Question

How do you find the vertex of a parabola `g(x) = x^2 - 4x + 2`?

Answer 1

I found (coordinates of the vertex):
`x_v=2`
`y_v=-2`

Explanation:

You have two main ways to find the coordinates of the vertex:
1] your parabola is in the form `ax^2+bx+c`
Where:
`a=1`
`b=-4`
`c=2`
The coordinates of the vertex are then:
`color(red)(x_v=-b/(2a))=-(-4)/(2*1)=2`
`color(red)(y_v=-(Delta)/(4a))=-(b^2-4ac)/(4a)=-(16-8)/4=-2`

2] Use the derivative.
At the vertex the derivative of your function must be ZERO;
So:
derivative `g'(x)=2x-4` set it equal to zero and solve for `x`:
`2x-4=0`
`x=4/2=2=x_v`
use this value into your original function to find `y_v`:
`g(2)=4-8+2=-2=y_v`