Question

What is the the vertex of `y = (x-1)^2+2x-12`?

Answer 1

`"vertex "=(0,-11)`

Explanation:

`"expand and rearrange into standard form"`

`•color(white)(x)y=ax^2+bx+c color(white)(x);a!=0`

`y=x^2-2x+1+2x-12`

`y=x^2-11`

`"A quadratic in the form "y=ax^2+c`

`"has it's vertex at "(0,c)`

`"this has it's vertex at "(0,-11)`
graph{x^2-11 [-40, 40, -20, 20]}

Answer 2

`y=(x-1)^2+2x-12`

Expand the brackets

`y=x^2-2x+1+2x-12`

`y=x^2-11`

The parabola `y=x^2` is a `uu` curve with the vertex (a minimum) at the origin (0,0)
`y=x^2-11` is this same curve but translated 11 units down the y axis so the vertex (again a minimum) is at (0,-11)

Another method:
To find the x coordinate of the vertex use `(-b)/(2a)` when the equation is in the form `y=ax^2+bx+c`

From `y=x^2-11 a=1 and b=0`

`-0/1=0` put `x=0` into the equation, `y=-11`
(0,-11) is your vertex