How do you find the max and min of #y=-x^2-8x+2# by completing the square?

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Answer 1 · 2015-05-31T11:28:19

#-x^2-8x+2 = -(x^2+8x-2)#

#=-((x+4)^2-16-2)#

#=-((x+4)^2-18)#

#=18-(x+4)^2#

Being the square of a real number #(x+4)^2 >= 0#

So the maximum value #18# occurs when #(x+4)^2 = 0# (that is when #x=-4#)

The value is unbounded in the other direction.

#y = -x^2-8x+2 -> -oo# as #x ->oo# or #x->-oo#