#### Answer

Yes, one-to-one.

#### Work Step by Step

We are given:
$r(x)=2+\sqrt{x+3}$
This function is one-to-one because it passes the horizontal line test (it is a shifted square root function). We can show this algebraically.
We start with the assumption:
$x_1 \neq x_2$
$x_1+3\neq x_2+3$
$\sqrt{x_1+3}\neq \sqrt{x_2+3}$
$2+\sqrt{x_1+3}\neq 2+\sqrt{x_2+3}$
Since square rooting a unique value produces a unique value. Thus the function will not have the same $y$ value for two different $x$ values. It is therefore one-to-one.