Equation of normal line?

Normal line and the curve share the point in common, so the point (which is required to find the equation of the normal line is: (- 1, 1 / e^2).

The normal's slope is the negative reciprocal of the tangent's slope:

If tangent's slope = m, then normal's slope = - 1 / m

d/dx e^(2x) = 2*e^(2x)

At the point where x = -1, the tangent s slope is 2/e^(2) and the normal's slope is - e^2 / 2

Normal line equation:

y - 1 / e^2 = - e^2 / 2 * (x + 1) Manipulate as you see fit.

dy/dx = 2e^(2x)
When x = -1
Slope of tangent = 2e^-2
Slope of normal = -1/[2e^-2] = -e^2/2
Point is (-1,e^-2)
y-e^-2 = (-e^2/2)(x+1)

y'=2e^(2x)
y'(-1)=2e^(-2)
y(-1)=e^(-2)
so y-e^(-2)=2e^(-2) (x+1)
y=2xe^(-2)+3e^(-2)