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find an equation of the tangent line to f(t) at a=-1

To find the equation of the tangent line to the function \( f(t) = \sqrt{3t + 8} \) at \( a = -1 \), we can follow these steps:

1. **Find the point of tangency**: This is \( f(-1) \).  
   \[
   f(-1) = \sqrt{3(-1) + 8} = \sqrt{5}.
   \]
   So, the point of tangency is \( (-1, \sqrt{5}) \).

2. **Find the slope of the tangent line**: This is given by \( f'(-1) \), which is already calculated in the problem as \( \frac{3\sqrt{5}}{10} \).

3. **Use the point-slope form of the equation of a line**: The point-slope form is:
   \[
   y - y_1 = m(t - t_1),
   \]
   where \( m \) is the slope and \( (t_1, y_1) \) is the point of tangency. Substituting \( m = \frac{3\sqrt{5}}{10} \), \( t_1 = -1 \), and \( y_1 = \sqrt{5} \), we get:
   \[
   y - \sqrt{5} = \frac{3\sqrt{5}}{10}(t + 1).
   \]

4. **Simplify the equation**:
   \[
   y = \sqrt{5} + \frac{3\sqrt{5}}{10}(t + 1).
   \]

This is the equation of the tangent line in exact form.

Would you like a more detailed explanation of any part of the process?

### Related Questions:
1. How do we find the derivative \( f'(t) \) using the definition of the derivative?
2. Why is the point-slope form used for tangent lines?
3. What is the difference between finding a secant line and a tangent line?
4. How does the chain rule apply when differentiating functions like \( \sqrt{3t + 8} \)?
5. What does the slope of the tangent line represent geometrically?

**Tip**: The point-slope form is very useful for writing the equation of a tangent line because it directly uses the slope and a known point on the curve.

Find an equation of the tangent line to f(t) = √(3t + 8) at a = -1.

Learn how to find the equation of the tangent line to the function f(t) = √(3t + 8) at t = -1, including the calculation of the derivative and the use of point-slope form. This problem is suitable for advanced calculus students, covering key concepts like derivatives and tangent lines.

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