What is the vertex of # y= (x-3)^2-5x^2-x-1#?

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Answer 1 · 2017-11-10T20:19:29

The vertex is at #(-7/8, 177/16)#

The equation given is a quadratic #y = ax^2 + bx +c#

The vertex is at #(h,k)# where #h = -b/(2a)#

First expand the equation
#y = x^2 - 6x + 9 -5x^2 -x -1#

Simplify
#y = -4x^2 -7x +8#

the #x# value of the vertex is #7/-8# or #-7/8#

plug the value for h back into the equation to get k

#y = -4*-7/8*-7/8 -7*-7/8 +8# = #177/16#

The vertex is at #(-7/8, 177/16)#