Posted by **Anonymous** on Tuesday, January 3, 2012 at 3:08pm.

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?

- Calc -
**Reiny**, Tuesday, January 3, 2012 at 3:21pm
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

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