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To find the value of \( a \), we start by using the information given in the problem.

The equation of the curve is:
\[
y = a + bx^{-1} - 9x^{-2}
\]
where \( a \) and \( b \) are constants. We're also told that the maximum point on this curve is \( (3, 4.4) \).

Since \( (3, 4.4) \) is a maximum point, two conditions hold:
1. The function value at \( x = 3 \) should be \( y = 4.4 \).
2. The derivative \( \frac{dy}{dx} \) at \( x = 3 \) should be zero (since it's a maximum point).

### Step 1: Differentiate \( y \) with respect to \( x \)
To find the derivative, \( \frac{dy}{dx} \), we differentiate each term in the equation:
\[
y = a + bx^{-1} - 9x^{-2}
\]

The derivative is:
\[
\frac{dy}{dx} = 0 - b x^{-2} + 18x^{-3}
\]
or equivalently,
\[
\frac{dy}{dx} = -\frac{b}{x^2} + \frac{18}{x^3}
\]

### Step 2: Set \( \frac{dy}{dx} = 0 \) at \( x = 3 \)
Since the maximum point occurs at \( x = 3 \), we set \( \frac{dy}{dx} = 0 \) when \( x = 3 \):
\[
-\frac{b}{3^2} + \frac{18}{3^3} = 0
\]
Simplify:
\[
-\frac{b}{9} + \frac{18}{27} = 0
\]
\[
-\frac{b}{9} + \frac{2}{3} = 0
\]

Multiply through by 9 to clear the fraction:
\[
-b + 6 = 0
\]
\[
b = 6
\]

### Step 3: Substitute \( x = 3 \), \( y = 4.4 \), and \( b = 6 \) into the original equation to solve for \( a \)
Now, substitute \( x = 3 \), \( y = 4.4 \), and \( b = 6 \) into the equation \( y = a + bx^{-1} - 9x^{-2} \):
\[
4.4 = a + 6 \cdot 3^{-1} - 9 \cdot 3^{-2}
\]

Calculate each term:
\[
4.4 = a + 6 \cdot \frac{1}{3} - 9 \cdot \frac{1}{9}
\]
\[
4.4 = a + 2 - 1
\]
\[
4.4 = a + 1
\]

Solve for \( a \):
\[
a = 4.4 - 1
\]
\[
a = 3.4
\]

### Final Answer
The value of \( a \) is:
\[
\boxed{3.4}
\]

Would you like further details on any of the steps?

---

Here are some related questions that might help deepen your understanding:

1. What is the significance of the derivative being zero at a maximum point?
2. How would you approach this problem if given only the minimum point instead of the maximum?
3. Why does substituting values of \( x \) and \( y \) into the original function help us solve for constants?
4. What role does the second derivative play in determining whether a point is a maximum?
5. Can this method be extended to solve for more complex functions?

**Tip:** When finding maxima or minima, remember to check both the first and second derivatives to confirm the nature of the critical points.

The equation of a curve is y = a + bx^{-1} - 9x^{-2} where a and b are constants. The maximum point on this curve is (3, 4.4). Find the value of a.

Discover how to find the value of the constant 'a' in the curve equation y = a + bx^{-1} - 9x^{-2} given the maximum point (3, 4.4). This problem involves key calculus concepts such as derivatives and the First Derivative Test. Suitable for students in Grades 11-12.

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