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##

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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