2x = 2

x = 2

dy/dx = 6x - 5

dy/dx = 4x - 5, slope at (1, 2) is 3

Try it now.

]]>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

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