Question

What is the vertex of `y= (2x-3)^2-x^2-2x+4`?

Answer 1

`(7/3, -10/3)`

Explanation:

First expand and simplify to get get one term for each power of x.
`y = 4x^2 -12x + 9 - x^2 - 2x + 4`
`y = 3x^2 -14x + 13`
`y = 3(x^2 -(14x)/3 +13/3)`
Use completing the square to put the expression into vertex form
`y = 3(x - 7/3)^2 -49/9 + 13/3) = 3((x-7/3)^2 -10/9)`
`y = 3(x-7/3)^2 -10/3`
Then the vertex occurs where the bracketed term is zero.
Vertex is `(7/3, -10/3)`