Answer
Show $n^2 + 2^n$ is $\Theta(2^n)$.
For all $n, 2^n \leq n^2 + 2^n$.
By property 11.4.10 ($x^r \leq b^x$ for b>1, r>0, and all sufficiently large real numbers x), $n^2 \leq 2^n$ for all $n>k$ where $k=4$.
$n^2 +2^n\leq 2^n +2^n$ (add $2^n$ to both sides)
$n^2 +2^n\leq 2(2^n)$
$|2^n|\leq |n^2 +2^n| \leq 2|2^n|$ (because all terms are positive)
Thus for A=1, B=2, k=4, $A|2^n|\leq |n^2 +2^n| \leq B|2^n|$. By the definition of $\Theta$-notation, $n^2 + 2^n$ is $\Theta(2^n)$.
Work Step by Step
Recall the definition of $\Theta$-notation: $f(x)$ is $\Theta(g(x))$ iff there exist positive real numbers A, B, k, such that $A|g(x)| \leq |f(x)| \leq B|g(x)|$ for all $x>k$.
