Question

求学霸帮忙讲下这道题，初三的数学，二次函数，化为顶点式

Answer 1

y=3(x² + 2x/3)
=3[x² + 2x/3 + (1/3)² - (1/3)²]
=3(x + 1/3)² - 3•(1/3)²
=3(x + 1/3)² - 1/3

追问

再帮帮我

4个题

追答

第6题？

追问

嗯嗯

可以吗？

追答

在另一题中答好了

Answer 2

y=3x^2+2x
=3*(x^2+2x/3)
=3*(x+1/3)^2-1/3

追问

亲，你再帮帮我吧

四个题

追答

1. y=-3x^2+12x-3

     =-3(x^2-4x+1)

     =-3(x^2-4x+4-3)

     =-3(x^2-4x+4)+9

     =-3(x-2)^2+9

2. y=4x^2-24x+26

     =4(x^2-6x+13/2)

     =4(x^2-6x+9-5/2)

     =4(x^2-6x+9)-10

     =4(x-3)^2=10

3. y=2x^2+8x-6

     =2(x^2+4x-3)

     =2(x^2+4x+4-7)

     =2(x^2+4x+4)-14

         =2(x+2)^2-14

   4. y=1/2x^2-2x-1

        =1/2(x^2-4x-2)

        =1/2(x^2-4x+4-6)

        =1/2(x^2-4x+4)-3

        =1/2(x-2)^2-3