Find if exists Supremum and Infimum for the set A: Hi im having difficulties with this problem

Let$\{({a_n})_{n \in N}\}$ be  a   sequence  defined  by 
  $${a_n}: = 4 - \frac{9}{n}$$
  Find  if  exists  infimum, supremum  for  the  set
  $$A: = \{ {a_n}^2 - 3{a_n} - 10:n \in N\} $$

I dont know how to start, i tried to replace the sequence $a_n$ on the set, and analize the set as a function. To find max, min, so i can detect critical points.  But i dont think is the way to do solve it.
 A: OUTLINE:
Substitute the expression $4-\frac 9n$ for $a_n$ into the expression $a_n^2-3a_n-10$. This will give you a function
$$f(n)=\left(4-\frac 9n\right)^2-3\left(4-\frac 9n\right)-10$$
Simplify that function expression.
Treat it as a function of real numbers $f(x)$ and take its derivative. Use that derivative to analyze the function $f$. You will find that it has a minimum at a particular positive non-integral number (let's call it $b$), is decreasing for $0<x<b$, and is increasing for $b<x$.
Therefore the minimum of that function over the positive integers is one of the two integers closest to $b$. Just calculate the function at those two numbers $\lfloor b\rfloor$ and $\lfloor b\rfloor+1$ to find the minimum value and its location. That is the desired infimum.
The function $f(x)$ is decreasing for $0<x<b$, so the maximum of $f$ for all integers in that interval is $f(1)$.
The function $f(x)$ is increasing for $b<x$, so the supremum of $f$ for all integers in that interval is the limit of $f(x)$ as $x$ approaches $+\infty$. That limit is easy to find.
The supremum of $f$ over all positive integers will be the larger of those last two numbers.
A: I have benefited from reading @RoryDaulton answer, but I prefer to do the same thing over without using derivatives (although the OP seems to be fine with derivatives, just not sure if this is the way to solve it ... why not). 
So $A: = \{ b_n:n \in N\}$ where $b_n={a_n}^2 - 3{a_n} - 10$ and $a_n=4-\dfrac9n$.
Note $a_{n+1}-a_n=\dfrac9{n(n+1)}$
Let $d_n=b_{n+1}-b_n=a_{n+1}^2-a_n^2-3(a_{n+1}-a_n)=$
$=(a_{n+1}+a_n)(a_{n+1}-a_n)-3(a_{n+1}-a_n)=$
$=(a_{n+1}+a_n-3)(a_{n+1}-a_n) = $
$=\bigl(4-\dfrac9{n+1}+4-\dfrac9n-3\bigr)\bigl(4-\dfrac9{n+1}-(4-\dfrac9n)\bigr)$.
After further simplification we get $d_n=\dfrac{5n^2-13n-9}{n^2(n+1)^2}$. 
We want to know when $d_n>0$ and when $d_n<0$. It is enough to look as the sign of the numerator $5n^2-13n-9$ (which as a function of $n$ is a parabola opening up, could not resist the word function after all, and involve calculus:) 
Solving $5n^2-13n-9=0$ we get $n_{1,2}=\dfrac{13\pm\sqrt{349}}{10}$. Ignore the negative root. The positive root, say $n_1$, is between $\dfrac{13+18}{10}$ and $\dfrac{13+19}{10}$ (since $18^2=324<349<361=19^2$), hence $\dfrac{31}{10}<n_1<\dfrac{32}{10}$, and in integers 
$3<n_1<4$. (Hope this agrees with the results you get when you compute the derivative as a function of a real argument, as suggested in the first posted answer.) 
We get that $d_n<0$ when $1\le n\le 3$ and $d_n>0$ for $n\ge4$ (ignoring what happens when $n\le0$).
Since $d_n=b_{n+1}-b_n$ it follows that $b_1>b_2>b_3>b_4$ but $b_4<b_5<b_6<b_7<\cdots$.  
It follows that $\inf_n b_n=\min_n b_n=b_4=a_4^2 - 3a_4 - 10=\dfrac{-195}{16}$. 
