Question

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

Answer 1

The vertex is at:`(4, -11)`

Explanation:

`y=(x−3)^2−2x−4` => expand to simplify:
`y=x^2-6x+9-2x-4` => simplify add/subtract like terms:
`y=x^2-8x+5` => quadratic function in standard/general form of :
`f(x)=y=ax^2+bx+c`=>where the x and y coordinates of the vertex are:
`(x, y)=[-b/(2a) , f(-b/(2a))]`
so in this case:
`f(x)=y=x^2-8x+5`=> where:`a=1, b=-8, c=5`, then:
`x=-(-8/(2))=4`, and:
`f(4)=4^2-8*4+5=-11`
hence the vertex is at:
`(4, -11)`