The point P lies on the curve C: y=f(x) where f(x)=x^3-2x^2+6x-12 and has x coordinate 1. Find the equation of the line normal to C which passes through P.
First we must find the y coordinate of the point P: We know the x-coordinate is x1=1 so the y coordinate must satisfy the equation y1=f(1) which gives y1=-7. So we now know P is at (1,-7).
We now need to find the gradient of C at P, we will call this a. We know a=f'(1)=5.
So the gradient, m, of the normal line at P will be: m=-1/a=-1/5.
So we know our normal line must have gradient m=-1/5 and must pass through P at (x1,y1)=(1,-7). Using the equation of straigt line y-y1=m(x-x1) gives our answer 5y+x+34=0.
KH
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