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

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

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

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