What is the equation of the normal line of f(x)= e^(x^2/(x-2)) at x = 3?

1 Answer
Nov 15, 2015

y=1/(3e^9)x-1/e^9+e^9

Explanation:

If tangent line is y=kx+n then k=f'(x_0) where x_0=3.

f'(x)=e^(x^2/(x-2)) * (2x(x-2)-x^2)/(x-2)^2

k = f'(3)=-3e^9

Equation of the normal is y-y_0 = k_1(x-x_0) where:

x_0=3 => y_0=f(x_0)=f(3)=e^9 and

k*k_1=-1 => k_1=-1/k = 1/(3e^9)

y-e^9=1/(3e^9)(x-3)

y=1/(3e^9)(x-3)+e^9

y=1/(3e^9)x-1/e^9+e^9