Question

How do find the vertex and axis of symmetry for a quadratic equation `y = x^2 - 4x + 1`?

Answer 1

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

We can convert the given equation: `y = x^2-4x+1` into vertex form:
`color(white)("XXXX")` `y= x^2-4xcolor(red)(+4) +1 color(red)(-4)`

`color(white)("XXXX")` `y=(x-2)^2+(-3)`

So the vertex is at `(2,-3)`

This is the equation of a parabola in normal position (vertical axis of symmetry through the vertex), so
the axis of symmetry is `x=2`

Answer 2

You can use the 'complete the square' method.

You're looking for `(x+a)^2=x^2+2ax+a^2`
Where `2ax=-4x->a=-2and a^2=4`

So we rewrite as:
`y=x^2-4x+4-3=(x-2)^2-3`

Now the axis of symmetry is `x=2`
(remember to reverse the sign with the `x`)
And the apex is `(2,-3)`
graph{x^2-4x+1 [-7.25, 12.75, -4.8, 5.2]}