Neal D. answered • 08/02/16

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y = x

^{2 }-6x + 14 this is standard form for a quadratic equation its graph will be a parabola

Change this to Graphing form: y = a ( x - h )

^{2}+ k(h, k) will be your vertex point, 'a' has same value in either form

y = x

^{2}-6x + 14 complete the square using only the x termsy = ( x -3 )

^{2 }this would make the constant term 9 and not 14 so add 5

y = ( x - 3 )

^{2}+ 5 is the equation of a parabola with vertex (3,5)Since a = 1, this parabola opens the same rate as y = x

^{2}**To find where the curve crosses the y-axis:**use either form and

letting x=0, solve for y

Neal D.

08/03/16