What are the local extema of #f(x)=x^2-4x-5#?

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Answer 1 · 2016-04-27T06:11:09

At #(2, -9)# There is a minima.

Given -

#y=x^2-4x-5#
Find the first two derivatives
#dy/dx=2x-4#

Maxima and Minima is to be determined by the second derivative.

#(d^2y)/(dx^2)=2>0#
#dy/dx=0 => 2x-4=0#
#2x=4#
#x=4/2=2#

At #x=2 ; y= 2^2-4(2)-5#

#y=4-8-5#
#y=4-13=-9#

Since the second derivative is greater than one.
At #(2, -9)# There is a minima.