What is the vertex of #y= 3x^2 - 4x - 12#?

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Answer 1 · 2018-08-10T08:40:46

Vertex is #(2/3,96/9)#

To find vertex convert the equation #y=3x^2-4x-12# to the form

#y=a(x-h)^2+k# and then vertex is #(h,k)#.

We have #y=3x^2-4x-12#

= #3(x^2-2xx2/3x+(2/3)^2-(2/3)^2)-12#

= #3(x^2-2xx2/3x+(2/3)^2)-3*4/9+12#

= #3(x-2/3)^2+96/9#

Hence, vertex is #(2/3,96/9)#