Question

What is the vertex form of `y= x^2 -6x+8`?

Answer 1

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

Explanation:

The general vertex form is
`color(white)("XXX")y=m(x-a)^2+b` for a parabola with vertex at `(a,b)`

To convert `y=x^2-6x+8` into vertex form, perform the process called "completing the square":

For a squared binomial `(x+k)^2 = color(blue)(x^2+2kx)+k^2`
So if `color(blue)(x^2-6x)` are the first two terms of an expanded squared binomial, then `k=-3` and the third term must be `k^2=9`

We can add `9` to the given expression to "complete the square", but we we also need to subtract `9` so that the value of the expression stays the same.

`y=x^2-6x color(red)(+9)+8 color(red)(-9)`

`y=(x-3)^2-1`
or, in explicit vertex form:
`y=1(x-3)^2+(-1)`

Typically I leave the value `m` off when it is `1` (the default anyway) but find that writing the constant term as `+(-1)` helps me remember that the `y` coordinate of the vertex is `(-1)`