1. **State the problem:** We are given the function $f(x) = (2x + 1)e^{1-2x}$ and asked to analyze its behavior, find the tangent line at $x_0 = \frac{1}{2}$, and understand the asymptote. 2. **Recall the derivative:** The derivative is given as $f'(x) = -4xe^{1-2x}$. 3. **Evaluate the function at $x_0 = \frac{1}{2}$:** $$f\left(\frac{1}{2}\right) = \left(2 \cdot \frac{1}{2} + 1\right) e^{1 - 2 \cdot \frac{1}{2}} = (1 + 1) e^{1 - 1} = 2 \cdot e^0 = 2$$ 4. **Evaluate the derivative at $x_0 = \frac{1}{2}$:** $$f'\left(\frac{1}{2}\right) = -4 \cdot \frac{1}{2} \cdot e^{1 - 2 \cdot \frac{1}{2}} = -2 \cdot e^{0} = -2$$ 5. **Equation of the tangent line $T$ at $x_0 = \frac{1}{2}$:** Using point-slope form: $$y = f\left(\frac{1}{2}\right) + f'\left(\frac{1}{2}\right)(x - \frac{1}{2}) = 2 - 2 \left(x - \frac{1}{2}\right) = 2 - 2x + 1 = 3 - 2x$$ 6. **Asymptotic behavior:** As $x \to +\infty$, note that $e^{1-2x} = e^1 \cdot e^{-2x}$ decays exponentially to 0 faster than $2x+1$ grows linearly. Hence, $$\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} (2x + 1) e^{1-2x} = 0$$ This means the $x$-axis ($y=0$) is a horizontal asymptote. 7. **Summary:** - Function: $f(x) = (2x + 1)e^{1-2x}$ - Tangent at $x=\frac{1}{2}$: $y = 3 - 2x$ - Horizontal asymptote: $y=0$