The function in the image is $f(x) = e^{-x^2} - 2x + 1$. The question asks us to analyze the behavior of the function and determine the asymptotes.

Step-by-step analysis:

1. Behavior for $x \to +\infty$ and $x \to -\infty$:

   • The term $e^{-x^2}$ approaches $0$ as $x \to \pm\infty$ because $e^{-x^2}$ decays exponentially fast.
   • The remaining terms, $-2x + 1$, dominate as $x \to \pm\infty$.

2. Oblique asymptote:

   • For oblique asymptotes, the dominant linear term $-2x + 1$ dictates the behavior:
     • As $x \to +\infty$, the function behaves approximately as $-2x + 1$.
     • As $x \to -\infty$, the function again behaves approximately as $-2x + 1$.

3. Asymptotic analysis:

   • The term $e^{-x^2}$ does not affect the linear asymptotic behavior because it vanishes exponentially fast as $|x| \to \infty$.

Conclusion:

• The correct answer is: "Per $x \to -\infty$ ammette come asintoto la retta di equazione $y = -2x + 1$."

Let me know if you'd like a detailed explanation of how oblique asymptotes are calculated or have additional questions!

Here are 5 related questions:

1. What are horizontal asymptotes, and how do they differ from oblique ones?
2. How do exponential terms like $e^{-x^2}$ impact the asymptotic behavior of a function?
3. How can we confirm the asymptotic behavior of a function using limits?
4. Are there cases where a function has both horizontal and oblique asymptotes?
5. How do the coefficients of polynomial terms affect the slope of oblique asymptotes?

Tip: When identifying oblique asymptotes, always focus on the dominant terms as $|x| \to \infty$.