Question

Find the equation of the normal to the curve y = (x² - x + 1) (x - 2) at the point where the curve cuts the x-axis.

Answer 1

First we need to have the x intercept
let y = 0
(x^2 - x + 1)(x-2) = 0
x = 2 is the only real solution
so it crosses the x-axis at (2,0)

y = x^3 - 2x^2 - x^2 + 2x + x - 2
= x^3 - 3x^2 + 3x - 2

dy/dx = 3x^2 - 6x + 3
at x = 2, dy/dx = 12 - 12 + 3 = 3
so at (2,0) the slope of the tangent is 3
which makes the slope of the normal = -1/3

equation of normal:
y-0 = (-1/3)(x - 2)
3y = -x + 2
x + 3y = 2