Question

What is the vertex of `y= 1/3(x/5+1)^2+4/15`?

Answer 1

The vertex form is `y = a(x-h)^2 +k` where `(h,k)` is the vertex. For our problem the vertex is `(-5,4/15)`

Explanation:

`y=1/3(x/5+1)^2 + 4/15`

`y=1/3((x+5)/5)^2 + 4/15`

`y=1/75(x+5)^2 + 4/15`

Compare with
`y=a(x-h)^2 + k`

`h=-5` and `k=4/15`

The vertex `(h,k)` is `(-5,4/15)`