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.