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We are given that the tangent line to the graph of the function \( g \) at the point \( (-6, -2) \) passes through the point \( (0, 2) \). We need to find the slope of the tangent line, which is also the derivative of the function \( g \) at \( x = -6 \), i.e., \( g'(-6) \).

### Steps:
1. **Find the slope of the line**: The slope \( m \) of the line that passes through two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
   \[
   m = \frac{y_2 - y_1}{x_2 - x_1}
   \]
   Here, the two points are \( (-6, -2) \) and \( (0, 2) \).

2. **Substitute the points**:
   \[
   m = \frac{2 - (-2)}{0 - (-6)} = \frac{2 + 2}{0 + 6} = \frac{4}{6} = \frac{2}{3}
   \]

Therefore, the slope of the tangent line is \( \frac{2}{3} \), and since this is also the derivative at \( x = -6 \), we conclude that:
\[
g'(-6) = \frac{2}{3}
\]

Would you like more details or have any further questions?

Here are some related questions for further practice:
1. What is the equation of the tangent line at \( (-6, -2) \)?
2. How do you compute the derivative of a function given its equation?
3. How do you find the slope of a secant line between two points on a curve?
4. Can the slope of a tangent line be negative? If so, under what conditions?
5. How do you interpret the meaning of a derivative in a real-world scenario?

**Tip**: The slope of the tangent line at any point on a curve represents the instantaneous rate of change of the function at that point.

The tangent line to the graph of function g at the point (-6, -2) passes through the point (0, 2). Find g'(-6).

Learn how to find g'(-6) by calculating the slope of the tangent line that passes through two points: (-6, -2) and (0, 2). This problem involves key calculus concepts such as the derivative and slope calculation.

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