Find the equation of the line tangent to the graph of  
f(x)=1/5x^3  at  x=1
Please submit your answer in slope-intercept form.

To find the equation of the tangent line to the graph of \( f(x) = \frac{1}{5}x^3 \) at \( x = 1 \), we follow these steps:

1. **Find \( f(1) \):**  
   Substitute \( x = 1 \) into \( f(x) \) to get the point where the tangent line touches the graph:
   \[
   f(1) = \frac{1}{5} \cdot 1^3 = \frac{1}{5}.
   \]
   So, the point on the graph is \( (1, \frac{1}{5}) \).

2. **Find \( f'(x) \), the derivative of \( f(x) \):**  
   The derivative \( f'(x) \) will give us the slope of the tangent line. Using the power rule:
   \[
   f(x) = \frac{1}{5}x^3,
   \]
   \[
   f'(x) = \frac{1}{5} \cdot 3x^2 = \frac{3}{5}x^2.
   \]

3. **Evaluate \( f'(1) \):**  
   Substitute \( x = 1 \) into \( f'(x) \) to find the slope of the tangent line at \( x = 1 \):
   \[
   f'(1) = \frac{3}{5} \cdot 1^2 = \frac{3}{5}.
   \]
   So, the slope of the tangent line is \( \frac{3}{5} \).

4. **Write the tangent line equation in slope-intercept form:**  
   Using the point-slope form of the line equation, \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) = (1, \frac{1}{5}) \) and \( m = \frac{3}{5} \):
   \[
   y - \frac{1}{5} = \frac{3}{5}(x - 1).
   \]

5. **Convert to slope-intercept form, \( y = mx + b \):**
   Distribute the slope on the right side:
   \[
   y - \frac{1}{5} = \frac{3}{5}x - \frac{3}{5}.
   \]
   Add \( \frac{1}{5} \) to both sides to solve for \( y \):
   \[
   y = \frac{3}{5}x - \frac{3}{5} + \frac{1}{5}.
   \]
   Simplify:
   \[
   y = \frac{3}{5}x - \frac{2}{5}.
   \]

**Final Answer:**
\[
y = \frac{3}{5}x - \frac{2}{5}.
\]

Would you like further clarification on any steps, or help with related problems?

Here are 5 related questions for further exploration:
1. How would the tangent line change if the function were \( f(x) = \frac{1}{5}x^4 \)?
2. What is the tangent line at \( x = -1 \) for the same function?
3. How do we find the normal line to the graph of \( f(x) \) at \( x = 1 \)?
4. How can we determine the second derivative of \( f(x) \) and interpret its meaning?
5. How does the derivative tell us about the behavior of \( f(x) \) near \( x = 1 \)?

**Tip:** For finding tangent lines, remember that the derivative at a point gives the exact slope at that point.

Find the equation of the line tangent to the graph of f(x)=1/5x^3 at x=1. Please submit your answer in slope-intercept form.

This guide shows how to find the tangent line to the function f(x) = 1/5x^3 at x = 1 in slope-intercept form. Includes a step-by-step solution using derivatives to determine the slope and point-slope form to write the equation. Ideal for students in Grades 11-12.

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