Let f be a function such that f(1)=2 , f′(1)=−1 , f′′(1)=4 . Which of the following is the second-degree Taylor polynomial of f centered at a=1 ? 
Question 1 Select one: 
T2(x)=2−(x−1)+4(x−1)^2 cross out 
T2(x)=2+(x−1)+4(x−1)^2 cross out 
T2(x)=1+(x−1)+2(x−1)^2 cross out 
T2(x)=2−(x−1)+2(x−1)^2 cross out

The second-degree Taylor polynomial of a function \(f(x)\) centered at \(a = 1\) is given by:

\[
T_2(x) = f(1) + f'(1)(x - 1) + \frac{f''(1)}{2!}(x - 1)^2
\]

### Step-by-Step Solution:
1. **Given:**
   - \(f(1) = 2\)
   - \(f'(1) = -1\)
   - \(f''(1) = 4\)

2. **Plugging in the values:**
   \[
   T_2(x) = 2 + (-1)(x - 1) + \frac{4}{2}(x - 1)^2
   \]

3. **Simplifying:**
   \[
   T_2(x) = 2 - (x - 1) + 2(x - 1)^2
   \]

4. **Final Expression:**
   \[
   T_2(x) = 2 - (x - 1) + 2(x - 1)^2
   \]

### Correct Answer:
\[
\boxed{T_2(x) = 2 - (x - 1) + 2(x - 1)^2}
\]

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Would you like me to provide more details or help with further clarification?  
Here are five related questions to explore further:  
1. What is the general formula for the \(n\)-degree Taylor polynomial?  
2. How do you derive Taylor series for functions centered at different points?  
3. Why is the second derivative divided by \(2!\) in the polynomial?  
4. What are the uses of Taylor polynomials in approximation?  
5. How does the Taylor polynomial behave as the degree increases?

**Tip:** Taylor polynomials provide increasingly accurate approximations near the center point \(a\).

Let f be a function such that f(1)=2, f′(1)=−1, f′′(1)=4. Which of the following is the second-degree Taylor polynomial of f centered at a=1? 
Options: 
T2(x)=2−(x−1)+4(x−1)^2 
T2(x)=2+(x−1)+4(x−1)^2 
T2(x)=1+(x−1)+2(x−1)^2 
T2(x)=2−(x−1)+2(x−1)^2

Find the second-degree Taylor polynomial for the function f(x) centered at a=1 with f(1)=2, f′(1)=−1, and f′′(1)=4. This problem covers key calculus concepts, including Taylor polynomials and derivatives. Suitable for Grades 11-12.

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