Final answer:
The given function s(x) = (x+2)^2 + 5 is an even function.
Step-by-step explanation:
A function is considered even if it satisfies the condition f(x) = f(-x) for all x in the domain. On the other hand, a function is considered odd if it satisfies the condition f(x) = -f(-x) for all x in the domain.
In the given function s(x) = (x+2)^2 + 5, let's check if it is even or odd:
Checking for evenness:
s(x) = s(-x)
(x+2)^2 + 5 = (-x+2)^2 + 5
x^2 + 4x + 4 + 5 = x^2 - 4x + 4 + 5
8x = -8x
16x = 0
x = 0
The function is even because it satisfies the condition for evenness, namely s(x) = s(-x) for all x in the domain. Therefore, s(x) = (x+2)^2 + 5 is an even function.