Find the x-coordinate of the stationary points of the curve and determine the nature of these stationary points. The equation of a curve is $y=x^2e^{-x}$.

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*Find the x-coordinate of the stationary points of the curve and determine the nature of these stationary points.

*Show that the equation of the normal to the curve at the point where $x=1$ is $e^2x+ey = 1+e^2$.


This is the full question I am having difficulty solving, I simply don't know where to begin. I moved back to my home country because of covid and now I am doing self-studying I don't know how to solve this any help wold be great and much appreciated.
 A: $y=x^2e^{-x} \implies y'(x)=2xe^{-x}-x^2e^{-x} \implies y'(1)=e^{-1}.$
So the slope of normal at $x=1$ is $m=-1/y'(1)=-e$. So the equation of line having slope $-e$ and passing through the point $(1,e^{-1})$ is
$$y-e^{-1}=-e(x-1) \implies ey+e^2 x=1+e^2$$
A: *

*The stationary point of a curve is simply the point where the derivative vanishes. You can see the behavior of these points from this link.
https://mathworld.wolfram.com/StationaryPoint.html
The first question asks for the $x$-coordinates of the stationary points. You can start by taking the derivative of the curve. Then, you should equate it to zero and solve the equation to find the roots.
$$\frac{dy}{dx}=2xe^{-x}-x^2e^{-x}=e^{-x}x(2-x)=0\space \Rightarrow \space x=0 \space\text{or}\space x=2$$
At $x=0$ and $x=2$, the derivative is zero which means they are the stationary points.


*We should verify the equation for the normal line of the curve at $x=1$. At this point $y$-coordinates of the normal line and the original curve is the same. Since tangent and normal lines to a curve at some particular point are perpendecular to each other, multiplication of their slopes yields $-1$. At $x=1$, the slope of the tangent line is $\frac{1}{e}$. This means the slope of the normal line is $-e$. Below you may see the normal line equation.  $$ax+b=y$$ $$x_0=1,\space \space a=-e, \space \space y_0=\frac{1}{e}$$ If we substitute these values into the line equation, we find $b$ to be $e+\frac{1}{e}$. $$-ex+e+\frac{1}{e}=y \space \rightarrow \space e^2+1=e^2x+ey$$
