1. The problem is to analyze and understand the function $y = xe^{-2x}$. 2. This function is a product of $x$ and an exponential decay $e^{-2x}$, which means it will grow initially and then decay for large $x$ due to the negative exponent. 3. To find critical points (where slopes are zero), we differentiate using the product rule: $$y' = e^{-2x} + x \cdot (-2)e^{-2x} = e^{-2x} - 2xe^{-2x} = e^{-2x}(1 - 2x)$$ 4. Set $y' = 0$ to find critical points: $$e^{-2x}(1 - 2x) = 0$$ Since $e^{-2x} \neq 0$ for any real $x$, we have: $$1 - 2x = 0 \implies x = \frac{1}{2}$$ 5. To determine if this point is a maximum or minimum, examine the second derivative or test values around $x = \frac{1}{2}$: The function changes from increasing to decreasing at $x=\frac{1}{2}$, so it is a local maximum. 6. The maximum value is: $$y\left(\frac{1}{2}\right) = \frac{1}{2} e^{-2 \cdot \frac{1}{2}} = \frac{1}{2} e^{-1} = \frac{1}{2e}$$ 7. The intercept with the y-axis is at $x=0$: $$y(0) = 0 \cdot e^0 = 0$$ 8. The intercept with the x-axis occurs where $y=0$: $$xe^{-2x} = 0 \implies x=0$$ Final answer: The function $y = xe^{-2x}$ has one critical point at $x=\frac{1}{2}$ which is a local maximum with value $\frac{1}{2e}$. The graph passes through the origin (0,0).