Question

Find the equation of the tangents to the curve $y=\left(x^3-1\right)\left(x-2\right)$ at the points where the curve cuts the x-axis.

Answer 1

Equation of curve : $y=(x^3-1)(x-2)\Rightarrow y=x^4-2x^3-x+2$

Slope of tangent$=\frac{dy}{dx}=\frac{d}{dx}(x^4-2x^3-x+2)=4x^3-6x^2-1$

At x-axis$,y=0\therefore (x^3-1)(x-2)=0\Rightarrow x=1,2$

When $x=1$, slope of tangent $=4\times 1^3-6\times 1^2-1=-3$

thus equation of tangent is $y-0=-3(x-1)\Rightarrow 3x+y-3=0$

When $x=2$, slope of tangent $=4\times 2^3-6\times 2^2-1=7$

thus equation of tangent is $y-0=7(x-2)\Rightarrow 7x-y-14=0$