Question

What is the vertex of `y=1/3(7x-2)^2-7`? Thank you so much, in advance.?

Answer 1

Compare with the vertex form and get the answer.

Explanation:

`y=1/3(7x-2)^2 - 7`

The vertex form would be `y = a(x-h)^2 + k` where (h,k) is the vertex.

We can write the given equation in the vertex form and get the vertex.

`y=1/3 (7(x-2/7))^2 - 7`
`y=1/3(7^2)(x-2/7)^2 - 7`
`y=49/3(x-2/7)^2 - 7`
Now we have got it to a form which we can recognize.
Comparing with `a(x-h)^2 + k` we can see `h=2/7 and k=-7`

The vertex is `(2/7, -7)`

Alternate Method.
The alternate method is when you put `7x-2 =0` and solve for x to find `x= 2/7` and get x-coordinate of the vertex. When you substitute `x=2/7` in the given equation you would get `y=-7` which would be the y-coordinate of the vertex and still you would get the vertex `(2/7,-7)`