The problem asks to sketch the graph of the following functions: 7. $f(x, y) = 6$ 8. $f(x, y) = 6 - x$ 9. $f(x, y) = 6 - x - 2y$ 10. $f(x, y) = 6 - x^2$ 11. $f(x, y) = \sqrt{16 - x^2 - y^2}$

1. Problem Description

The problem asks to sketch the graph of the following functions:

7. $f(x, y) = 6$

8. $f(x, y) = 6 - x$

9. $f(x, y) = 6 - x - 2y$

1

0. $f(x, y) = 6 - x^2$

1

1. $f(x, y) = \sqrt{16 - x^2 - y^2}$

2. Solution Steps

7. $f(x, y) = 6$

Let $z = f(x, y)$. Then $z = 6$. This is a plane parallel to the $xy$-plane, passing through the point $(0, 0, 6)$.

8. $f(x, y) = 6 - x$

Let $z = f(x, y)$. Then $z = 6 - x$. This is a plane. We can rewrite it as $x + z = 6$. The intersection with the $xz$-plane ($y = 0$) is the line $x + z = 6$. The intersection with the $xy$-plane ($z = 0$) is the line $x = 6$. The intersection with the $yz$-plane ($x = 0$) is the line $z = 6$.

9. $f(x, y) = 6 - x - 2y$

Let $z = f(x, y)$. Then $z = 6 - x - 2y$. We can rewrite it as $x + 2y + z = 6$. This is a plane. The intersection with the $xy$-plane ($z = 0$) is the line $x + 2y = 6$. The intersection with the $xz$-plane ($y = 0$) is the line $x + z = 6$. The intersection with the $yz$-plane ($x = 0$) is the line $2y + z = 6$.

1

0. $f(x, y) = 6 - x^2$

Let $z = f(x, y)$. Then $z = 6 - x^2$. This is a parabolic cylinder. The intersection with the $xz$-plane ($y = 0$) is the parabola $z = 6 - x^2$. The graph is the same parabola extended along the $y$-axis.

1

1. $f(x, y) = \sqrt{16 - x^2 - y^2}$

Let $z = f(x, y)$. Then $z = \sqrt{16 - x^2 - y^2}$. Since $z$ is a square root, $z \ge 0$. Squaring both sides, we get $z^2 = 16 - x^2 - y^2$, or $x^2 + y^2 + z^2 = 16$. Since $z \ge 0$, this is the upper hemisphere of a sphere with radius $\sqrt{16} = 4$ centered at the origin.

3. Final Answer

7. A plane $z = 6$ parallel to the $xy$-plane.

8. A plane $x + z = 6$.

9. A plane $x + 2y + z = 6$.

1

0. A parabolic cylinder $z = 6 - x^2$.

1

1. The upper hemisphere $x^2 + y^2 + z^2 = 16, z \ge 0$.