To find the degree 2 Taylor polynomial of the function $f(x) = \ln(-2x - 5)$ centered at $x = -5$, we will follow these steps:

Step 1: Function and its Derivatives

We need the function and its first two derivatives evaluated at $x = -5$.

Function at $x = -5$:

$f(x) = \ln(-2x - 5)$ Substitute $x = -5$: $f(-5) = \ln(-2(-5) - 5) = \ln(10 - 5) = \ln(5)$

First derivative:

The derivative of $f(x) = \ln(-2x - 5)$ using the chain rule is: $f'(x) = \frac{-2}{-2x - 5}$ Substitute $x = -5$: $f'(-5) = \frac{-2}{-2(-5) - 5} = \frac{-2}{10 - 5} = \frac{-2}{5}$

Second derivative:

Differentiate $f'(x) = \frac{-2}{-2x - 5}$ again using the quotient rule: $f''(x) = \frac{(-2)(-2)}{(-2x - 5)^2} = \frac{4}{(-2x - 5)^2}$ Substitute $x = -5$: $f''(-5) = \frac{4}{(10 - 5)^2} = \frac{4}{25}$

Step 2: Taylor Polynomial Formula

The formula for the degree 2 Taylor polynomial is: $P_2(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2}(x - a)^2$ Here, $a = -5$, and the values we calculated are: $f(-5) = \ln(5), \quad f'(-5) = \frac{-2}{5}, \quad f''(-5) = \frac{4}{25}$ Substitute into the Taylor polynomial: $P_2(x) = \ln(5) + \frac{-2}{5}(x + 5) + \frac{4}{25} \cdot \frac{(x + 5)^2}{2}$ Simplifying: $P_2(x) = \ln(5) - \frac{2}{5}(x + 5) + \frac{2}{25}(x + 5)^2$

Thus, the degree 2 Taylor polynomial for $\ln(-2x - 5)$ centered at $x = -5$ is: $P_2(x) = \ln(5) - \frac{2}{5}(x + 5) + \frac{2}{25}(x + 5)^2$

Would you like more details or have any questions?

Here are some related questions:

1. How do we determine the degree of a Taylor polynomial?
2. Why is the Taylor series centered around a specific point, and how is this point chosen?
3. How can we approximate natural logarithmic functions using Taylor polynomials?
4. What happens if we use a higher-degree Taylor polynomial for this function?
5. How can we generalize Taylor series expansions for more complex functions?

Tip: Always double-check the function and its derivatives at the chosen center to avoid small calculation errors in Taylor series.