Graphing multivariate functions by hand? Impossible? My question here is derivative of my last one but for increased brevity of both questions I've separated them. I'm exploring the following function for learning:$$f(x,\ y) = (x-2)^2 + (y-2)^2$$
And a graphic (courtesy of GeoGebra):

Now, for my question. How would I graph $f(x,\ y) = x^2 + y^2$ by hand? Would I have to assume $x$ and $y$ both as inputs, and then evaluate them for $z$? I surely can't approach it the way I do with a function of $x$ only, where I can evaluate everything at different $x$. 
I'm leaning towards it being extremely hard or near impossible to graph them hence why we use computers, but knowing exactly why is what I'm most keen to understand. Since there are two inputs, any arbitrary pair of alike or unalike inputs of $x$ and $y$ respectively have an output $z$, i.e. there is a $z$ for $x = 2$ with $y = 0.22222$ just as there is a $z$ for $x = 1$ with $y = 1$, and thus too many points to plot. Is this the correct thinking? If not, why, and what is the correct thinking?
 A: There are a few useful tricks when it comes to drawing the graph of a function $f(x,y)$ of two variables by hand: 


*

*Analyze the level sets $f(x,y) = c$ of your function. This is typically a curve or a collection of curves so it is easier to draw. Hopefully, this implicit equation will be familiar to you (or you can try and isolate one of the variables) and since the function is constant on each such level set, you can try and draw a few level sets and then a few values of $f(x,y)$ above each level set.

*Analyze the intersection of the graph $f(x,y)$ with planes that pass through the origin (or other point) and are orthogonal to the $xy$-plane. For example, the intersection of the graph of $f(x,y)$ with the $xz$ plane is given by $f(0,y)$ which is a one-variable function.


To see how those tricks are useful, consider for example the functions $f(x,y) = x^2 + y^2$ and $g(x,y) = \sqrt{x^2 + y^2}$. Both functions are non-negative, zero at the origin and the level sets on which they are constant are circles. The intersection of the graph of $f(x,y)$ with the $xz$ plane (or any other plane, as the function $f$ has rotation symmetry) is given by $f(0,y) = y^2$ so it looks like a parabola. Hence, the graph of $f$ will look like a paraboloid (obtained by rotating a parabola around an axis). The intersection of the graph of $g$ with the $xz$ plane is given by $g(0,y) = \sqrt{y^2} = |y|$ which looks like a two lines intersecting at the origin at 90 degrees. Hence, the graph of $g$ will be a cone (again, obtained by rotating the slice).
To practice, you can try and draw $x^2 - y^2, x^2, x^3 + x^2$ and so on using the tricks above.
A: How do we graph normally? In the beginning we pick an x, get out a y. Then as you get better at graphing certain graphs become very familiar and you don't have to pick a ton of x values. For example $x^2+y^2=1$ you can recognize that graph is a circle with practice and without picking points. For 3D it's a similar process. Start by holding a variable constant. When $x=0$ we have $z=y^2$ which is a parabola. This translates to "along the plane $x=0$ we have a parabola". Repeating this process for both x and y we realize that we have all these parabolas and we can recognize our graph as a paraboloid. 
A: Graphing a 3D graph by hand is not entirely feasible; you can, however, draw what is known as a contour-plot.
I.e for z=(x-2)^2+(y-2)^2, first plot the graph for y=n, where n is any desired number, lets say 2. This would be the parabola (x-2)^2 on the x vs z graph. Then do the same, but for y=1, y=3, etc, on the same axes, depending on how much you want to show.
If my explanation lacks depth, you can find plenty on the subject just by googling "contour plot"
