Question

What is the the vertex of `x = (y - 6)^2 - y+1`?

Answer 1

Vertex is `(-5 1/4,-6 1/2)`

Explanation:

We can write `x=(y-6)^2-y+1` as

`x=y^2-12y+36-y+1`

= `y^2-13y+(13/2)^2-169/4+37`

= `(y-13/2)^2-(169-148)/4`

= `(y-13/2)^2-21/4`

Hence vertex is `(-21/4,-13/2)` or `(-5 1/4,-6 1/2)`