quadratic functions

yoshua

English
1.let f(x)=-(x-a)^2+b,where a and b are real.Point P is the vertex of the graph of y=f(x)
the coordinates of point P is (a,b)
2.let g(x) be aquadratic function such that the coefficient of x^2 is 1 and the vertex of the graph of y=g(x) is point Q(b,a).
It is given that the graph of y=f(x) passes through point Q
Question:
g(x)=(x-b)^2+a,show that the graph of y=g(x) passes through point P

中文
1.設f(x)=-(x-a)^2+b，其中a、b為實數。點p是y=f(x)的頂點。
p座標為數(a,b)
2.設g(x)為二次[咸]數，x^2項係數是1，y=g(x)頂點為q(b,a)。已知y=f(x)通過q
g(x)=(x-b)^2+a，證明y=g(x)通過p

望各位詳細些解阿，thx~~~:smilie_:):

36661124

target :g(a)=b
we have f(x)=-(x-a)^2+b
               g(x)=(x-b)^2+a
and f(b)=a
      ie  - (b²  + a² - 2ab)+b=a
                   b=a +(b²  + a² - 2ab)
prove : g(a)=a²+ b² -2ab +a
                   =b