What is the vertex of # y= -8x^2+8x-(x+9)^2#?

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Answer 1 · 2017-05-19T11:31:56

A sort of cheat method (not really)

#color(blue)("Vertex"->(x,y)=(-5/9,-704/9)#

Expanding the brackets we get:

#y=-8x^2+8x" "-x^2-18x-81#

#y=-9x^2-10x-81" ".......................Equation(1)#

As the coefficient of #x^2# is negative the graph is of form #nn#
Thus the vertex is a maximum.

Consider the standardised form of #y=ax^2+bx+c#

Part of the process of completing the square is such that:

#x_("vertex")=(-1/2)xxb/a" "=>" "(-1/2)xx((-10)/(-9)) = -5/9#

Substitute for #x# in #Equation(1)# giving:

#y_("vertex")=-9(-5/9)^2-10(-5/9)-81#

#y_("vertex")=-78 2/9->-704/9#

#color(blue)("Vertex"->(x,y)=(-5/9,-704/9)#

Note that #-5/9~~0.55555... -> -0.56# to 2 decimal places
Tony B