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

Answer 1

Eva Abramson

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

Let's begin with the provided 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[-30,30,60,30]}

May God bless you all. I hope this explanation helps.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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