Let *f* (*x*) = −9 sin(6 *x*).

Answer the following. (Exact answers only: decimal approximations will be marked incorrect.)


(a) Determine the coordinates of the first maximum turning point on the graph of *f(x)* in the interval (0,2𝜋).
First Max Turning Point:      *x* = *y* = 
(b) Determine the coordinates of the first minimum turning point on the graph of *f(x)* in the interval (0,2𝜋).
First Min Turning Point:      *x* = *y* = 

Given the function \( f(x) = -9 \sin(6x) \), we need to determine the first maximum and minimum turning points within the interval \( (0, 2\pi) \).

### Step 1: Determine the Period
The period of \( f(x) = -9 \sin(6x) \) is given by:
\[
\text{Period} = \frac{2\pi}{|k|}
\]
where \( k = 6 \). Therefore,
\[
\text{Period} = \frac{2\pi}{6} = \frac{\pi}{3}
\]
This means that the function completes one full cycle in the interval \( \left(0, \frac{\pi}{3}\right) \) and repeats this behavior within \( (0, 2\pi) \).

### Step 2: Analyzing the Turning Points for \( f(x) = -9 \sin(6x) \)

The sine function \( \sin(6x) \) oscillates between \( -1 \) and \( 1 \). Therefore:
- The **maximum** of \( f(x) = -9 \sin(6x) \) occurs when \( \sin(6x) = -1 \), resulting in \( f(x) = -9 \cdot (-1) = 9 \).
- The **minimum** of \( f(x) = -9 \sin(6x) \) occurs when \( \sin(6x) = 1 \), resulting in \( f(x) = -9 \cdot 1 = -9 \).

### (a) First Maximum Turning Point
To find the first maximum, we solve for \( x \) when \( \sin(6x) = -1 \).

1. The value \( \sin(6x) = -1 \) occurs at \( 6x = \frac{3\pi}{2} \) within the interval \( (0, 2\pi) \).
2. Solving for \( x \):
   \[
   x = \frac{3\pi}{2 \cdot 6} = \frac{\pi}{4}
   \]
3. Substituting \( x = \frac{\pi}{4} \) into \( f(x) \):
   \[
   f\left(\frac{\pi}{4}\right) = -9 \sin\left(6 \cdot \frac{\pi}{4}\right) = -9 \sin\left(\frac{3\pi}{2}\right) = -9 \cdot (-1) = 9
   \]

Thus, the coordinates of the first maximum turning point are:
\[
(x, y) = \left(\frac{\pi}{4}, 9\right)
\]

### (b) First Minimum Turning Point
To find the first minimum, we solve for \( x \) when \( \sin(6x) = 1 \).

1. The value \( \sin(6x) = 1 \) occurs at \( 6x = \frac{\pi}{2} \) within the interval \( (0, 2\pi) \).
2. Solving for \( x \):
   \[
   x = \frac{\pi}{2 \cdot 6} = \frac{\pi}{12}
   \]
3. Substituting \( x = \frac{\pi}{12} \) into \( f(x) \):
   \[
   f\left(\frac{\pi}{12}\right) = -9 \sin\left(6 \cdot \frac{\pi}{12}\right) = -9 \sin\left(\frac{\pi}{2}\right) = -9 \cdot 1 = -9
   \]

Thus, the coordinates of the first minimum turning point are:
\[
(x, y) = \left(\frac{\pi}{12}, -9\right)
\]

### Final Answers:
- **(a) First Max Turning Point:** \( x = \frac{\pi}{4}, \, y = 9 \)
- **(b) First Min Turning Point:** \( x = \frac{\pi}{12}, \, y = -9 \)

Would you like further details on the solution? 

---

1. What is the period of \( f(x) = -9 \sin(6x) \)?
2. How would the maximum and minimum points change if the amplitude were different?
3. Why does the function reach its maximum and minimum at these specific points?
4. Can this method be applied to find turning points in \( f(x) = -9 \cos(6x) \) as well?
5. How would a phase shift affect the location of maximum and minimum points?

**Tip:** For trigonometric functions, turning points can be determined by setting the argument of sine or cosine to known values that yield maximum or minimum outputs.

Let f(x) = −9 sin(6 x).

Answer the following. (Exact answers only: decimal approximations will be marked incorrect.)

(a) Determine the coordinates of the first maximum turning point on the graph of f(x) in the interval (0,2π).
First Max Turning Point:      x = y = 
(b) Determine the coordinates of the first minimum turning point on the graph of f(x) in the interval (0,2π).
First Min Turning Point:      x = y = 

Determine the first maximum and minimum turning points of the function f(x) = -9 sin(6x) in the interval (0, 2π). This trigonometric problem includes steps to find turning points, sine function behavior, and period calculation. Suitable for Grades 10-12.

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