Calculus: Early Transcendentals, Chapter 3, 3.2, Section 3.2, Problem 33

The equation of the tangent to a curve y = f(x) at the point (x_0, y_0) is given by the equation: (y - y_0)/(x-x_0) = f'(x_0)
Here, the equation of the curve is y = 2x*e^x
The derivative of y = 2*x*e^x is
y' = 2*(e^x + x*e^x)
For x = 0, y' = 2
The equation of the tangent at the point (0, 0) is y/x = 2 or y = 2x
A line perpendicular to the tangent with slope m has slope -1/m.
The equation of the normal is y/x = -1/2
y = -x/2
For the curve y = 2x*e^x , the equation of the tangent at (0,0) is y = 2x and the equation of the normal is y = -x/2