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### Problem:
Let \( f \) be a differentiable function with \( f(3) = 2 \) and \( f'(3) = -5 \). Let the function \( g(x) = (x^2 - 5)f(x) \). Write the equation of the line tangent to the graph of \( g \) at the point where \( x = 3 \).

### Solution:

1. **Find \( g(3) \):**
   \[
   g(3) = (3^2 - 5)f(3)
   \]
   \[
   g(3) = (9 - 5) \times 2
   \]
   \[
   g(3) = 4 \times 2 = 8
   \]

2. **Find \( g'(x) \):**
   Use the product rule for differentiation. If \( g(x) = u(x) \cdot v(x) \), then \( g'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \).

   Here, let:
   \[
   u(x) = x^2 - 5 \quad \text{and} \quad v(x) = f(x)
   \]
   Then:
   \[
   u'(x) = 2x \quad \text{and} \quad v'(x) = f'(x)
   \]
   So, applying the product rule:
   \[
   g'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)
   \]
   \[
   g'(x) = 2x \cdot f(x) + (x^2 - 5) \cdot f'(x)
   \]

3. **Find \( g'(3) \):**
   \[
   g'(3) = 2 \times 3 \cdot f(3) + (3^2 - 5) \cdot f'(3)
   \]
   \[
   g'(3) = 6 \cdot 2 + (9 - 5) \cdot (-5)
   \]
   \[
   g'(3) = 12 + 4 \cdot (-5)
   \]
   \[
   g'(3) = 12 - 20 = -8
   \]

4. **Equation of the Tangent Line:**
   The equation of the tangent line at \( x = 3 \) is given by:
   \[
   y - g(3) = g'(3) (x - 3)
   \]
   Substituting \( g(3) = 8 \) and \( g'(3) = -8 \):
   \[
   y - 8 = -8(x - 3)
   \]
   \[
   y - 8 = -8x + 24
   \]
   \[
   y = -8x + 32
   \]

So, the equation of the tangent line is:
\[
y = -8x + 32
\]

Would you like further details or have any questions?

### Related Questions:
1. How do you find the derivative of a product of two functions using the product rule?
2. What is the geometric interpretation of the tangent line to a curve?
3. How would the tangent line equation change if \( f(3) \) and \( f'(3) \) had different values?
4. Can you explain why the derivative \( g'(x) \) at a point gives the slope of the tangent line at that point?
5. How do you determine the y-intercept of a tangent line?

**Tip:** When using the product rule, carefully differentiate both functions separately and then combine using the rule \( u'(x)v(x) + u(x)v'(x) \).

Let f be a differentiable function with f(3) = 2 and f'(3) = -5. Let the function g(x) = (x^2 - 5)f(x). Write the equation of the line tangent to the graph of g at the point where x = 3.

Explore how to calculate the tangent line to the function g(x) = (x^2 - 5)f(x) at x = 3 using the product rule for differentiation. Learn the steps to apply differentiation rules effectively with provided values f(3) = 2 and f'(3) = -5.

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