Question

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

Answer 1

Maximum: `oo` , Minimum: `-3`

Explanation:

`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]