dont use the quotient rule. use the tangent line formula

Sure! To solve the problem using the tangent line formula without using the quotient rule, we'll use the following method:

### **1. Tangent Line Formula**

The formula for the tangent line at a point \(x = a\) on the curve \(y = f(x)\) is given by:

\[
y - y_1 = f'(a)(x - a)
\]

Where:
- \(f'(a)\) is the derivative of \(f(x)\) evaluated at \(x = a\).
- \(y_1 = f(a)\) is the y-coordinate at \(x = a\).

### **2. Find the Value of the Function at the Point**

The function is given by:

\[
f(x) = \frac{x + 2}{x - 4}
\]

Substitute \(x = 3\) to find \(f(3)\):

\[
f(3) = \frac{3 + 2}{3 - 4} = \frac{5}{-1} = -5
\]

So, the point is \((3, -5)\), which matches the given point.

### **3. Find the Derivative \(f'(x)\) Using the Tangent Line Formula**

We can find the derivative at \(x = 3\) using the definition of the derivative (difference quotient):

\[
f'(3) = \lim_{h \to 0} \frac{f(3+h) - f(3)}{h}
\]

Let's compute \(f(3+h)\):

\[
f(3+h) = \frac{(3+h) + 2}{(3+h) - 4} = \frac{h + 5}{h - 1}
\]

Now, plug this into the difference quotient:

\[
f'(3) = \lim_{h \to 0} \frac{\frac{h + 5}{h - 1} - (-5)}{h} = \lim_{h \to 0} \frac{\frac{h + 5 + 5(h - 1)}{h - 1}}{h}
\]

Simplify the expression inside the limit:

\[
f'(3) = \lim_{h \to 0} \frac{\frac{h + 5 + 5h - 5}{h - 1}}{h} = \lim_{h \to 0} \frac{\frac{6h}{h - 1}}{h} = \lim_{h \to 0} \frac{6h}{h(h - 1)} = \lim_{h \to 0} \frac{6}{h - 1}
\]

When \( h \) approaches 0:

\[
f'(3) = -6
\]

### **4. Use the Tangent Line Formula**

Now that we have the slope \(f'(3) = -6\), and the point \((3, -5)\), use the point-slope form:

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

Simplifying this gives:

\[
y + 5 = -6x + 18
\]
\[
y = -6x + 13
\]

### **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. What is the geometric interpretation of the tangent line to a curve?
2. Can you find the tangent line at a different point on the same curve?
3. How does the method of finite differences relate to the tangent line formula?
4. Can you apply this approach to find a normal line to the curve?
5. How would the problem change if we used implicit differentiation?

**Tip:** Using the tangent line formula is especially useful when you want to bypass more complicated differentiation techniques like the quotient rule.

Learn how to solve a differentiation problem using the tangent line formula instead of the quotient rule. Detailed step-by-step solution provided with the application of differentiation concepts and tangent line formulas. Suitable for advanced high school students.

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