SOLUTION: The curve y = x^2 - 3x + 4 passes through the points P(1,2) and Q(3,4). Find a) the equation of tangent at P b) the equation of the normal at Q

Question 1202600: The curve y = x^2 - 3x + 4 passes through the points P(1,2) and Q(3,4). Find
a) the equation of tangent at P
b) the equation of the normal at Q
Answer by ikleyn(52776) (Show Source): You can put this solution on YOUR website!
.
The curve y = x^2 - 3x + 4 passes through the points P(1,2) and Q(3,4). Find
(a) the equation of tangent at P
(b) the equation of the normal at Q
~~~~~~~~~~~~~~~~~

(a)  Take the derivative.   It is  y' = 2x - 3.

     Calculate the value of the derivative at P(1,2).  It is

         y'(1) = 2*1 - 3 = 2 - 3 = -1.


     Hence, the slope of the tangent line at P is -1.

     So, the tangent line at  P(1,2)  is  

         y-2 = (-1)*(x-1),  (1)

     or any other equivalent form.



(b)  Calculate the value of the derivative at Q(3,4).  It is

         y'(3) = 2*3 - 4 = 6 - 4 = 2.


    Hence, the slope of the tangent line at Q is 2.

    Therefore, the slope of the normal to the curve at Q is  = -0.5.

    Thus the normal to the curve at  Q(3,4)  is

        y-4 = -0.5(x-3)

    or any other equivalent form.

Solved.

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