Function 問題!_-

e.g. y=f(x)
=2x^2+4x+3
=2(x+1)^2+1
vertex=(-1,1)

y=f(x-3)
=2(x-3)^2+4(x-3)+3
=2x^2-8x+9
=2(x-2)^2+1
vertex=(2,1)
第一:因為(x-3)呢個值只影響到x唔會影響到y(substitude x-3 into x)
所以首先f(x-3)只係沿住x-axis走
第二:當你substitude x-3into f(x)既時候,,,
你會發現 x-?既?值大左或者x+?既?值細左
eg. y=x^2+2x+1
=(x+1)^2
y=f(x)=[(x-3)+1]^2
=(x-2)^2
咁x既值咪大左囉!
因此f(x-3) 係向住positive direction of x-axis 走ga!!

原因係因為我地completing square o個陣係in “(x-h)²+k” form

咁x-h=0係axis of symmetry 架麻...

所以h越細，x就越大
亦即是話我地計到f(x-1), complet出黎個square個h就一定係細左
所以x就大
x大就即是向x-axis positive走。


舉個例：

f(x) = x²+2x+1
= (x+1)²

By consider ax²+bx+c into the form (x-h)²+k, axis of symmetry is x=-1


f(x-1) = (x-1)²+2(x-1)+1
= x²-2x+1+2x-2+1
= x²

By consider ax²+bx+c into the form (x-h)²+k, axis of symmetry is x=0


Compare the two cases above, f(x-1) is translated 1 unit rightwards.


(比較佢地axis of symmetry就已經足夠去證明架啦，
因為佢地都係x²，唔會話條curve個弧度大細唔同)

假設 y1=f(x)=x, 咁 y2=f(x-3)=x-3
如果 x=0,
y1=f(x)=f(0)=0
y2=f(x-3)=f(-3)=-3
如果 x=-3;
y1=f(x)=f(-3)=-3
y2=f(x-3)=f(-6)=-6
由此可以睇到,如果y唔變,即係y1=y2,
原本 y1=f(x)=-3 係(-3,-3), 之後變咗 y2=f(x-3)=-3 係(0,-3),
咁個x-coordinate就由-3 向右去咗 0, 咁就睇到佢向住positive direction of x-axis 走ga!!