My aim is to compose a tangent equation for the line y=(x-1)*(x+1)*(x+2) provided that tangent points are intersections with x-axis.
Hello,Originally Posted by totalnewbie
you know already the intersections between x-axis and curve:
(-2,0), (-1,0), (1,0).
Now you need the slope of the tangent lines:
Your function is $\displaystyle y=f(x)=(x-1)(x+1)(x+2)=x^3+2x^2-x-2$
The gradient of the function gives the slope of the tangents:
$\displaystyle f'(x)=3x^2+4x-1$
First slope: $\displaystyle m_1=f'(-2)=3 \cdot 4 +4 \cdot(-2) - 1=3$
2nd slope: $\displaystyle m_2=f'(-1)=3 \cdot 1 +4 \cdot(-1) - 1=-2$
3rd slope: $\displaystyle m_3=f'(1)=3 \cdot 1 +4 \cdot(1) - 1=6$
Now use the point-slope-formula and you'll get:
$\displaystyle t_1:y=3 \cdot (x+2)$
$\displaystyle t_2:y=-2 \cdot (x+1)$
$\displaystyle t_3:y=6 \cdot (x-1)$
Greetings
EB