Graph y< = square root of 4x+8

$y \leq \sqrt{4 x + 8}$

Simplify $\sqrt{4 x + 8}$ .

Factor $4$ out of $4 x + 8$ .

Factor $4$ out of $4 x$ .

$y \leq \sqrt{4 (x) + 8}$

Factor $4$ out of $8$ .

$y \leq \sqrt{4 x + 4 \cdot 2}$

Factor $4$ out of $4 x + 4 \cdot 2$ .

$y \leq \sqrt{4 (x + 2)}$

Rewrite $4$ as $2^{2}$ .

$y \leq \sqrt{2^{2} (x + 2)}$

Pull terms out from under the radical.

$y \leq 2 \sqrt{x + 2}$

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

Not Linear

Graph a solid line, then shade the area below the boundary line since $y$ is less than $2 \sqrt{x + 2}$ .

$y \leq 2 \sqrt{x + 2}$