Question

What is the vertex form of `y=9x^2-48x+64`?

Answer 1

You can see a more in-depth build approach example at https://socratic.org/s/aCybisPL

`y=9(x-8/3)^2`

Explanation:

`color(blue)("Preamble")`

If you can do so it is worth committing to memory the standardised form.

Using `y=ax^2+bx+c` as the bases we have the vertex form format of:

`y=a(x+b/(2a))^2+k+c`

The extra `k` is a correction that 'gets rid' if the error introduced by squaring the `+b/(2a)` part of `(x+b/(2a))^2`
The `(b/(2a))^2` part is not in the original equation.

Do not forget about the whole bracket being multiplied by a
So to get rid of it we set: `" "a(b/(2a))^2+k=0`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Building the vertex form")`

Write as `y=9(x-48/(2(9)))^2+k+64`

`9(-48/18)^2+k=0`

`k=-64`

Thus we have

`y=9(x-8/3)^2-64+64`

`y=9(x-8/3)^2`