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

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Answer 1 · 2017-11-23T12:15:52

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

#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]