find the vertex and the axis of symmetry
y=2(2-x)(x+2)-4x
THZ,
未訓就幫下手啦
2007-10-11 7:57 am
回答 (4)
2007-10-11 8:05 am
✔ 最佳答案
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
2007-10-11 8:07 am
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
= 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
2007-10-11 8:05 am
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
= 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
2007-10-11 8:04 am
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
= - 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
參考: My Maths Knowledge
收錄日期: 2021-04-13 19:30:57
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071010000051KK04875