What is the vertex of #y=-(x + 2)^2 - 3x+9#?

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Answer 1 · 2018-06-19T18:32:57

get the equation into the standard form of a quadratic

#y=ax^2+bx+c#

Expand the brackets

#y=-(x^2+4x+4)-3x+9#

Remove the brackets

#y=-x^2-4x-4-3x+9#

Collect like terms

#y=-x^2-7x+5#

Now use #(-b)/(2a)# to find the x coordinate of the vertex.

#(- -7)/(2xx -1)=7/(-2)#

Put this into the equation

#y=-(7/(-2))^2-7xx7/(-2)+5#

#y=-49/4+49/2+5#

#y=69/4#

The maximum is #(-7/2,69/4)#