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How do you find the local max and min for #f(x)= x^2 + 8x -12#?

1 Answer

Jun 22, 2016

#y_{min} = -28# at #x = -4#

Explanation:

calculus is OTT for this

So

#y = x^2 + 8x - 12 = (x+4)^2 - 16 - 12#
#= (x+4)^2 - 28#

#(x+4)^2 \ge 0# so #y_{min} = -28# at #x = -4#

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