$\sup_n b_n$ would be the larger of $b_1=\bigl(4-\dfrac91\bigr)^2-3\bigl(4-\dfrac91\bigr)-10=25+15-10=30$ and $\lim_{n\to\infty}b_n=\lim_{n\to\infty}\Bigl(\bigl(4-\dfrac9n\bigr)^2-3\bigl(4-\dfrac9n\bigr)-10\Bigr)=16-12-10=-6$.
That is, $\sup_n b_n=\max\{30,-6\}=30$. 
A: excuse my delay. I tried to solve it with the tools i know.
\begin{array}{l}
Let{({a_n})_{n \in N}}\,\,{a_n}: = 4 - \frac{9}{n}\,\,A: = \{ {a_n}^2 - 3{a_n} - 10:n \in N\} \\
\,4 - \frac{9}{n} > 0\,\,\,\,\,\,\,\,4 > \frac{9}{n}\,\,\,\,\,\,\,\,\,4n > 9\,\,\,n > \frac{9}{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,n\, > 0\,\,\forall \,n > 2\\
4 - \frac{9}{{n + 1}} > \,4 - \frac{9}{n}\,\,\,\,\,\,\,\,\, - \frac{9}{{n + 1}} + \frac{9}{n} > 0\,\,\,\,\frac{9}{{(n + 1)n}} > \,0\,\, \to \,\,{a_n} \nearrow \\
Supr({a_n}),\,Inf({a_n})\\
\mathop {\lim }\limits_{n \to \infty } \,\,\,4 - \frac{9}{n}\, = 4\,\,\, \to \,\,{a_n} < 4\,\,\,\,\,\,\,\,\,{a_1} = 4 - \frac{9}{1} =  - 5\,\\
Supr({a_n}) = 4\,\,\,\,\,\,\,Inf({a_n}) =  - 5\\
{a_n} \in [ - 5,4)\\A: = \{ {a_n}^2 - 3{a_n} - 10:n \in N\} \\
f(x) = {x^2} - 3x - 10 = (x - 5)(x + 2)\\
f'(x) = 2x - 3 = (x - 3/2)\\
Min = 3/2\,\,\,\,1 \le 3/2 \le 2\, \to \,1 \le {a_{n\,}} \le 2\\
\,1 \le 4 - \frac{9}{n} \le 2\,\,\,\,\, \to \,\,\,\,\,3 \le n \le 9/2\,\,\,\,\,n \in N\, \to n = \left\{ {3,4} \right\}\,\,\\
 - 5 < {a_n} < 4\,\,\,\,\,\,f( - 5) = 30\,\,\,\,\,\,\,f(4) =  - 6\,\,\,{a_3} = 1\,\,{a_4} = \frac{7}{4}\\
 - 6 < f({a_n}) < 30\\
f({a_3}) = \,f(1) =  - 12\,\,\,f({a_4}) = \,f(\frac{7}{4}) =  - 12,81\,\,\,f(1)\, \in \,N\\
Supr(A) = 30\,\,Inf(A) =  - 12\,\,
\end{array}
Using that the sequence is increasing and bounded between [-5,4).
I just had to analize critical points in the set between f(-5),f(4).
I tried to solve for, \begin{array}{2}f(x) = {(4 - \frac{9}{n})^2} - 3(4 - \frac{9}{n}) - 10\end{array}
I tried this in the exam, but i made a mistake, somewhere.
Thanks, for all the answers. 
Mirko approach is new to me, and much more simpler.
Update 12/03
I asked in class for some justification questions. My analysis is valid only for the increasing  of the set $$A: = {\rm{\{ }}{{\rm{a}}_n}^2{\rm{ - 3}}{{\rm{a}}_n}{\rm{ - 10:n}} \in {\rm{N\} }}$$ So i have to find n in ${{\rm{a}}_n}$ using the critical point value, in that case both ${{\rm{a}}_n}$ and $A$ are increasing, and can be its behaviour analyzed. Before the critical point i have to get values by hand for $$A({{\rm{a}}_1}),A({{\rm{a}}_2}),A({{\rm{a}}_3}),A({{\rm{a}}_4})$$Since one function is decreasing, and the other increasing. I also have to do lim->inf  A(an),to get the extreme of the composition.
