Graph y> = square root of 3x+1+2

$y \geq \sqrt{3 x + 1} + 2$

The equation is not linear, so a constant slope does not exist.

Not Linear

Graph a solid line, then shade the area above the boundary line since $y$ is greater than $\sqrt{3 x + 1} + 2$ .

$y \geq \sqrt{3 x + 1} + 2$