I've been working out the gradient (and whole equation) of the tangent of certain points on a curve but this new set of questions wants the equation of the normal. I figure I'd just work as I had been and get the equation of the tangent, then simply flip over the gradient and reverse the sign for the normal i.e.

tangent; y = 7x - 13

normal; y = -1/7x - 13

The full sum is as follows:

y = 2x^2 - 5x + 3 at (2, 1)

dy/dx = 6x - 5

12 - 5 = 7

(y - 1)/(x - 2) = 7

y - 1 = 7(x - 2) = 7x - 14

y = 7x - 13

and thus the normal; y = -1/7x - 13.

The book claims x + 3y = 5