Question

Tangents to $y=(x^3-1)(x-2)$ at the points where the curve cuts the $x-$axis.

Answer 1

$y=(x^3-1)(x-2)$

y=0 where curve cuts x axis

$\therefore 0=(x^3-1)(x-2)$

$\Rightarrow x=1$ or x=2

points are (2,0) & (1,0)

slope of tangent = $y^1$ at the point

$y^1=(x^3-1)\left[1\right]+\left[3x^2\right](x-2)$

$=x^3-1+3x^3-6x^2$

$=4x^3-6x^2-1$

For (2,0)

$y^1=4(2{)}^3-6(2{)}^2-1=7$

For (1,0)

$y^1=4(1{)}^3-6(1{)}^2-1=-7$

y = mx + c $\Rightarrow y=7x-14$

$0=7\times 2+c$ $\therefore c=-14$

$0=-3\times 1+c$ $\therefore c=3$ $\Rightarrow y=-3x+3$