We are asked to describe the graph of the given functions. Specifically, we will focus on problems 7, 9, 11, 12, and 13. Problem 7: $f(x, y) = 6$ Problem 9: $f(x, y) = 6 - x - 2y$ Problem 11: $f(x, y) = \sqrt{16 - x^2 - y^2}$ Problem 12: $f(x, y) = \sqrt{16 - 4x^2 - y^2}$ Problem 13: $f(x, y) = 3 - x^2 - y^2$

1. Problem Description

We are asked to describe the graph of the given functions. Specifically, we will focus on problems 7, 9, 11, 12, and

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Problem 7: $f(x, y) = 6$

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

Problem 11: $f(x, y) = \sqrt{16 - x^2 - y^2}$

Problem 12: $f(x, y) = \sqrt{16 - 4x^2 - y^2}$

Problem 13: $f(x, y) = 3 - x^2 - y^2$

2. Solution Steps

Problem 7: $f(x, y) = 6$. This is equivalent to $z = 6$.

This represents a horizontal plane at $z = 6$.

Problem 9: $f(x, y) = 6 - x - 2y$. This is equivalent to $z = 6 - x - 2y$, or $x + 2y + z = 6$.

This is a plane. We can find the intercepts to help visualize it.

When $y = z = 0$, $x = 6$.

When $x = z = 0$, $2y = 6$, so $y = 3$.

When $x = y = 0$, $z = 6$.

Problem 11: $f(x, y) = \sqrt{16 - x^2 - y^2}$. This is equivalent to $z = \sqrt{16 - x^2 - y^2}$, where $z \ge 0$.

Squaring both sides, we get $z^2 = 16 - x^2 - y^2$, so $x^2 + y^2 + z^2 = 16$.

Since $z \ge 0$, this is the upper half of a sphere with radius $\sqrt{16} = 4$ centered at the origin.

Problem 12: $f(x, y) = \sqrt{16 - 4x^2 - y^2}$. This is equivalent to $z = \sqrt{16 - 4x^2 - y^2}$, where $z \ge 0$.

Squaring both sides, we get $z^2 = 16 - 4x^2 - y^2$, so $4x^2 + y^2 + z^2 = 16$.

Dividing by 16, we get $\frac{x^2}{4} + \frac{y^2}{16} + \frac{z^2}{16} = 1$.

Since $z \ge 0$, this is the upper half of an ellipsoid.

Problem 13: $f(x, y) = 3 - x^2 - y^2$. This is equivalent to $z = 3 - x^2 - y^2$, so $z = 3 - (x^2 + y^2)$.

This is a paraboloid opening downwards with its vertex at $(0, 0, 3)$.

3. Final Answer

Problem 7: The graph is a horizontal plane at $z = 6$.

Problem 9: The graph is a plane defined by $x + 2y + z = 6$.

Problem 11: The graph is the upper half of a sphere with radius 4 centered at the origin.

Problem 12: The graph is the upper half of an ellipsoid defined by $\frac{x^2}{4} + \frac{y^2}{16} + \frac{z^2}{16} = 1$.

Problem 13: The graph is a paraboloid opening downwards with its vertex at $(0, 0, 3)$.