Question

What is the vertex of `y= 2|x|^2 – 4x+1`?

Answer 1

`y_"vertex" = (1,-1)`

Explanation:

`y= 2abs(x)^2-4x+1`

First note that `absx^2 =x^2`

Hence, `y= 2x^2-4x+1`

`y` is a parabolic function of the form `y=ax^2+bx+c` which has a vertex at `x=-b/(2a)`

`x= - (-4)/(2*2) =1`

`y(1) = 2-4+1 =-1`

Hence, `y_"vertex" = (1,-1)`

We can see this result from the graph of `y` below:

graph{2abs(x)^2-4x+1 [-5.55, 6.936, -2.45, 3.796]}