The coordinates of the point where the normal to the curve y= 1/3 x^3 + 1/2 x^2 + x at x=1 intersects the y-axis are what?

Answer 1

dy/dx = x^2 + x + 1
at x=1 ,slope of tangent is 1+1+1 = 3
so slope of normal is -1/3
when x=1, y = 1/3 + 1/2 + 1 = 11/6

so equation of normal:
y = (-1/3)x + b, at (1, 11/6)
11/6 = -1/3 + b
b = 3/2
y = (-1/3)x + 13/6
for y-intercept, let x = 0
y = 13/6

it cuts at (0,13/6)

check my arithmetic