# How to plot the domain of $f(x,y)=\sqrt {x+y}$?

I know how to find the domain of functions with one or more variables, but I have some difficulties understanding how to plot it on a chart.

The domain of $f(x,y)=\sqrt {x+y}$, for instance, is $D(f) = \{(x,y) \in \mathbb{R} | x + y \geq 0 \}$, but how do I plot this? I mean, how do I define the line position on chart and find the areas that are part of the domain?

• Can you plot $x+y=0$? And can you then figure out what part of the plane corresponds to $x+y \ge 0$? Mar 25, 2018 at 1:52
• But how do I choose x and y values? Random numbers that equals zero? Mar 25, 2018 at 1:56

$$x+y \geq 0 \implies y \geq -x.$$ Plot $$y=-x$$ on the plane and pick “the right-hand side” of the plane.

If the Range of the function is real numbers, then $x+y\ge0$. In order to check the position of the region, the region is either left or right to the straight line. consider the 3rd quadrant. Can the sum of two negative number be positive?

Generally I would traces (also used in quartic shapes) They're easy to deduce, you set one variable to 0 and see the shape. In your question:

$$f(x, y) = z =\sqrt{x + y}$$

• XY plane; Set z = 0 \begin{align*}0 &= \sqrt{x + y} \\ &= x + y \\ -x &= y \end{align*}
• XZ plane; set y = 0

$$etc...$$ Construct a vision of whats in the 3D shape. I'm sure you've got it from here