中四A.Maths..help(Quadratic Equation)

let y=f(x)=a(x-b)(x-c)
when y=0, x=1, 3
i.e. 0=a(x-b)(x-c)
........(x-b)(x-c)=0
........x=b or c
therefore, when b=1, c=3---------------(1)
..............when b=3, c=1 ---------------(2)

when x=0, y=k
i.e. k=a(0-b)(0-c)------------------------(3)
substituting (1) into (3)
.......k=3a
......a=k/3
substituting (2) into (3)
.......k=3a
.......a=k/3
therefore, overall a=k/3

f(x)=k/3(x-1)(x-3) or f(x)=k/3(x-3)(x-1)
therefore, overall f(x)=k/3(x-1)(x-3)

任何二次方程都可以表達為a(x-b)(x-c)的形式。其中b及c就是方程的兩個根。

根據factor therom，對於所有Polynomial F(x),如果F(a)=0，F(x)=(x-a)Q(x)，其中Q(x)是另外一個polynomial。

因為f(1)=0,f(3)=0，所以b及c都分別等於1及3這兩個數。任何二次方程都有及只有兩個根。將f(x)展開，常數項為abc=3a，f(0)=k,所以3a=k，a=k/3

故此可以不需經過繁複的運算也能得到答案了。

當然，個別地代入然後運算也能得到上述答案的，但這就是因為不太熟知根與方程的關係的情況了。

The graph of a quadratic function y=f(x) which intersects the x-axis at(1,0) and (3,0) and cuts the y-axis at P(0,k) with k larger than 0.

Express f(x) in the form a(x-b)(x-c) where a,b and c are constants.
a(1-b)(1-c)=a(3-b)(3-c)=0
1-b-c+bc=9-3b-3c+bc
8-2b-2c=0
b+c=4
b=4-c....(1)
2bc=6
bc=3
c(4-c)=3
4c-c²=3
c²-4c+3=0
c=1 or c=3
b=3 or b=1

abc=k
3a=k
a=k/3

so answer is k/3(x-1)(x-3)