Question

What is the vertex of `y=(x-4) (x-2)+x`?

Answer 1

`(5/2,7/4)`

Explanation:

First expand the equation to get it into standard form, then convert into vertex form by completing the square.

`y = (x^2 - 4x - 2x +8) +x`
`y = x^2-5x +8`

`y = (x-5/2)^2 -25/4 +8`

`y = (x-5/2)^2 +7/4`

The vertex is `(5/2,7/4)` which is the point where the bracketed term is zero and therefore the expression is at its minimum.

Answer 2

A related but very slightly different approach

`color(green)("Vertex " ->" "(x,y)" "->" "(5/2,7/4)`

Explanation:

An alternative approach. It does in fact incorporate part of the process of constructing the vertex equation.

Multiply out the brackets

`y=x^2-6x+8+x`

`y=x^2-5x+8`

Consider the `-5` from `-5x`

Apply`(-1/2)xx(-5) = +5/2`

`color(blue)(x_"vertex"=5/2)`

By substitution

`color(blue)(y_("vertex")=(5/2)^2-5(5/2)+8= +7/4)`

`color(green)("Vertex " ->" "(x,y)" "->" "(5/2,+7/4)`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(red)("A word of caution")`

given that the standard form is`y=ax^2+bx+c`

When applying this approach you must have
`" "y=a(x^2+b/ax)+c`

So in fact`" "y_("vertex") = (-1/2)xx(b/a)`

In your question `a=1` so for that question

`" "color(brown)(y_("vertex") = (-1/2)xx(b/a))color(green)( -> (-1/2)xx(-5/1))`