To solve this problem, we need to use the equation of the tangent line to approximate the value of $f(-2.85)$.

Step 1: Find the slope of the tangent line

The slope of the tangent line at $x = -3$ is given by the derivative of $f(x)$ evaluated at $x = -3$. The derivative is given as: $f'(x) = -5x - 6$ So, at $x = -3$: $f'(-3) = -5(-3) - 6 = 15 - 6 = 9$ The slope of the tangent line at $x = -3$ is 9.

Step 2: Write the equation of the tangent line

The tangent line at $x = -3$ can be written as: $y = f(-3) + f'(-3)(x + 3)$ Given $f(-3) = 0$ and $f'(-3) = 9$, the equation becomes: $y = 0 + 9(x + 3) = 9(x + 3)$ Simplifying: $y = 9x + 27$

Step 3: Approximate $f(-2.85)$

Substitute $x = -2.85$ into the equation of the tangent line: $f(-2.85) \approx 9(-2.85) + 27$ $f(-2.85) \approx -25.65 + 27 = 1.35$

Final Answer:

The approximation for $f(-2.85)$ using the tangent line at $x = -3$ is 1.35.

Would you like further details or have any questions? Here are 5 related questions that build on this concept:

1. How would the approximation change if we used a different point close to $x = -3$?
2. What does the tangent line tell us about the behavior of the function $f(x)$ near $x = -3$?
3. How could you use higher derivatives to get a better approximation of $f(-2.85)$?
4. What is the geometric significance of the slope of the tangent line?
5. How does the approximation method used here compare with using a Taylor series?

Tip: The closer the point of approximation is to the point where the tangent line is calculated, the more accurate the approximation will be.