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The problem asks to find the slope of the tangent line to the graph of the function \( g(x) = \frac{8}{7}x + 5 \) at the point \( (-7, -3) \).

Since \( g(x) = \frac{8}{7}x + 5 \) is a linear function, its graph is a straight line. The slope of the tangent line to a linear function at any point is simply the slope of the line itself, which is constant.

From the equation \( g(x) = \frac{8}{7}x + 5 \), we can see that the slope of the line is the coefficient of \( x \), which is \( \frac{8}{7} \).

Thus, the slope of the tangent line at the point \( (-7, -3) \) is \( \frac{8}{7} \).

Would you like further explanation, or do you have any questions?

Here are 5 related questions for further exploration:
1. What is the y-intercept of the line \( g(x) = \frac{8}{7}x + 5 \)?
2. How would the slope of the tangent line change if the function was quadratic?
3. How do you find the equation of the tangent line at the point \( (-7, -3) \)?
4. What is the geometric interpretation of the slope of a line?
5. What would the tangent line be if the function was nonlinear?

**Tip:** For linear functions, the slope is constant, meaning the tangent line is the line itself at any point.

Find the slope of the tangent line to the graph of the function at the given point. g(x) = (8/7)x + 5, point (-7, -3).

Learn how to find the slope of the tangent line to the linear function g(x) = (8/7)x + 5 at the point (-7, -3). The problem involves understanding the slope of a line and its tangent properties, suitable for students in Grades 8-10.

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