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To solve the problem of finding the equation of the tangent line to the curve \( y = \frac{x + 2}{x - 4} \) at the point \( (3, -5) \) without using the power rule, we will follow these steps:

### 1. **Find the derivative of the function** \( y = \frac{x + 2}{x - 4} \) using the quotient rule: 

The quotient rule is given by:

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

Here, \( u(x) = x + 2 \) and \( v(x) = x - 4 \).

First, compute the derivatives \( u'(x) \) and \( v'(x) \):

\[
u'(x) = 1
\]
\[
v'(x) = 1
\]

Now, apply the quotient rule:

\[
y' = \frac{(x - 4)(1) - (x + 2)(1)}{(x - 4)^2}
\]

Simplify the numerator:

\[
y' = \frac{x - 4 - x - 2}{(x - 4)^2} = \frac{-6}{(x - 4)^2}
\]

### 2. **Evaluate the derivative at the given point \( x = 3 \):**

Substitute \( x = 3 \) into the derivative \( y' \):

\[
y'(3) = \frac{-6}{(3 - 4)^2} = \frac{-6}{1} = -6
\]

So, the slope of the tangent line at \( (3, -5) \) is \( -6 \).

### 3. **Use the point-slope form to find the equation of the tangent line:**

The point-slope form of a line is given by:

\[
y - y_1 = m(x - x_1)
\]

Where \( m \) is the slope and \( (x_1, y_1) \) is the point on the line. Plugging in the values:

\[
y - (-5) = -6(x - 3)
\]

Simplify the equation:

\[
y + 5 = -6x + 18
\]

Finally, rearrange the equation to the slope-intercept form:

\[
y = -6x + 13
\]

### Final Answer:

The equation of the tangent line to the curve at the point \( (3, -5) \) is:

\[
\boxed{y = -6x + 13}
\]

Would you like more details or have any questions?

Here are some related questions:
1. How would the solution change if the point of tangency was different?
2. What are the applications of the quotient rule in calculus?
3. Can you derive the tangent line using implicit differentiation?
4. How does the point-slope form compare to the slope-intercept form?
5. What happens to the tangent line if the function was different, say \( y = \frac{x^2 + 2x}{x - 4} \)?

**Tip:** Understanding the derivation of the quotient rule can give you more confidence when applying it to complex functions.

Discover how to find the equation of the tangent line to a curve without using the power rule. This detailed solution covers the application of the quotient rule in calculus, deriving the tangent line equation at a specific point, and transforming it into point-slope and slope-intercept forms. Ideal for advanced high school students and calculus enthusiasts.

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