Injection/Surjection of a quadratic function 
Consider the function $f: \mathbb{N} \to \mathbb{N}$ (where $\mathbb{N}$ is the set of all natural numbers, zero included) defined as follows $$f(x) = (x+3)^{2} - 9.$$ Is the function injective and/or surjective? 

How can I prove it? 
I know that a function is injective if for all $x,y\in\mathbb{N}$ s.t. $f(x)=f(y)$ then $x=y$. So, if I put $(x+3)^2-9=(y+3)^2-9$, how can I obtain $x=y$? Furthermore, how can I find the inverse of $f(x)$?
 A: You can easily verify that it is injective but not surjective.
Indeed 
$f: \mathbb{R^+} \to \mathbb{R^+}$ is injective and strictly increasing
$f(1)=7$ and $f(2)=16$ thus $\nexists x$ such that $f(x)=8$
A: I like using $n,m$ for naturals.  Here $f: \mathbb{N} \to \mathbb{N}$ such that $n \to (n+3)^2-9$


*

*Injective $\implies$ no two naturals have the same image $\implies$ $f(n_1) \neq f(n_2)$ for any DISTINCT $n_1$ and $n_2$, meaning $n_1 \neq n_2$


Can two different inputs produce the same output?  Are there two distinct members of $\mathbb{N}$, $\ $ $n_1$ and $n_2$ $\ $ such that $(n_1+3)^{2} - 9=(n_2+3)^2-9 \ $?  It takes one counter example to show if it's not.


*Surjective $\implies$ every natural is contained in the range of this function$\implies$ $f(n)$ takes on all values of $\mathbb{N}$ 


Does the range of this function contain every natural number with only natural numbers as input?  Is there an $m \in \mathbb{N}$ such that $(m+3)^2-9=2 \ $for instance?  It takes one counter example to show if it's not.
A: $$
f(x) = f(y) \iff \\
(x+3)^2 - 9 = (y+3)^2 -9 \iff \\
(x+3)^2 = (y+3)^2 \iff \\
x+3 = y+3 \quad \vee \quad x+3 = -(y+3)
$$
As $x$ and $y$ are non-negative, what holds for $x+3$ and $y+3$? And what can be inferred?
A: Note that the function $f\colon \mathbb{N} \to \mathbb{N}$ is not surjective. Indeed, there does not exist $x\in\mathbb{N}$ such that 
$$
f(x)= (x+3)^{2} - 9=2.
$$
If there was such an $x$, then $\sqrt{11}$ would be an integer a contradiction. It is injective. Indeed
$$
(x+3)^{2} - 9=(y+3)^{2} - 9\implies |x+3|=|y+3| \implies x=y
$$ 
since $x,y\geq 0$.
A: 1) Injective:
$y = (x+3)^2 -9  = x(x+6)$ , $x \in \mathbb{N}$.
This function is strictly increasing , hence injective.
For $x_1 < x_2$ : $y_1 = x_1(x_1+6) \lt x_2(x_2+6) =y_2.$
2) Not surjective:
$y(0) = 0$;  $y(1) = 7$.
$1,2,3,4,5,6 $ are not image points of f.
