Question

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

Answer 1

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

Explanation:

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)`