Find the equation of the line passing through the point 
P(3, 3)
 and tangent to the graph of 
f(x) = 
6
x − 1
.
y = 
 

To find the equation of the tangent line to the graph of the function \( f(x) = 6x - 1 \) at the point \( P(3, 3) \), we need to follow these steps:

1. **Find the derivative of \( f(x) \)**: The derivative represents the slope of the tangent line at any point \( x \).
   \[
   f'(x) = \frac{d}{dx}(6x - 1) = 6
   \]
   This means the slope of the tangent line at any point on the graph is always 6, as the derivative is constant.

2. **Use the point-slope form of the equation of a line**: The point-slope form is given by
   \[
   y - y_1 = m(x - x_1)
   \]
   where \( m \) is the slope, and \( (x_1, y_1) \) is the point of tangency.

   We are given that the point is \( P(3, 3) \) and the slope is \( m = 6 \). Substituting these into the point-slope formula gives:
   \[
   y - 3 = 6(x - 3)
   \]

3. **Simplify the equation**: Expand and simplify the equation:
   \[
   y - 3 = 6x - 18
   \]
   \[
   y = 6x - 18 + 3
   \]
   \[
   y = 6x - 15
   \]

Thus, the equation of the tangent line is:
\[
y = 6x - 15
\]

Would you like further explanation on any step?

Here are five related questions:
1. How is the derivative of a function used to find the slope of a tangent line?
2. What does it mean geometrically when the derivative is constant?
3. How can you verify if a point lies on a given function or line?
4. Can a tangent line be horizontal? If so, under what conditions?
5. How would the equation change if the slope were variable?

**Tip**: The point-slope form of a line is very useful when you know a point on the line and the slope, simplifying the process of finding the equation.

Find the equation of the line passing through the point P(3, 3) and tangent to the graph of f(x) = 6x - 1.

Learn how to find the equation of the line that is tangent to the graph of f(x) = 6x - 1 at the point P(3, 3). This problem involves key concepts such as differentiation and the point-slope form of a line, suitable for students in Grades 10-12.

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