Clarification regarding Branching Process So the question I am trying to solve is given below:

Find the extinction probability of the branching process generated by $ξ$ ∼ Bin$(2, p)$.

So I saw various approach has been used to solve this type of problem but I have few doubts. How I approached it is this way:
Let the extinction probability obtained by the generating function $g(z)$ is given as $g(z)=\sum^{\infty}_{k=0}z^kP(ξ =k)$ which after substituting the values I get as follows:
$= 2(1-p)^2 + z2p(1-p)+z^2p^2$. After this, we have to equate it to zero to find the solution(That is what I am guessing). But I am not understanding how to proceed and I saw a lot of people are plotting graph which I do not understand properly. Moreover, I think the generating function I got is wrong. I saw the PGF for binomial distribution on the Wikipedia page, but I cannot implement it.
Please any help is appreciated. Thank you in advance.
 A: You are asking multiple questions, please try to avoid this next time.
1.) The generating function of the binomial  distribution ($\xi \sim Binom(n, p)$)
$$g(z) = \sum_{k=0}^{\infty} z^k \mathbb{P}(\xi=k) = \sum_{k=0}^{n} \bigg(\begin{matrix} n \\ k \end{matrix}\bigg) (zp)^k (1-p)^{n-k} = (pz + (1-p))^n$$
2.) The branching process is defined as follows:
$$X_0 = 1$$
$$X_{t+1} = \sum_{i=0}^{X_t} \xi_{i, t} ~\text{ , where }~ \xi_{i, t} \text{ is }\mathbb{N} \text{-valued i.i.d}$$
This means, that the generator function
$$ g_{X_{t+1}}(z) = \sum_{k=0}^{\infty} z^k \mathbb{P}(X_{t+1} = k) = \mathbb{E}(z^{X_{t+1}}) = \mathbb{E}\big(z^{\sum_{i=0}^{X_t} \xi_{i, t}}\big) = \mathbb{E}\big[\mathbb{E}(z^{\xi} | X_t)^{X_t}\big] = g_{X_t}(g_{\xi}(z))$$
So
$$ g_{X_{t}}(z) = (\underbrace{g_{\xi} \circ .. \circ g_{\xi}}_{t\text{-many}}) (z) \text{ , where $\circ$ is the composition operator}$$
What you want to know is the extinction probability, which is $\mathbb{P}(lim_{t \to \infty} X_t = 0) = ?$ 
First note that by the definition of the PGF $g_{X_{t}}(0) = \mathbb{P}(X_{t} = 0)$. Then using this identity
$$\mathbb{P}(lim_{t \to \infty} X_t = 0) = lim_{t \to \infty} \mathbb{P}(X_t = 0) = lim_{t \to \infty} (\underbrace{g_{\xi} \circ .. \circ g_{\xi}}_{t\text{-many}}) (0)$$
So basically you have an infinite composition at the end. The graph which you mentioned is presumably the plot of the PGF function of $\xi$ on the $[0, 1]$ interval. We know that the PGF function is convex, which means the plot will look like this.

Where the orange line is the PGF function, the black line is $y = x$. And $p_0 = \mathbb{P}(\xi = 0)$. The blue line shows iteratively how the infinite composition starting at zero behaves. If we have a fix point below 1 ($g_\xi(z) = z$, s.t. $z<1$), then
$a_n = (\underbrace{g_\xi \circ ...\circ g_\xi}_{n\text{-many}})(0)$ converges to that fix point, which is the probability of extinction. If you don't have such fixed point, than it converges to $1$, so the population dies out in a finite amount of steps with probability 1.
You an apply this to your problem with binomially distributed $\xi$.
$$$$
Edit 1: Adding some additional steps to the generating function part.
Because of the law of total probability and independence of $\xi_{i,t}$, we have:
$$\mathbb{E}\big(z^{\sum_{i=0}^{X_t} \xi_{i, t}}\big) = \sum_{n=0}^{\infty} \mathbb{E}\big(z^{\sum_{i=0}^{X_t} \xi_{i, t}} | X_t=n\big) \mathbb{P}(X_t=n)= \sum_{n=0}^{\infty} \mathbb{E}\big(z^{\sum_{i=0}^{n} \xi_{i, t}}\big) \mathbb{P}(X_t=n)$$
And because of $\xi_{i,t}$ being i.i.d, we have
$$\mathbb{E}\big(z^{\sum_{i=0}^{n} \xi_{i, t}}\big) = \prod_{i=0}^{n}\mathbb{E}\big(z^{\xi_{i, t}}\big) = \mathbb{E}\big(z^\xi\big)^n$$
You put these together, and you get the result:
$$\mathbb{E}\big(z^{X{t+1}}\big) = \sum_{n=0}^{\infty} 
 \mathbb{E}\big(z^\xi\big)^n \mathbb{P}(X_t=n) = g_{X_t}(g_{\xi}(z))$$
