Graph each function:

Solution (a) The graph of the equation $z=1-x-y$, or $x+y+z=1$, is a plane. The intercepts are the points $(1,0,0),$ $(0,1,0)$, and $(0,0,1)$. See Figure 6.

(b) The graph of the equation $z=x^{2}+4y^{2}$ is an elliptic paraboloid whose vertex is at the origin. See Figure 7.

(c) The equation $z=f(x,y) =\sqrt{x^{2}+y^{2}}$ is equivalent to $z^{2}=x^{2}+y^{2}$, where $z\geq 0$. The graph of the equation is part of a circular cone whose vertex is at the origin. Since $z\geq 0,$ the graph of $f$ is the upper nappe of the cone. See Figure 8.

Figure 6 $z=f (x,y) =1-x-y$

Figure 7 $z=f(x, y)=x^{2} + 4y^{2}$

Figure 8 $z=f (x,y) = \sqrt{x^2 +y^2}$