider expression y=2x^2+3x+1
y-coordinates of vertex=-3/2(2)=-3/4
method 2: by completing square
y=2x^2+3x+12
  =2(x^2+3x/2)+1                        facrotize the expression until the coefficient of x^2 is 1
  =2[x^2+3x/2+(3/4)^2-(3/4^2)]  divide 3/2 by 2 and obtain the number in red
  =2(x+3/4)^2-2(3/4)^2+1
  =2(x+3/4)^2-1/8
  =2[x-(-3/4)]^2-1/8
so y-coordinates of vertex=-3/4
x-coordinates of vertex=-1/8
其實兩種方法都可以求 (h, k), 但method 2 比較考技術, 另外它亦是一個比較高level的做法, 建議用method 2 做這類題目

reflection: consider y=f(x)
y=f(x)+k means translate tha graph y=f(x) k units upward
y=f(x)-k means translate tha graph y=f(x) k units downwards
y=f(x+h) means translate the graph y=f(x) h units to the left
y=f(x-h) means translate the graph y=f(x) h units to the right
y=f(-x) means reflect the graph y=f(x) along y-axis
y=-f(x) means reflect the graph y=f(x) along x-axis

that's all for function



[ 本帖最後由 inthomaspy 於 7/12/2007 10:10 PM 編輯 ]

超夢夢的夢境

用d英文解釋,有問題再問
樓上的quadratic graph解得唔錯,加d補充
quadratic functions(completing square):
y=ax^2+bx+c
y=a[x+(b/2a)]^2+c-(b^2)/2a

上面o既h係-(b/2a)
(記住個負號)
k就係c-(b^2)/2a

當 x=h 果陣, y 達至minimum value=kifa>0
當 x=h 果陣, y 達至maximum value=k if a<0
原因係一個square大過/等如0

原帖由 冰火 於 7/12/2007 05:24 PM 發表

                               
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1.一元二次方程=Quadratic equation with one unknown
2.函數及其圖像=Function
3.續多項式=?,(8X^3-4X^2+3X-2)+(3X^2-X+1)之類
4.續方程=Equation(?),X^3-2=0之類

函數及其圖像=Function and its image(直譯...)

f(ax)係縮窄
af(x)係拉長
(a>1)(0<a<1係掉轉)

-f(x)係reflect about x-axis
f(-x)係reflect about y-axis

f(x+b)係向左shift
f(x)+b係向上
(b>0)(b<0係掉轉)

總之o係括號入面,個transformation就係打橫
出面就係打直

多項式=polynomials
教remainder theorem?
remainder when f(x) is divided by (x-a):f(a)
Since f(x)=(x-a)Q(x)+r(x)
f(a)=(a-a)Q(x)+r(x)=r(x)
f(a)=0即係divisible

equation?有d乜?唔記得左...
都可以用subtraction(相減)或者叫elimination(消去)
x^2+xy=2...................................(1)
y-3x=7.......................................(2)
(1) - x*(2), x^2+3x^2=2-7x
4x^2+7x-2=0
之後同樓下o既做法一樣,省略
不過用subtraction o既話,觀察力要比較好,而且唔係每一條都可以用subtraction做到
用substitution比較保險

EDIT:橙字係後加o既(quadratic graph度都加左d)

Quadratic equation:discriminant=b^2-4ac
discriminant>0  implies 2 real roots
discriminant=0 implies 1 repeated real roots
discriminant<0  implies no real root

no real root 的話,考試就搵discriminant出黎,話佢細過0,所以no real root

[ 本帖最後由 超夢夢的夢境 於 8/12/2007 09:35 AM 編輯 ]

梓

原帖由 超夢夢的夢境 於 7/12/2007 09:09 PM 發表

                               
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equation?有d乜?唔記得左...

有聯立方程式
? x^2+xy=2
?
? y-3x=7

雪貓

原帖由 冰火 於 7/12/2007 09:18 PM 發表

                               
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有聯立方程式
? x^2+xy=2
?
? y-3x=7

用x/y做subject的時候一定要用沒有次方那條調位,不然可能會有開方~.~
x^2+xy=2 ----(1)
y-3x=7-------(2)
from (2)
y=7+3x -----(3)
sub(3)into(1)
x^2+(7+3x)x=2
4x^2+7x-2=0
(4x-1)(x+2)=0
x=1/4 or -2
再每個sub回去..但一定要sub去沒有2次的那條,不然又多答案- -
when x=1/4,(sub數..)
when x= -2,又sub
就會得出x=1/4,y=? or x= -2,y=?

[ 本帖最後由 雪貓 於 7/12/2007 10:31 PM 編輯 ]

鴨

話說啦...
題外話...
Quadratic Equation的精髓不就是懂如何用計數機嗎...""

梓

原帖由 RX78-2 於 7/12/2007 10:28 PM 發表

                               
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話說啦...
題外話...
Quadratic Equation的精髓不就是懂如何用計數機嗎...""

雖說是...但是有d數係Math Erorr..計唔到的...

sapphire

原帖由 冰火 於 7/12/2007 11:28 PM 發表

                               
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雖說是...但是有d數係Math Erorr..計唔到的...

那些是complex roots
CE的話寫 no real roots 就可以了