How do you find the maximum and minimum of $y=(x+2)^2-3$ ?

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Answer 1 · 2016-12-21T11:41:45

Maximum: $oo$ , Minimum: $-3$

$y=(x+2)^2-3 = x^2+4x+1$ Comparing with quadratic equation $y= ax^2+bx+c$
Here $a=1 , b= 4 ,c =1$ .Discriminant $D= b^2-4ac=4^2-4*1*1 =12$

If $a>0$ ; maximum is $oo$ and minimum is $y=-D/(4a)=-12/4=-3$ graph{x^2+4x+1 [-10, 10, -5, 5]}[Ans]

How do you find the maximum and minimum of y=(x+2)^2-3y=(x+2)^2-3?