The problem is: at what point does the normal to at intersect the parabola a second time?

$\displaystyle y ' = 8x +3 $

Then can I just take the slope of 8, inverse it to be $\displaystyle -1/8 $ and then put it in point slope form, $\displaystyle (y-9)=-1/8(x-1)$

This ends up being $\displaystyle y = -1/8x + 73/8 $.

Then do I make it $\displaystyle -1/8x + 73/8 = 4x^2 + 3x + 2$ ? The math gets pretty hairy after that, and I just would like to know if I'm going in the right direction!