Question

What is the vertex of `f(x) = x^2-6x+13`?

Answer 1

Vertex`->(x,y)=(3,4)`

Explanation:

`color(blue)("A sort of cheat method")`

Set as `y=x^2-6x+13`

as the coefficient of `x^2` is 1 we have:

`color(blue)(x_("vertex")=(-1/2)xx(-6) = +3`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
By substituting `x=3` we have

`color(blue)(y_("vertex")=(3)^2-6(3)+13 = 4)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The true format is

Given that `y=ax^2+bx+c`

Write as `y=a(x^2+b/a x)+c`

`x_("vertex")=(-1/2)xxb/a`

In your question `a=1`