未訓就幫下手啦

y = 2(2-x)(x+2) - 4x
= 2(4 - x^2) - 4x
= -2x^2 - 4x + 8
= -2(x^2 + 2x) + 8
= -2(x^2 + 2x + 1) + 2 + 8
= -2(x + 1)^2 + 10
vertex = (-1, 10)
axis of symmetry : x = -1

2007-10-11 00:06:28 補充：
咁搶手, 幫幫忙, 投我一票啦唔該

y = 2(2 - x)( 2 + x) - 4x
= 2(4 - x^2) - 4x
= 8 - 2x^2 - 4x
= -2(x^2 + 2x - 4)
= -2(x^2 + 2x + 1) + 10
= -2(x +1)^2 + 10
the vertex is (-1, 10),
axis of symmetry is x = -1

y = 2(2 - x)( 2 + x) - 4x
= 2(4 - x^2) - 4x
= 8 - 2x^2 - 4x
= -2(x^2 + 2x - 4)
= -2(x^2 + 2x + 1) + 10
= -2(x +1)^2 + 10
So the vertex is (-1, 10), and the axis of symmetry is x = -1

y = 2 ( 2 - x )( x + 2 ) - 4x

= - 2 ( x^2 - 4 ) - 4x

= - 2x^2 - 4x + 8

= - 2 ( x^2 + 2x + 1 ) + 8 + 2 ( 1 )

= - 2 ( x + 1 )^2 + 10

Vertex: ( - 1 , 10 )

Axis of symmetry: x = -1


2007-10-11 00:06:51 補充：
The Q can be solved by completing square, i.e., rewrite the function in the form ofy = a ( x - h )^2 + k