What is the vertex form of #y= 6x^2-9x+3 #?

The axis of the parabola is in the positive y-direction.
The focus is at ( 3/4, #-#3/8 #-#1/24.) = ( 3/4, #-#5/12.), below the vertex on the axis x = 3/4.

Given:#" "y=6x^2-9x+3# ................................(1)

This process introduces an error. This error is dealt with by introducing a correction constant.

Let the correction constant be #color(green)(k)#

#color(blue)("Step 1")#

Write as:#" "y=6(x^2-9/6x)+3#

Note that#" " 6xx(-9/6)x = -9x#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 2")#

#color(brown)("Move the index of 2 from "x^2" to outside the brackets.")#

We have now changed the value of the RHS so unable at this stage to equate it to #y#

Write as:#" "6(x^(color(magenta)(2))-9/6x)+3" " ->" "6(x-9/6x)^(color(magenta)(2))+3 #
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 3")#

#color(brown)("Remove the right hand side "x" from inside the brackets.")#

#6(x-9/6color(magenta)(x))^2+3 " "->" "6(x-9/6)^2+3 #

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 4")#

#color(brown)("Add correction constant of "color(green)(k))#

#6(x-9/6)^2+3" " ->" " 6(x-9/6)^2+3 +color(green)(k)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 5")#

#color(brown)("Halve the "9/6" inside the brackets")#

#=> y=6(x-9/12)^2+3 +k#.........................(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Solving for "k)#

Expand the brackets of equation (2)

#y=6x^2-9x+81/144+3+k#

The #81/144# is the error that #k# is correcting

So to get rid of #81/144# we make k # -81/144#

so our constant become: #" "3-81/144 = 2 7/16 = 39/16#

Equation (2) becomes: #" "y=6(x-9/12)^2+39/16" ".....(2_a)#

But #9/12 = 3/4# so we now have

#color(magenta)(" "y=6(x-3/4)^2+39/16)" ".....(2_b)#