Posted by **Amarjeet** on Thursday, June 21, 2012 at 2:40pm.

use the second derivate test to locate the maxima and minima of y = x^2 + 2x - 3

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**Steve**, Thursday, June 21, 2012 at 3:35pm
since y'' = 2 any max/min will be a min.

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**Amarjeet**, Thursday, June 21, 2012 at 3:42pm
what do you mean by that? :/

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**Steve**, Thursday, June 21, 2012 at 3:50pm
Take a look at the explanation of the 2nd derivative test. If y'' < 0 you have a max, and if y'' > 0 you have a min where y'=0.

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**Amarjeet**, Thursday, June 21, 2012 at 3:54pm
if y' = 0 doesn't that mean there is no max or min? im sorry but i really don't understand this.. please help me

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