Question

Find the equation (s) of the tangent (s) to the curve $y=(x^3-1)(x-2)$ at the points where the curve intersects the x-axis.

Answer 1

When the curve cuts x-axis, y = 0 i.e., $(x^3-1)(x-2)=0$
$\Rightarrow (x-1)(x^2+x+1)(x-2)=0\therefore x=1,2(as{x}^2+x+1\ne 0\forall xϵR.$
Hence points of contact are A(1, 0) and B(2, 0).
Now $\frac{dy}{dx}=(x^3-1)+3(x-2)x^2=4x^3-6x^2-1\therefore (\frac{dy}{dx}{)}_{atA}=-3,(\frac{dy}{dx}{)}_{atB}=7,$
Now equation of tangent at A : y - 0 = - 3(x-1) i.e., 3x + y = 3.
And equation of tangent at B : y - 0 = 7(x-2) i.e., 7x - y = 14.