Show that for any real constants $$a$$ and $$b$$, where $$b > 0$$,

$(n + a)^b =\Theta(n^b)$

Note that, $$n + a \le 2n$$, when $$\vert a \vert \le n$$

Also note, $$n + a \ge n/2$$, when $$\vert a \vert \le n/2$$

Therefore, when $$n \ge 2 \vert a \vert$$, both of the above inequalities hold true, and we can write,

$0 \le \frac n 2 \le n + a \le 2n$

As $$b > 0$$, we can raise all the terms of the previous inequality to the power of $$b$$ without breaking the inequality:

\begin {aligned} 0 \le (\frac n 2)^b & \le (n + a)^b \le (2n)^b \\ 0 \le \frac 1 {2^b}n^b & \le (n + a)^b \le 2^bn^b \end {aligned}

So, $$(n + a)^b = \Theta(n^b)$$ because there exists $$c_1 = 1/{2^b}$$, $$c_2 = 2^b$$, and $$n_0 = 2 \vert a \vert$$.