Question

Find an equation of the tangent line to the curve at the given point y = x^2e^-x (1,1/e)

Answer 1

y'=2xe^(-x)+x^2*(-1)*e^(-x)=2xe^{-x}-x^2e^{-x}=xe^{-x}*[2-x]

y'(1)=1*e^{-1}*[2-1]=e^{-1} is slope of the tangent line

y=mx+b
1/e=e^{-x}*1+b
solve for b to find y-intercept