h(x)=(x+1)^2-3 What is the minimum value of h(x)?

h(x) = (x+1)^2-3. To find the
minimum.

Since (x+1)^2 > = 0, for all x, as (x+1)^2
is a square.

(x+1)^2  - 3 > = 0
-3.

Theefore h(x) > = -3 for all
x.

Therefore minimum of h(x) = -3, when (x+1) = 0, or when
x= -1.

or h(-1) = -3 is the minimum of h(x)
.