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The problem asks for the equation of the tangent line to the graph of \( y = g(x) \) at \( x = 3 \), given that \( g(3) = -5 \) and \( g'(3) = 6 \).

To find the equation of the tangent line, we use the point-slope form of a line:

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

Where:
- \( (x_1, y_1) \) is the point of tangency,
- \( m \) is the slope of the tangent line (which is the derivative \( g'(x_1) \)).

### Step-by-step solution:

1. **Point of Tangency**: We know that at \( x = 3 \), the value of \( y = g(3) = -5 \). So, the point of tangency is \( (3, -5) \).

2. **Slope of the Tangent Line**: The slope of the tangent line is given by the derivative of \( g(x) \) at \( x = 3 \), which is \( g'(3) = 6 \).

3. **Equation of the Tangent Line**:
   Using the point-slope form:
   \[
   y - (-5) = 6(x - 3)
   \]
   Simplifying this equation:
   \[
   y + 5 = 6(x - 3)
   \]
   Expanding the right side:
   \[
   y + 5 = 6x - 18
   \]
   Finally, subtract 5 from both sides:
   \[
   y = 6x - 23
   \]

### Final Answer:
The equation of the tangent line is:
\[
y = 6x - 23
\]

Would you like more details or have any questions?

Here are five related questions for deeper understanding:
1. What is the geometric meaning of the derivative at a point?
2. How do we find the equation of a tangent line if the slope is unknown?
3. Can the point-slope form of a line be used for secant lines as well?
4. How would this problem change if \( g'(3) \) were negative?
5. What if \( g(3) = 0 \)? How would the tangent line equation be different?

**Tip**: The slope of a tangent line represents how steep the graph is at a specific point, and it is given by the derivative at that point.

Find an equation of the tangent line to the graph of y = g(x) at x = 3 if g(3) = -5 and g'(3) = 6.

This problem asks for the equation of the tangent line to the graph of y = g(x) at x = 3. Given g(3) = -5 and g'(3) = 6, the solution involves using the point-slope form of a line. Learn how to find the tangent line and understand key calculus concepts.

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