What is the vertex of # y= (x-3)^2-x-2#?

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Answer 1 · 2016-06-23T15:36:33

Vertex#->(x,y)=(7/2, -45/2)#

Multiply out the bracket so that you combine terms as appropriate.

#y=x^2-6x+3" "-x-2#

#y=x^2-7x+1#

As the coefficient of #x^2# is 1 we can apply directly

#x_("vertex")=(-1/2)xx(-7)# where the -7 is from #-7x#

#x_("vertex")=+7/2#

Substitute in equation giving

#y_("vertex")=(7/2)^2-7(7/2)+1#

#y_("vertex")=-11 1/4 ->-45/4#

Tony B