Question

What is the vertex form of `y=-3x^2-2x+1`?

Answer 1

The vertex form is the following,
`y=a*(x-(x_{vertex}))^2+y_{vertex}`
for this equation it is given by:
`y=-3*(x-(-1/3))^2+4/3`.

It is found by completing the square, see below.

Explanation:

Completing the square.

We begin with
`y=-3*x^2-2x+1`.

First we factor the `3` out of `x^2` and `x` terms
`y=-3*(x^2+2/3 x)+1`.
Then we separate out a `2` from in from of the linear term (`2/3x`)
`y=-3*(x^2+2*1/3 x)+1`.

A perfect square is in the form

`x^2 + 2*a*x+a^2`,

if we take `a=1/3`, we just need `1/9` (or `(1/3)^2`) for a perfect square!

We get our `1/9`, by adding and subtracting `1/9` so we don't change the value of the left hand side of the equation (because we really just added zero in a very odd way).

This leaves us with
`y=-3*(x^2+2*1/3 x+1/9-1/9)+1`.

Now we collect the bits of our perfect square
`y=-3*((x^2+2*1/3 x+1/9)-(1/9))+1`
Next we take the (-1/9) out of the bracket.

`y=-3*(x^2+2*1/3 x+1/9) + (-3)*(-1/9)+1`

and neaten up a bit

`y=-3*(x^2+2*1/3 x+1/9)+(3/9)+1`
`y=-3*(x+1/3)^2+4/3`.
Remember the vertex for is
`y=a*(x-(x_{vertex}))^2+y_{vertex}`
or we turn the plus sign into two minus signs producing,
`y=-3*(x-(-1/3))^2+4/3`.
This is the equation in vertex form and the vertex is `(-1/3,4/3)`.