Find the equation of the normal line to the curve y=2x-3x^2 at the point (-3,-33)
is it a. x-16y+339=0 b. x+20y+339=0
c. x+20y+663=0 d. x-16y+663=0
The tangent at (x,y) has slope y' = 2-6x
So, at x = -3, y' = 20
The normal therefore has slope -1/20
Now we have a point and a slope:
(y+33)/(x+3) = -1/20
(C)