<s>[INST] Show that $\frac{1}{2}n^2-3n=\Theta{(n^2)}$ Show that $$\frac{1}{2}n^2-3n=\Theta{(n^2)}$$
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$\displaystyle{\frac{1}{2}n^2-3n=\Theta{(n^2)}: \\ \exists c_1, c_2 >0 , \ \ \exists n_0 \geq 1 \text{ such that } \forall n \geq n_0 \\ 0<c_1 n^2 \leq \frac{1}{2}n^2-3n \leq c_2 n^2}$
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That means that I have to find the values of $n_0, c_1, c_2$ such that 
$$0<c_1 n^2 \leq \frac{1}{2}n^2-3n \leq c_2 n^2$$
$$\Rightarrow 0<c_1 \leq \frac{1}{2}-\frac{3}{n} \leq c_2$$


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*$\displaystyle{0<c_1 \leq \frac{1}{2}-\frac{3}{n}} :$


It must be $\displaystyle{\frac{1}{2}-\frac{3}{n}>0 \Rightarrow \frac{3}{n} < \frac{1}{2} \Rightarrow n>6 \Rightarrow n \geq 7}$
So, $\displaystyle{n_1=7}$.
$\displaystyle{n \geq 7 \Rightarrow \frac{1}{7} \geq \frac{1}{n} \Rightarrow \frac{3}{7} \geq \frac{3}{n} \Rightarrow -\frac{3}{n} \geq -\frac{3}{7} \Rightarrow \frac{1}{2}-\frac{3}{n} \geq \frac{1}{2}-\frac{3}{7} \Rightarrow \frac{1}{2}-\frac{3}{n} \geq \frac{1}{14}}$
So $\displaystyle{c_1=\frac{1}{14}}$.


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*$\displaystyle{\frac{1}{2}-\frac{3}{n} \leq c_2} :$


How can I find a $\displaystyle{n_2}$ such that $\displaystyle{\forall n \geq n_2:\frac{1}{2}-\frac{3}{n} \leq c_2}$ ??
And how can I find this $\displaystyle{c_2}$ ??
 [/INST] $$n\geqslant12\implies\frac14n^2\leqslant\frac12n^2-3n\leqslant\frac12n^2$$
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