Given $f ( x , y ) = x e ^ { - 2 y }$ , at the point $( 4,0 )$ find the equation of the tangent plane to the graph of $f$ . Also, find a normal vector to the graph of $f$ .

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The Answer of Given $f ( x , y ) = x e ^ { - 2 y }$ , at the point $( 4,0 )$ find the equation of the tangent plane to the graph of $f$ . Also, find a normal vector to the graph of $f$ .

Given f(x,y)=xe−2yf ( x , y ) = x e ^ { - 2 y }f(x,y)=xe−2y

Given $f ( x , y ) = x e ^ { - 2 y }$