Question

How do you write the Vertex form equation of the parabola `y=x^2-6x+5`?

Answer 1

Please see the explanation.

Explanation:

Given `y = x^2 - 6x + 5" [1]"`

The vertex form of a parabola of this type is:

`y = a(x - h)^2 + k" [2]"`

where `(h,k)` is the vertex and "a" is the coefficient of the `x^2` term in standard form.

The first step is to add 0 to equation [1] in the form `ah^2 - ah^2`. Because `a = 1`, we add `h^2 - h^2` to equation [1]:

`y = x^2 - 6x + h^2 - h^2 + 5" [3]"`

If we expand the square in equation [2], `(x - h)^2 = x^2 - 2hx + h^2`, we can see that `-2hx` must equal `-6x` in equation [3]. Write this equation:

`-2hx = -6x`

Solve for h:

`h = 3`

Substitute `(x - h)^2` for `x^2 - 6x + h^2` in equation [3]:

`y = (x - h)^2 - h^2 + 5" [4]"`

Substitute 3 for every h:

`y = (x - 3)^2 - 3^2 + 5`

Simplify the constant terms:

`y = (x - 3)^2 - 4`

This is the vertex form.