What is the vertex form of #y= 6x^2+16x-12 #?

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Answer 1 · 2016-07-10T14:26:01

Vertex form
#(x+4/3)^2=1/6(y+68/3)" "#with Vertex at #(-4/3, -68/3)#

Let us start from the given equation

#y=6x^2+16x-12#

#y=6(x^2+16/6x)-12#

#y=6(x^2+8/3x+16/9-16/9)-12#

#y=6(x^2+8/3x+16/9)-((6*16)/9)-12#

#y=6(x+4/3)^2-68/3#

#y+68/3=6(x+4/3)^2#

#1/6(y+68/3)=(x+4/3)^2#

#(x+4/3)^2=1/6(y+68/3)#

Kindly see the graph of #(x+4/3)^2=1/6(y+68/3)" "#with Vertex at #(-4/3, -68/3)#

graph{y=6x^2+16x-12[-60,60,-30,30]}

God bless....I hope the explanation is useful.