Find an equation for a line that is normal to the graph of y=xe^x and goes through the origin

Y = xe^x

dy/dx (e^x x) = >use the product rule, d/dx (u v) = v * (du) / (dx) + u * (dv) / (dx), where u = e^x and v = x:

= e^x (d/dx (x)) + x (d/dx (e^x))

y' = e^x x + e^x

y' (0) = 1 = > slope of the tangent

slope of the normal = - 1

y - 0 = - 1 (x - 0)

y = - x = > normal at origin