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

Answer

`7 x-y-14=0`

Explanation

Given equation of curve: `y=(x^{3}-1)(x-2) \Rightarrow y=x^{4}-2 x^{3}-x+2`


Slope of tangent `=\frac{d y}{d x}=\frac{d}{d x}(x^{4}-2 x^{3}-x+2)=4 x^{3}-6 x^{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 3 x+y-3=0`


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


So, equation of tangent is `y-0=7(x-2) \Rightarrow 7 x-y-14=0`