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
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(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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