Question

What is the vertex form of `y= 6x^2 + 20x + 6`?

Answer 1

Vertex form of equation is `y=6(x +5/3)^2-96/9`

Explanation:

Vertex form of equation is `y=a(x-h)^2+k ; (h.k)` being vertex.

`y= 6x^2+20x+6 or y=6(x^2+20/6x)+6` or

`y=6(x^2+10/3x)+6` or

`y=6{x^2+10/3x +(5/3)^2}+6-150/9` [`150/9` is added and

subtracted simultaneously to make a square]

`:. y=6(x +5/3)^2-96/9` , here `h= -5/3 and k= -96/9`

So vertex is at `(-5/3 ,-96/9)` and vertex form of equation is

`y=6(x +5/3)^2-96/9` [Ans]

Answer 2

`y=6(x-(-5/3))^2+(-32/3)`

Explanation:

Let's start by recognizing the general vertex form which will be our target:
`color(white)("XXX")y=color(green)m(x-color(red)a)^2+color(blue)bcolor(white)("xxx")` with vertex at `(color(red)a,color(blue)b)`

Given
`color(white)("XXX")y=6x^2+20x+6`

We will first separate the `x` terms and the constant:
`color(white)("XXX")y=6x^2+20xcolor(white)("xxxxx")+6`
then extract the `color(green)m` factor from the `x` terms:
`color(white)("XXX")y=color(green)6(x^2+10/3x)color(white)("xxxxx")+6`

To "complete the square" fro the `x` terms, remember that
`color(white)("XXX")(x+k)^2=(x^2+2kx+k^2)`
In this case since we already have `x^2+10/3x`
the value of `k` must be `10/6=5/3`
and
we will need to add `k^2=(5/3)^2=25/9` to "complete the square".

Obviously, if we are going to add an amount someplace we will need to subtract it someplace else to keep everything equal to the original expression.
...but how much do we need to subtract?
If we look carefully we see that we will not just be adding `25/9` but we will be adding this amount times the `color(green)m=color(green)6` factor.
So we will need to subtract `color(green)6xx25/9=50/3`

We now have:
`color(white)("XXX")y=color(green)6(x^2+20xcolor(magenta)(+25/9))color(white)("xxxx")+6color(magenta)(-50/3)`

If we re-write the parenthesized component as a squared binomial and simplify the constants we get
`color(white)("XXX")y=color(green)6(x+5/3)^2color(white)("xxx")-32/3`
or, in explicit vertex form
`color(white)("XXX")y=color(green)6(x-color(red)(""(-5/3)))^2+color(blue)(""(-32/3))`
`color(white)("XXXXXXXXXXXXXXX")` with vertex at `(color(red)(-5/3),color(blue)(-32/3))`

The graph below of the original equation indicates that this answer is "reasonable" (although I haven't figured out how to capture it with the vertex coordinates displayed)
graph{6x^2+20x+6 [-5.582, 2.214, -11.49, -7.593]}