+130081 f(x) = x^3 - 4
The derivative gives the slope of a tangent line at any x value
So
f ' ( x) = 3x^2
At x = 2.... the slope of the tangent line is 3(2)^2 = 12
The slope of a normal line = -1/12
So...the equation of the normal line is
y = - (1/12) ( x - 2) + 4 simplify
y = (-1/12)x + 1/6 + 4
y = (-1/12)x + 25 / 6
