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.

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