Question

What is the vertex form of `5y = 11x^2-15x-9`?

Answer 1

`y=11/5(x-15/22)^2-621/220`

Explanation:

Vertex form of such equation is `y=a(x-h)^2+k`, with `(h,k)` as vertex.

Here we have `5y=11x^2-15x-9`

or `y=11/5x^2-3x-9/5`

or `y=11/5(x^2-3xx5/11x)-9/5`

`=11/5(x^2-2xx15/22 x+(15/22)^2-(15/22)^2)-9/5`

`=11/5(x-15/22)^2-(15/22)^2xx11/5-9/5`

`=11/5(x-15/22)^2-45/44-9/5`

`=11/5(x-15/22)^2-(45xx5+44xx9)/220`

`=11/5(x-15/22)^2-(225+396)/220`

`=11/5(x-15/22)^2-621/220`

and vertex is `(15/22,-621/220)`

graph{5y=11x^2-15x-9 [-4.667, 5.333, -4.12, 0.88]}