Question

What equation is `y=(x+3)^2 + (x+4)^2` rewritten in vertex form?

Answer 1

`y=2(x+7/2)^2+1/2`

Explanation:

This is a bit of a sneaky question. It isn't immediately obvious that this is a parabola, but "vertex form" is a form of equation specifically for one. It is a parabola, a closer look reveals, which is fortunate... It's the same thing as "completing the square" - we want the equation in the form `a(x-h)^2+k`.

To get there from here, we first multiply out the two brackets, then collect up terms, then divide through to make the `x^2` coefficient 1:
`1/2y=x^2+7x+25/2`

Then we find a square bracket that gives us the correct `x` coefficient. Note that in general
`(x+n)^2=x^2+2n+n^2`
So we choose `n` to be half our existing `x` coefficient, i.e. `7/2`. Then we need to subtract off the extra `n^2=49/4` that we've introduced. So
`1/2y=(x+7/2)^2-49/4+25/2=(x+7/2)^2+1/4`

Multiply back through to get `y`:
`y=2(x+7/2)^2+1/2`