Question

What is the vertex form of `y= x^2-6x+5`?

Answer 1

`y=(x-3)^2+(-4)` with vertex at `(3,-4)`

Explanation:

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

Given `y=x^2-6x+5`

We can "complete the square"
`color(white)("XXX")y=x^2-6xcolor(red)(+3^2)+5color(red)(-3^2)`

`color(white)("XXX")y=(x-3)^2-4`

Answer 2

`y=(x-3)^2-4`

Explanation:

To find the vertex form of the equation, we have to complete the square:

`y=x^2-6x+5`

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

When completing the square, we must make sure the bracketed polynomial is a trinomial. So `c` is `(b/2)^2`.
`y=(x^2-6x+(6/2)^2-(6/2)^2)+5`

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

`y=(x^2-6x+9-9)+5`

Multiply `-9` by the `a` value of `1` to bring `-9` outside of the brackets.
`y=(x^2-6x+9)+5-(9*1)`

`y=(x-3)^2+5-(9)`

`y=(x-3)^2-4`

`:.`, the vertex form is `y=(x-3)^2-4`.