Question

What is the the vertex of `x = (y -3)^2 - 5y-5`?

Answer 1

`11/2,-105/4`

Explanation:

Let `f(y)=(y-3)^2-5y-5`
then we get by using

`(a-b)^2=a^2-2ab+b^2`

`f(y)=y^2-6y+9-5y-5`
combining like terms

`f(y)=y^2-11y+4`

we compute the coordinates of the vertex:_

`f'(y)=2y-11`
so

`f'(y)=0` if `y=11/2`

and
`f(11/2)=-105/4`