Question

Solving fp Algebra 2 I do not understand how to solve this, could someone please explain?

Answer 1

$f(x) = \sqrt{x + 2}$
What's The Fixed Point? Well, Let's Assume X=0, this is our point.

$f(0) = \sqrt{0 + 2} \\ f(0) = \sqrt{2} \\ this \: is \: our \: fixed \: point \: \\ (0, \sqrt{2} )$
OR

$f(p) = \sqrt{p + 2} \\ \\ or \: \sqrt{x + 2} = 0 \\ ( \sqrt{x + 2} ) ^{2} = {0}^{2} \\ x + 2 = 0 \\ x = - 2$
I'm unaware of the nature of the question, so here are some different ways a fixed point is found based on the merits of the question.