Question 1026208
{{{y = x^2e^x+5}}} ==> y' = {{{(x^2+2x)e^x}}} ==> y" = {{{(x^2 +4x +2)e^x}}} (Verify!!)

y' = 0 ==> x = 0, -2

Use the 2nd derivative test:

At x = 0:  y" = {{{(0^2 +4*0 +2)e^0 = 2 > 0}}} ==> local min at x = 0.

At x = -2:  y" = {{{((-2)^2 +4*(-2) +2)e^(-2) = -2/e^2 < 0 }}}==> local max at x = -2.