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

1 Answer
Mar 29, 2017

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#