Question

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

Answer 1

minimum vertex at `-49/24` and symetry at `x = - 1/12`

Explanation:

it can be solve by using completing a square.

`y = 6 x^2 + x - 2`
`y = 6(x^2 + 1/6 x) -2`

`y = 6(x + 1/12)^2 - 6(1/12)^2 -2`

`y = 6(x + 1/12)^2 - 1/24 -48/24`

`y = 6(x + 1/12)^2 - 49/24`

since coefficient of `(x + 1/12)^2` is +ve value, it has a minimum vertex at `-49/24` and it symetry at `x = - 1/12`