Answer
$ y $ is continuous on
$$[-\frac{3}{2},\infty)$$
Work Step by Step
Given $$ y=\sqrt{2x+3}$$
Since the square root is always non-negative
$2x+3 \geq 0 \ \ \ \Rightarrow x\geq-\frac{3}{2}$
So, $ y $ is continuous on
$$[-\frac{3}{2},\infty)$$
