Determine the equation of the normal line to f(x)= e^(2x) at x= − 1

Answer 1

f(x)= e^(2x)
f(-1) = e^-2
so the tangent touches the curve at (-1, e^-2)
f'(x) = 2e^(2x)
at x = -1, f'(-1) = 2e^-2 = 2/e^2, which is the slope of the tangent at that point
Then the slope of the normal is -e^2/2

equation of normal :
y - e^-2 = (-e^2/2)(x + 1)

convert this to whatever form of the equation was expected.