How to find the turning points from these quadratic equations?

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1 Answer
Feb 12, 2018

d) #-2(x-2)^2 + 2#
#(2,2)#
#{y: y<=2}#
f) #6(x+0.5)^2 - 18.375#
#(-0.5, -18.375)#
#{y:y>=-18.375}#

Explanation:

#-2x^2+8x -6:#

factorise:

#-2(x^2-4x+3)#

#(-4)/2 = -2#

#x^2-4x + 4 = (x-2)^2#

#x^2 - 4x + 3 = (x-2)^2 - 1#

#-2(x^2-4x+3) = -2((x-2)^2-1)#

#=-2(x-2)^2 + 2#

#a(x-h)^2 + k = -2(x-2)^2 + 2#

turning point: #(-h,k)#, where #x=h# is the axis of symmetry.

#(-h, k) = (2,2)#

#x= 2# is the axis of symmetry.

since the coefficient of #x^2# is negative #(-2)#, the graph opens to the bottom.

the point #(-h, k)# is therefore a maximum point.

since the maximum point is the highest possible, the range is equal to or below #2#.

#{y: y<=2}#

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#6x^2+3x-18:#

factorised:

#6(x^2+0.5x-3)#

#x^2+0.5x+0.0625 = (x+0.25)^2#

#x^2+0.5x-3 = (x+0.25)^2-3.0625#

#6((x+0.25^2)-3.0625) = 6(x+0.25^2) - 18.375#

#a(x-h)^2 + k = 6(x+0.25)^2 - 18.375#

turning point: #(-h,k)#, where #x=-h# is the axis of symmetry.

#(-h, k) = (-0.25,-18.375)#

#x= -0.25# is the axis of symmetry.

since the coefficient of #x^2# is positive #(6)#, the graph opens to the top.

#(-h, k)#, therefore, is a minimum point.

#(-h, k) = (-0.25, -18.375)#

since the minimum point is the lowest possible, the range is equal to or above #-18.375#.

#{y:y>=-18.375}#

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