dy/dx = e^x
at the point x=3 sub it into the derivative to get the gradient:
f'(3) = e^3
since we need the normal, find the perpendicular gradient, ie -1/e^3
at the point x=3, sub it into y=e^x to find the y coordinate.
ie y=e^3
:. The point is (3, e^3)
eqn of tangent can be found using...