# How to determine a in r - in a function of relations

I'm pretty stuck on the following question

$f$ on $\mathbb{R}$ given by $xfy\Leftrightarrow (y(2x-3)-3x=y(x^2-2x)-5x^3)$ is a function.

Let $g$ be the restriction of $f$ to $\mathbb{Z}^+$, implying $g(n) = f(n),\,n \in\mathbb{Z}^+$

Determine $a\in R\,$, so $g\in \Theta(n^a)$

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Have you expressed $f(x)$ (by expressing $y$ in terms of $x$'s)? – Berci Oct 28 '12 at 22:12
In your last statement: "Determine $a \in R$...do you mean $a \in \mathbb{R}$? – amWhy Oct 28 '12 at 22:22
@Berci: Could u eloborate it? amWhy Yes :) – Alek Oliver Oct 28 '12 at 22:30
y=(-5x^3+5x)/(-x^2+4x-3) is that what u mean :)? – Alek Oliver Oct 28 '12 at 22:37
If the question is solved, you indicate that by accepting an answer, not by changing the title to include the word solved. – Graphth Oct 30 '12 at 15:26

Let's express $y=f(x)$: $$(x^2-2x-2x+3)y=5x^3-3x \implies y =\frac{5x^3-3x}{x^2-4x+3}$$ So, $a=3-2=1$ by the leading exponents.
Update: Formally, you have to prove that there are constants $A,B>0$ such that $An\le f(n)\le Bn\$ (for $n\ge n_0$).