Question

What is the vertex form of `y= (6x-6)(x+2)+4x^2+5x`?

Answer 1

Vertex form of equation is `y= 10(x+0.55)^2-15.025`

Explanation:

`y=(6x-6)(x+2)+4x^2+5x` or

`y=6x^2+12x-6x-12+4x^2+5x` or

`y=10x^2+11x-12 or y= 10(x^2+11/10x)-12` or

`y= 10{x^2+11/10x+(11/20)^2}-10*(11/20)^2-12` or

`y= 10(x+11/20)^2-3.025-12` or

`y= 10(x+0.55)^2-15.025` .Comparing with standard vertex form

of equation `f(x)=a(x-h)^2+k ; (h,k)` being vertex we find

here `h= -0.55 , k=-15.025` So vertex is at

`(-0.55,-15.025)` and vertex form of equation is

`y= 10(x+0.55)^2-15.025` [Ans]