Question

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

Answer 1

( x `-` 3/4 )^2 = 1/6 ( y + 3/8 ).
The vertex of the parabola is ( 3/4, `-`3/8).

Explanation:

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.

Answer 2

`y=6(x-3/4)^2+39/16`

The solution method has been shown in a lot of detail

Explanation:

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)`