Question

How do find the vertex and axis of symmetry, and intercepts for a quadratic equation `f(x)=2x^2-4x+1`?

Answer 1

Rewrite `f(x) = 2x^2 -4x+1` in vertex form to determine vertex is at `(1,-1)` and the axis of symmetry `x=1`.
The `f(x)` intercept is at `f(x) = 1`; the x intercepts are at `x= 2+-sqrt(2)/2`

Explanation:

The vertex form of a quadratic is
`color(white)("XXXX")` `f(x) = m(x-a)^2+b` with a vertex at `(a,b)`

`f(x) = 2x^2-4x+1`
extract `m`
`color(white)("XXXX")` `f(x) = 2(x^2-2x) +1`
complete the square
`color(white)("XXXX")` `f(x) = 2(x^2-2x+1) +1 -2`
`color(white)("XXXX")` `f(x)=2(x-1)^2 +(-1)` with a vertex at `(1,-1)`

The given equation is a parabola in standard orientation, so the axis of symmetry is a vertical line through the vertex; that is `x=1`

The `f(x)` and `x` intercepts are determined by solving the equations for `f(x)` with `x=0` and for `x` with `f(x)=0` respectively. (Use the quadratic formula to determine the `x` solution).