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

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

You keep two deep ways to confront the coordinates of the vertex: 1] your parabola is in the make ##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 character must be ZERO; So: derivative ##g'(x)=2x-4## set it similar to naught and work-out for ##x##: ##2x-4=0## ##x=4/2=2=x_v## use this appreciate into your ancient character to confront ##y_v##: ##g(2)=4-8+2=-2=y_v##

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