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The given problem asks to find the derivative:

\[
\frac{d}{dx} \left( \frac{h(x)}{x} \right) \Bigg|_{x = 2}
\]

Given the values \( h(2) = 8 \) and \( h'(2) = -9 \), we can proceed by differentiating \( \frac{h(x)}{x} \).

### Step 1: Differentiate \( \frac{h(x)}{x} \)

Using the quotient rule, which states:

\[
\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}
\]

Let \( u(x) = h(x) \) and \( v(x) = x \). Then, the derivative becomes:

\[
\frac{d}{dx} \left( \frac{h(x)}{x} \right) = \frac{h'(x) \cdot x - h(x) \cdot 1}{x^2}
\]

Simplifying:

\[
\frac{d}{dx} \left( \frac{h(x)}{x} \right) = \frac{h'(x) \cdot x - h(x)}{x^2}
\]

### Step 2: Substitute \( x = 2 \)

Now substitute \( h'(2) = -9 \), \( h(2) = 8 \), and \( x = 2 \) into the derivative:

\[
\frac{-9 \cdot 2 - 8}{2^2} = \frac{-18 - 8}{4} = \frac{-26}{4} = -6.5
\]

Thus, the value of the derivative at \( x = 2 \) is:

\[
-6.5
\]

Would you like further details on any step? Here are five related questions to explore:

1. How do you apply the quotient rule in other types of problems?
2. What happens if the denominator in the quotient rule is more complex?
3. Can you explain how the quotient rule compares to the product rule?
4. What are common mistakes when using the quotient rule?
5. How do second derivatives work with quotient functions?

**Tip:** Always make sure to simplify your expressions as much as possible before plugging in values for easier calculations.

If h(2) = 8 and h'(2) = -9, find the derivative d/dx (h(x)/x) at x = 2.

Find the derivative of h(x)/x using the quotient rule, given h(2) = 8 and h'(2) = -9, and evaluate it at x = 2. This calculus problem applies the quotient rule step-by-step.

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