2010 Paper 1 Q11 - tangents and normals

A curve is such that  

 

(i) Given that the curve passes through the point (1, 5), find the equation of the curve

(ii) Find the x-coordinates of the stationary points of the curve

(iii) Obtain an expression for d²y/dx² and hence, or otherwise, determine the nature of each stationary point.

(i) We all know that to get the equation of the curve from the differential, we need to integrate the differential. But don’t forget the constant of integration.

At (1, 5), then

5 = -8 – 2 + c

c = 15

Therefore the equation of the curve is:

(ii) At the stationary points, dy/dx = 0

Therefore,

Hence, the x-coordinates of the stationary points are 2 and -2

(iii) 

Hence, at x = 2, d²y/dx² is negative, Þ it is a max point

and when x = -2, it is a min point.

And that's all for today....

Tomorrow, we will do the last question of the paper 1 and move into paper 2.