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

1 Answer
Grimalkyne
Dec 4, 2017

the vertex is at #(-0.875, 9.0625)#

Explanation:

#y = −3x^2 −x −3 −(x −3)^2#

Simplify the RHS
#y = -3x^2 -x -3 - x^2 - 6x +9#
#y = -4x^2 -7x +6#

The general quadratic form is #y = ax2 + bx + c#
The vertex can be found at #(h,k)#
where #h = -b/2a#

Substitute in what we know
#h = -(-7)/(2*-4) = -7/8 = -0.875#

Substititue the value of #h# for #x# in the original equation
#y = -4(-7/8)^2 -7(-7/8) +6 = 9.0625#

the vertex is at #(-0.875, 9.0625)#

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