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To find the slope of the tangent line to the graph of a function at a given point, we need to calculate the derivative of the function and then evaluate it at the specified \( x \)-value.

For part (a), the function is given as:
\[
f(x) = \frac{-5x}{2x + 3}
\]
with the point of interest being \( x = 2 \).

### Steps to find the slope

1. **Differentiate \( f(x) \) with respect to \( x \)**: We’ll use the quotient rule for differentiation, which states that if \( f(x) = \frac{g(x)}{h(x)} \), then
   \[
   f'(x) = \frac{g'(x) h(x) - g(x) h'(x)}{[h(x)]^2}.
   \]

   Here:
   - \( g(x) = -5x \), so \( g'(x) = -5 \).
   - \( h(x) = 2x + 3 \), so \( h'(x) = 2 \).

2. **Apply the quotient rule**:
   \[
   f'(x) = \frac{(-5)(2x + 3) - (-5x)(2)}{(2x + 3)^2}.
   \]

3. **Simplify the expression**:
   Expanding and simplifying the numerator:
   \[
   f'(x) = \frac{-10x - 15 + 10x}{(2x + 3)^2}.
   \]
   Combine like terms in the numerator:
   \[
   f'(x) = \frac{-15}{(2x + 3)^2}.
   \]

4. **Evaluate \( f'(x) \) at \( x = 2 \)**:
   Substitute \( x = 2 \) into the derivative to find the slope of the tangent line at that point:
   \[
   f'(2) = \frac{-15}{(2 \cdot 2 + 3)^2} = \frac{-15}{(4 + 3)^2} = \frac{-15}{7^2} = \frac{-15}{49}.
   \]

Therefore, the slope of the tangent line at \( x = 2 \) is \( \frac{-15}{49} \).

### Finding the Value of \( x \) Where There is No Tangent Line
To determine the \( x \)-values where there is no tangent line, we need to find where the derivative is undefined, which occurs when the denominator \( (2x + 3)^2 = 0 \). Solving this equation:

\[
2x + 3 = 0 \Rightarrow x = -\frac{3}{2}.
\]

Thus, there is no tangent line at \( x = -\frac{3}{2} \) because the function has a vertical asymptote there.

Would you like further details on any of these steps?

---

1. What is the quotient rule, and how is it derived?
2. How would the slope change if \( f(x) \) were modified slightly, say by a different numerator or denominator?
3. What does it mean for a function to have no tangent line at a certain point?
4. How can vertical asymptotes affect the existence of tangent lines?
5. How would you find the equation of the tangent line at \( x = 2 \) using the slope we calculated?

**Tip:** When applying the quotient rule, always ensure to carefully simplify each part to avoid errors in combining terms.

Determine the slope of the line that is tangent to the graph of each function at the given point. Then determine the value of x at which there is no tangent line.

For part (a), f(x) = -5x / (2x + 3), where x = 2.

Determine the slope of the tangent line to the function f(x) = -5x / (2x + 3) at x = 2, and identify where there is no tangent line due to a vertical asymptote. This problem involves calculus concepts, including the derivative of a rational function using the quotient rule and finding where the derivative is undefined.

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