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To find the slope of the line tangent to the graph of \( f(x) = \frac{e^{2x}}{-x+3} \) at \( x = -3 \), we need to compute the derivative of \( f(x) \) and evaluate it at \( x = -3 \).

### Steps:
1. **Differentiate** \( f(x) \) using the quotient rule.
   - The quotient rule states that for \( f(x) = \frac{g(x)}{h(x)} \), the derivative is:
   \[
   f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{h(x)^2}
   \]
   Here, \( g(x) = e^{2x} \) and \( h(x) = -x + 3 \).

2. **Evaluate the derivative** at \( x = -3 \) to find the slope of the tangent line.

Let me go ahead and calculate this for you.The slope of the tangent line to the graph of \( f(x) = \frac{e^{2x}}{-x + 3} \) at \( x = -3 \) is \( \frac{13 e^{-6}}{36} \).

Would you like any further details or have any additional questions?

Here are 5 related questions for practice:
1. How do you apply the quotient rule in differentiation?
2. What is the derivative of \( e^{2x} \)?
3. How would you compute the slope of the tangent line at a different point \( x = 0 \)?
4. What happens to the slope if the numerator and denominator of \( f(x) \) are both linear functions?
5. Can you use the product rule to verify part of the quotient rule?

**Tip:** Always simplify your expressions to avoid unnecessary complexity when calculating derivatives!

What is the slope of the line tangent to the graph of f(x) = e^(2x)/(-x+3) at x = -3? Express your answer in simplest form with no negative exponents.

Learn how to find the slope of the tangent line to the graph of f(x) = e^(2x)/(-x+3) at x = -3 using the quotient rule and exponential function differentiation. This problem provides a detailed solution suitable for advanced high school or college calculus students.

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