# asymptotic tight bounds for quadratic functions

In Introduction to Algorithms by CLRS, it's said

For any quadratic function $$f(n)=an^2+bn+c$$, where $$a$$, $$b$$ and $$c$$ are constants and $$a>0$$, $$f(n)=\Theta (n^2).$$ Formally, to show the same thing, we take constants $$c_1=a/4, c_2=7a/4$$ and $$n_0 = 2 \cdot max(|b|/a, \sqrt{|c|/a}).$$ You may verify that $$0\leq c_1n^2\leq an^2+bn+c \leq c_2n^2$$ for all $$n\geq n_0.$$

They didn't specify how values of these constants came? I tried to prove it but couldn't.

I think I can suggest more easy way. Having $$a>0$$ in $$f(n)=an^2+bn+c$$ we can take any $$a_1 and then show that $$a_1n^2 < an^2+bn+c$$ for some $$N_1,n>N_1$$ , because it is same as $$a_1 < a + \frac{bn+c}{n^2}$$.
Same we can make for any $$a for some $$N_2,n>N_2$$ because it is same as $$a + \frac{bn+c}{n^2}< a_2$$.
At last we will have $$a_1n^2 .