https://file.aiquickdraw.com/mathbot/file/17269607275061z3awo4c.jpg

To solve this problem, we need to find the slope of the tangent line to the graph of the function \( f(x) = \frac{x}{5} - 1 \) at the point \( x = -10 \).

The slope of the tangent line at any point \( x = a \) is given by \( f'(a) \), where \( f'(x) \) is the derivative of the function.

### Steps:

1. **Differentiate the function \( f(x) = \frac{x}{5} - 1 \)**:
   \[
   f'(x) = \frac{d}{dx} \left( \frac{x}{5} - 1 \right)
   \]
   The derivative of \( \frac{x}{5} \) is \( \frac{1}{5} \), and the derivative of the constant \( -1 \) is 0. So,
   \[
   f'(x) = \frac{1}{5}
   \]

2. **Evaluate the derivative at \( x = -10 \)**:
   Since the derivative is a constant \( f'(x) = \frac{1}{5} \), this means that \( f'(-10) = \frac{1}{5} \).

Thus, the slope of the tangent line at the point \( (-10, -3) \) is:

\[
f'(-10) = \frac{1}{5}
\]

Would you like more details or further clarification?

Here are 5 related questions:
1. What is the geometric interpretation of a derivative in terms of tangent lines?
2. How does the slope of the tangent line relate to the behavior of a function at a point?
3. Can the slope of the tangent line ever be zero? If so, under what conditions?
4. How would the problem change if the function were non-linear, say quadratic?
5. What is the general process for finding the equation of a tangent line at a given point?

**Tip**: When differentiating linear functions, their slope is constant, meaning the derivative is the same everywhere!

Find the slope of the tangent to the graph of the given function f(x) = x/5 - 1 at the indicated point (-10, -3).

This problem shows how to find the slope of the tangent to the linear function f(x) = x/5 - 1 at the point (-10, -3). Learn how to differentiate and interpret the slope of a tangent line at a given point.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation