find a formula for the function whose graph is the given curve. the top half of a circle x^2 +(y-3)^2=9

To find the formula for the function whose graph is the top half of a circle given by the equation:

\[
x^2 + (y - 3)^2 = 9
\]

### Step-by-step Solution:

1. **Identify the Circle's Center and Radius:**

   The given equation is in the standard form of a circle:

   \[
   (x - h)^2 + (y - k)^2 = r^2
   \]

   where \((h, k)\) is the center of the circle and \(r\) is the radius. For the given circle:

   - Center \((h, k) = (0, 3)\)
   - Radius \(r = \sqrt{9} = 3\)

2. **Rewrite the Equation to Isolate \(y\):**

   The equation \(x^2 + (y - 3)^2 = 9\) can be rewritten as:

   \[
   (y - 3)^2 = 9 - x^2
   \]

3. **Take the Square Root of Both Sides:**

   Solving for \(y\), we take the square root of both sides:

   \[
   y - 3 = \pm \sqrt{9 - x^2}
   \]

   Thus,

   \[
   y = 3 \pm \sqrt{9 - x^2}
   \]

4. **Select the Top Half of the Circle:**

   The top half of the circle corresponds to the positive square root, so:

   \[
   y = 3 + \sqrt{9 - x^2}
   \]

### Conclusion:

The formula for the function whose graph is the top half of the circle \(x^2 + (y - 3)^2 = 9\) is:

\[
\boxed{y = 3 + \sqrt{9 - x^2}}
\]

Would you like further details or have any questions?  

Here are 5 related questions that might interest you:

1. What is the formula for the bottom half of the same circle?
2. How do you find the domain and range of this function?
3. How would you find the area of the top half of the circle?
4. How can we graph this function to visually confirm the result?
5. How would you find the derivative of this function?

**Tip:** When dealing with circles, remember that the general form is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.

Learn how to find the formula for the function representing the top half of a circle with the equation x^2 + (y - 3)^2 = 9. Step-by-step solution includes identifying the circle's center, radius, and isolating the function. Suitable for high school students.

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