What is the vertex form of $y=6x^2+17x+12$ ?

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Answer 1 · 2016-02-16T16:35:50

$6(x+17/32)^2 + 5277/512$ This is the required vertex form. Vertex is $(-17/32, 5277/512)$

It is $y= 6(x^2 +(17x)/6) +12$

= $6(x^2 + (17x)/16 + 289/1024 -289/1024)+12$

= $6(x + 17/32)^2 + 12 -6(289/1024)$

= $6(x+17/32)^2 + 5277/512$ This is the required vertex form. Vertex is $(-17/32, 5277/512)$

What is the vertex form of y=6x^2+17x+12 y=6x^2+17x+12 ?